DIFFERENTIAL EQUATIONIntroduction

In the sciences and engineering, mathematical models are developed to aid in the understanding of physical phenomena. These models often yields an equation that contains some derivatives of an unknown function. Such an equation is called a differential equation.Examples: y’= cosx, y’’+3y’=0

DEFINITION: A differential equation is an equation that contains one or several derivatives of an unknown function. The unknown function may be

a) function of one variableb) function of more variables

In a) the differential equation is called ordinary differential equationIn b) the differential equation is called partial differential equationExamples: a) y’’+2y’+3y=0, y’= 2x+3y , y=y(x)b) ux−uy=x−2 y , uxx+uyy=0 , u=u(x,y)

Word problems.

1. Find the curve through the point (1,1) in the XY-plane having at each of its points the slope -y/x. The curve must be a solution of the differential equation y

'=− y /x .

2. If we drop a stone, then its acceleration y ' '=d2 y

dx2 (x=time) is equal to the acceleration of

gravity g. Hence the model of this problem of free fall is y ' '=g , since the air resistance will not matter too much in this case.

3. (exponential growth) The growth rate y '=dy

dx (x=time) of a population equals the

population y(x) present. The population model is y'= y , solutions are y ( x )=cex

.a) What is the particular solution satisfying y(0)=3 ?b) What initial amount y(0) is necessary to get y=100 after x=2 ?4. If the rate of growth is proportional to the population present at time x, say, y '=ky , verify

that y ( x )=cekx.

If y doubles in 1 day, how much can be expected after 1 week at the same rate of growth?5. (exponential decay) Observations show that the rate of change of the atmospheric pressure y with altitude x is proportional to the pressure. Assuming that the pressure at 6000 meters is

half its value y0 at sea level. Find the formula for the pressure at any height.

6. (interest rate) Let y(x) be the investment resulting from a deposit y0 after x years at an interest rate r.y ( x )= y0(1+r )x

(interest compounded annually)y ( x )= y0(1+r / 4 )4 x

(interest compounded quarterly)y ( x )= y0(1+r /365 )365 x

(interest compounded daily)

1

y ( x )= y0erx(interest compounded continuously), the differential equation of

this problem is y'=ry .

7. Find the curve y(x) that passes through (1,1/2) and is such that at each point (x,y) the

intercept of the tangent on the Y-axis is equal to 2 xy2.

8. Find the equations of the orthogonal trajectories of the family of curves y=cx2.

( If the differential equation of the family of curves is y '=f ( x , y ) , then the differential

equation of the orthogonal trajectories is y '=− 1

f (x , y ) . )9. Find the curve passing through the point (0,-2) such that the slope of the tangent at any of its points is equal to the ordinate of that point multiplied by a factor 3.10. Find the curve passing through the point (0,-2) such that the slope of the tangent at any of its points is equal to the ordinate of that point increased by 3 units. 11. Prove that a curve such that at any point the slope of its tangent is proportional to the abscissa of the point of tangency is a parabola.

Definition

The order of a differential equation is the order of the highest order derivative that appears in the equation.

The first order ordinary differential equations

We will omit the “ordinary” word, because in this semester will be only this type of differential equations.

These equations can be givena) F(x,y,y’)=0 implicit formb) y’=f(x,y) explicit form.

A solution of a given first order differential equation on some open interval (a,b) is a function y=h(x) that has a derivative and satisfies the diff. eq. for all x in (a,b).

Example: 1. Verify that y= x2 is a solution of xy’=2y for all x. y’=2x, substitute into the diff. eq. x2x=2x2 , that is x2 is a solution. 2. Sometimes a solution will appear as an implicit function. x2 +y2 -1=0 (y ix positive) implicit solution of yy’=-x.Use the implicit differentiation 2x+2yy’=0. From we get yy’=-x.

GENERAL solution of y’=f(x,y) is a sety=y(x,c). The range of the constant may have to be restricted in some cases to avoid

imaginary expressions.PARTICULAR solution : If we choose a specific c we obtain one solution , called particular solution. Some extra cases of the solutions.

2

1. A differential equation may sometimes have an additional solution that can not be obtained from the general solution and is the called a singular solution. 2. There exist more than one solution through the point (a,b).

INITIAL Value Problem A differential equation together with an initial condition is called an initial value problem. y’=f(x,y), y(x0 )=y0 (initial condition) Some examples: 1. y '2−x y '+ y=0General solution: y=cx – c2 (lines), y= x2 /4 is a solution which can not be obtained from the gen. solution (singular solution)2. y '2=−1 does not have a solution for real y. 3. |y '|+|y|=0 does not have a general solution, because its only solution y=0.

Let us consider the very important theorem for initial value problem.

Existence and uniqueness problem of solutions. Given the initial value problem y '=f (x , y ) , y ( x0 )= y0 . (1) Let us consider the following three cases:1. |y '|+|y|=0 , y(0)=1 has no solution , because y=0 is the only solution of differential equation.2. y’=x, y(0)=1 has exactly one solution y= x2/2+1.3. xy’=y-1, y(0)=1 has infinitely many solution . y=1+cx, where c is arbitrary real numbers.

There are two fundamental questions:1. Problem of existence: under what conditions does an initial value problem of the form (1) have at least one solution?2. Problem of uniqueness: Under what conditions does an initial value problem of the form (1) have at most one solution? There are theorems separately for the first and for the second problem. I will give one theorem which gives a sufficient condition for the first and the second problem.

CAUCHY-LIPSCHITZ existence and uniqueness theorem.

Given the initial value problemy’=f(x,y) , y(x0 )=y0 (1)

If f and fy are continuous and bounded on R={( x , y ) :|x−x0|<a ,|y− y0|<b } , |f|≤ K , |f y|≤ M , then the initial value problem (1) has a unique determine solution which is defined on |x−x0|<α=min ¿ ) . The solution can be obtained by Picard’s method: the sequence of

functions y0 , y1 ,….→ y (x ), yn= y0+∫x0

x

f (t , yn−1 (t ) )dt , n=1,2,…. limn → ∞

yn (x )= y (x).

(Instead the continuity of partial derivative of f with respect to y there is a weaker condition.)

Separable differential equation

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Definition: If the y’=f(x,y)= h(x)g(y) is a product of two one variable function , then the differential equation is called separable differential equation.If g(y)=0 for example y=y1 , y=y2 , then these lines are solutions. There are books in which they are called singular solutions.

Theorem: Given a differential equation y’=h(x)g(y). If h and g are continuous on T={( x , y ):a< x<b , c< y<d } , g ≠0 on T, (x0 , y0)∈T , then there exists only one solution satisfying the initial condition y(x0)=y0 .

∫y0

y 1g(u)

du=∫x0

x

h (t ) dt

How we can find the general solution and the particular solution?

1

g ( y )y '=h(x ) , Integrate the functions on both sides. From the conditions follows that

the antiderivatives exist

∫ 1g( y )

y ' dx=∫ h ( x )dx , use y(x)=y, y’(x)dx=dy we get ∫ 1g ( y )

dy=∫ h (x ) dx ,

G(y)=H(x)+C . The value of C can be found from the initial condition: G ( y0 )=H ( x0 )+C is an equation for C.

We get the solution in implicit form.

Examples:Solve the following equations:1. y’2 +y2 +3=0 , has no real-valued solution.2. y '=3 y2 /3

a) Give the general solution. y=0 is a solution.

If y≠0, separate the equation , we get 13

y−2 /3 dy=dx.

y1 /3=x+c , y=(x+c)3

b) If given the initial condition y(2)=0, then we get two solutions y=0 and y=(x-2)3 . There are more than one solution satisfying the initial condition. If we consider arbitrary point on the line y=0, then there are more than one solution through this point. There are books in which this solution is called (y=0) singular solution. c) If the initial condition is y (0 )=10−7, then there exits exactly one solution satisfying this

condition y=(x+10−7

3 )3.

3. y '=(x−3)( y+1)2 /3

y= -1 is a solution. If y ≠-1, then separate ( y+1)−2/3 dy=( x−3 )dx and integrate we get

3( y+1)1/ 3= x2

2−3 x+c, y=−1+( x2

6−x+c1)

3

,where c1=c /3 .

From the general solution you can see, that there is no choice the constant c 1 that will make the solution y= -1. Then we lost the solution y= -1 when we divided by ( y+1)2/3 .4. y '=√ y3

y ≥0, y=0 is a solution.

4

y≠0, y−3 /2 dy=dx ,−2 y−12 =x+c , y=( 2

x+c )2

, these curves are increasing , y=0 is the

horizontal asymptote.5. y '=3√ y , y=0 is a solution. If y≠0, then separate the differential equation and integrate. We

get y−1 /3 dy=dx , 32

y2/3=x+c , y=[(x+c) 23 ]

3 /2

.

6. Find a first order differential equation for which the given function is a solution.a) y=x3−4 , the diff . eq . y '=3x2

b) x2+9 y2=9 ,2 x+18 y y '=0 , y y '=−x /9

c) y=c x2 is the set of curves. y'=c 2x , c= y

x2 substitute this into y ' we get y '=2 yx .

7. y '=√1− y2, y=1 and y= -1 are solutions. If y≠1 or -1, then separate the equation and

integrate. 1

√1− y2dy=dx , arcsiny=x+c , y=sin (x+c ) .

8. ( x2+4 x+3 ) y y '=( x+2 ) ey2

. y≠0, x≠-1, x≠-3 .

After separation y e− y2

dy= x+2x2+4 x+3

dx ,integration −12

e− y2

=12

ln|x2+4 x+3|+c

Exercise set1. 9yy’+4x=0 2. y’=1+y2 3. y’= x/y , y(1)=34. y’= - y/x , y(1)=1 5. y '=−5 x4 y2 , y (0 )=1

Reduction to separable differential equation.

1. y '=f ( yx) , use the new variable u=y/x. y=ux and y’= u’x+u.

After the substitution we get u' x+u=f (u ) , du

f (u )−u=dx

x if f(u)-u≠0. The solution of the

equation f(u)-u=0 are solutions.2. y '=f (ax+by+c ), use u= ax+by+c, u’=a+by’.

y’= (u’-a)/b, after the substitution we get u'−ab

=f (u ) , dubf (u)+a

=dx if bf(u)+a≠0.

Some examples:

1. y '= yx+ 2 x3 cos x2

y , u=y/x. We get u' x+u=u+ 2 x2 cos x2

u , u u'=2 xcos x2,

after integration we get u2

2=sin x2+c , y2=2 x2 (sin x2+c ).

2. y '=( y+x)2 , use the transformation u=y+x, u’=y’+1

u’-1=u2 , u’=u2 +1, separate and integrate du

1+u2 =dx , arctanu=x+c ,

y+x=tan(x+c), y= -x+tan(x+c)

3. y '= y2−x2

2 xy4. x y '= y+√x2+ y2 , x>0 5. y '= y+x

y−x

6. y '=tan (x+ y )−1 7. y '= y−x+1y−x+5

5

Exact Differential Equation

The function f(x,y) has continuous first order partial derivative, its differential is df =f x dx+ f y dy . If f=c, then df=0. If df=0, then we can show (Lagrange theorem) that f(x,y)=cDefinition: A first order differential equation of the form M(x,y)dx+N(x,y)dy=0 (M(x,y)+N(x,y)y’=0) is called exact differential equation if its left side is the (total) differential df =f x dx+ f y dy of some function f(x,y). f x=M , f y=N . The general solution is f(x,y)=c.

Theorem: If M(x,y) and N(x,y) are continuously differentiable , M(x,y)dx+N(x,y)dy=0 is exact, then M y=N x . (to show this you can use the Young theorem)Theorem: If M(x,y) and N(x,y) are continuously differentiable in a region in the XY-plane whose boundary is a closed curve having no self-intersection (the curve is simple closed curve, the region is simply connected ) and M y=N x , then M(x,y)dx+N(x,y)dy=0 is exact differential equation. Finding f(x,y)a) by guessing, in this case you have to show that the function satisfies the equation.b) by systematic waySolve the following system: f x=M , f y=N .Let us start with the first equation

f ( x , y )=∫Mdx+k ( y ) , f y=∂

∂ y (∫Mdx+k ( y ))=N ,

∂

∂ y (∫Mdx )+k ' ( y )=N , express k’(y) from this equation and integrate .

Substitute instead k(y), you will get f(x,y). The general solution is f(x,y)=c.

Example: 1. Solve the equation

( x3+3 x y2 ) dx+(3 x2 y+ y3 ) dy=0 M=x3+3 x y2 and M y=6 xy N=3 x2 y+ y3 and N x=6 xy from we get M y=N x.M and N are continuously differentiable and this equation determines a differential equation if coefficient of dy is not 0. We have to find a function whose partial derivatives are given.

f x=M , from f ( x , y )=∫ ( x3+3 x y2 ) dx= x4

4+ 3

2x2 y2+k ( y )

f y=3 x2 y+ y3=3x2 y+k ' ( y ) from k ' ( y )= y3∧k ( y )= y4

4+c1

f ( x , y )= x4

4+ 3

2x2 y2+ y4

4+c 1∧the general solutionf ( x , y )=c 2

c2-c1= c (it is enough to give one k(y)), we get x4

4+ 3

2x2 y2+ y4

4=C . It is the general solution.

Find the particular solution satisfying the initial condition y(1)=1. Substitute the coordinates

of the point (1,1) we get c=2, the particular solution x4

4+ 3

2x2 y2+ y4

4=2 .

2. (2 x+e y) dx+x e y dy=03. (e x cosy ) dx−(ex siny ) dy=0

6

INTEGRATING FACTOR

P(x,y)dx+Q(x,y)dy=0 is not exact differential equation.

Definition: The differentiable function m(x,y)≠ 0 is called an integrating factor if mPdx+mQdy=0 is an exact differential equation. How we find an integrating factor? (mP )y=(mQ)x from m y P+m P y=mx Q+m Qx which is a partial differential equation.We will find the integrating factor of one variable

1. m=m(x), mx=m' , the differential equation for m is m'm

=P y−Qx

Q . If this expression does

not depend on y , then m(x) exists. It is enough to give one integrating factor.

2. m=m(y), m y=m' , the differential equation for m is m'm

=Q x−P y

P. If this expression does

not depend on x, then m(y) exists. It is enough to give one integrating factor.

Examples:1. (2 x2+ y ) dx+( x2 y−x ) dy=0 P=2 x2+ y ,Q=x2 y−x P y=1≠ Qx=2 xy−1 , Q=x (xy−1)

From this you can see that P y−Q x

Q=

2(1−xy)x (xy−1)

=−2x

does not depend on y . There exists

m(x).

m'm

=−2x , ln(m)= -2lnx, m (x )= 1

x2 , (2+ yx2 )dx+( y− 1

x )dy=0is an exact differential

equation.

f x=2+ yx2 , f ( x , y )=2 x− y

x+k ( y )

f y= y−1x=−1

x+k ' ( y) from k’(y)=y , one function for k(y)=y2 /2

The general solution: 2 x− yx+ y2

2=c .

2. 2 xydx+ ( y2−3 x2) dy=0 P y=2 x≠ Q x=−6 x

Qx−P y

P=−4

y , does not depend on y. There exists m(y).

m'm

=−4y , After integration we get ln(m)= -4lny, m¿ 1

y4 , 2 xy3 dx+( 1

y2−3 x2

y4 )dy=0 is an

exact differential equation.

f y=1y2 −

3 x2

y4 , f ( x , y )=−1y

+ x2

y3 +l(x)

f x=2 xy3 =2 x

y3 +l ' (x) from l’(x)=0, l(x)=constant =0

The general solution −1y

+ x2

y3 =C

7

3. 2sin ( y2 ) dx+xycos ( y2 ) dy=0 (m(x))4. 2 yxdx+ (4 y+3 x2 ) dy=0 (m(y))

First order linear differential equation

Definition: A differential equation of the form y '+p ( x ) y=f (x)

is called the first order linear differential equation. It is linear y, y’. y '+p ( x ) y=0 homogeneous differential equationy '+p ( x ) y=f (x) nonhomogeneous differential equation

Let us consider at first the linear homogeneous differential equationy '+p ( x ) y=0 separable differential equation .y=0 is a solution.

If y≠0 , then separate the equation dyy

=−p ( x ) dx and integrate.

Theorem: If p(x) is continuous on the interval I y '+p ( x ) y=0 , y ( x0 )= y0 , x0∈ I , then the initial value problem has a unique determined solution.

y= y0 ∙ e−∫

x0

x

p ( x ) dx

y=0 is called the trivial solution.

Corollary: If p(x) is continuous on I y '+p ( x ) y=0 has a non-trivial solution φ(x) , then all solutions can be given in the form cφ(x), c∈R. Proof: a) cφ(x) is a solution

(cφ ( x ) )'+ p ( x ) cφ ( x )=c [φ' ( x )+ p ( x ) φ ( x ) ]=c ∙ 0=0b) If μ ( x ) is a solution, then μ ( x )=c¿φ (x)

φ '+ p ∙φ=0 multiplyby μμ' + p∙ μ=0multiply byφ, subtract the first one from the second one, we get

μ' φ−μφ '=0 ,( μφ )

'

=0 implies that μφ=c¿ , μ ( x )=c¿φ (x)

Definition: The general solution of y '+p ( x ) y=0 is the set of functions H= {cφ ( x ) , φ ( x ) ≠0 , x∈ I , c∈R } .

You can see from the previous corollary and the definition of general solution, that general solution =all solution (linear homogeneous differential equation).

Nonhomogeneous linear differential equation.

Theorem: If y '+p ( x ) y=f (x ) , p ( x ) , f ( x ) are continuous on theinterval I , y1(x) is a nontrivial solution of the corresponding homogeneous differential equation , then there exists c(x) continuously differentiable function on I such that y p ( x )=c (x) y1(x) satisfies y '+py=f nonhomogeneous differential equation, y p is a particular solution and

8

c ( x )=∫x0

x f (t)y1(t )

dt .

Proof: Substitute into the nonhomogeneous differential equation y p

' + p ( x ) y p= f ( x ) , c ' ( x ) y1+c ( x ) y1' + p ( x ) c ( x ) y1=f (x )

c ' ( x ) y1+c ( x ) ( y1' +p (x ) y1 )=c' ( x ) y1= f (x ) from c ' ( x )= f (x )

y1(x) after integration we get

c(x).

The name of the method to find a particular solution is variation of parameter.y1 is a nontrivial solution of the homogeneous differential equation, yh is the general solution of homogeneous differential equation.yp is a particular solution of nonhomogeneous differential equation.yh=c y1 , c∈R , y p=c ( x ) y1

Theorem: p(x) and f(x) are continuous functions on the interval I.All solutions of y '+p ( x ) y=f (x ) can be given in the form y (x )= y p(x )+c y1(x) where y p(x ) is a particular solution of nonhomogeneous differential equation and y1(x)≠ 0 solution of homogeneous differential equation, c∈R .

Proof: a) y p+c y1 , c∈R , solution ( y p+c y1 )'+ p ( y p+c y1 )=( y p

' + p y p)+c ( y1' +p y1 )=f +0=f

b) if φ is a solution , then there is c∗¿ such that φ= y p+c∗ y1 y p

' + p y p= f φ '+ pφ=f , subtract these equations we get (φ− y p )'+ p (φ− y p )=0

that means φ− y p is a solution of homogeneous differential equation that is there is a c∗¿ such that φ= y p+c∗ y1 .

Definition: Given y '+py=f , p and f are continuous on the interval I.General solution is the set of functions H '= {y p+c y1 , c∈R } other notation ynh= y p+ yh.

Theorem: If p and f are continuous on the interval I , x0∈ I , then y '+py=f , y ( x0 )= y0initial value problem has a unique determined solution

φ ( x )=e−∫

x0

x

p (t ) dt [ y0+∫x0

x ( f (t )e∫t 0

t

p ( u) du)dt ] , x∈ I

General properties of linear differential equations

The linear differential equations (homogeneous, nonhomogeneous) have certain important properties. In the next time we shall see that the same is true for linear differential equations of higher order.

The homogeneous linear differential equations have the following properties (theorems)y '+py=0 (1 )

9

y '+py=f (2)

1. y=0 is a solution of (1) called the trivial solution.2. If y1 is a solution of (1) and c∈R , then cy1 is a solution of (1). Proof: (c y1 )'+ p ( c y1 )=c ( y1

' + p y1 )=c0=03. If y1∧ y2 are solutions of (1), then y1+ y2 is a solution of (1). Proof: ( y1+ y2 )'+p ( y1+ y2 )=( y1

' + p y1)+( y2' + p y2 )=0+0=0

The expressions in the first and second parenthesis are 0 according to the condition.

Using the corollary we get that the set of the solutions of homogeneous linear differential equation is a one-dimensional linear space.

Nonhomogeneous linear differential equation

1. If y1∧ y2 are solutions of (2), then y1− y2 is a solution of (1).Proof: We know that y1

' + p y1=f ∧ y2' + p y2=f . Using these we get

( y1− y2 )'+ p ( y1− y2 )=( y1' + p y1)−( y2

' + p y2 )=f −f =02. If y1 is a solution of (1), y2 is a solution of (2) , then y1 +y2 is a solution of (2).Proof: We know that y1

' + p y1=0 , y2' + p y2=f .

( y1+ y2 )'+p ( y1+ y2 )=( y1' + p y1)+( y2

' + p y2 )=0+f =f3. If y1 is a solution of y’+py=f1 and y2 is a solution of y’+py=f2 , then y1 +y2 is a solution of y’+py= f1 +f2 .Proof: ( y1+ y2 )'+p ( y1+ y2 )=( y1

' + p y1)+( y2' + p y2 )=f 1+f 2

The next problem is: How we can use the integrating factor to solve first order linear differential equation? Given y’+py=f, from we get (py-f)dx+dy=0P=py-f, P y= p , Q x=0Q=1

There exists m(x): m'm

=P y−Qx

Q=p , m ( x )=e∫ p ( x ) dx

Multiply by this the equation e∫ pdx ( py− f ) dx+e∫ pdx dy=0e∫ pdx y '+e∫ pdx py=e∫ pdx f

The left side is derivative a product (e∫ pdx y ) ' , integrate the function on both sides we get

e∫ pdx y=∫ (e∫ pdx f )dx+c from

y=e−∫ pdx [∫ (e∫ pdx f ) dx+c ]or F x=e∫ pdx ( py−f ) , F y=e∫ pdx , from the first we get

F ( x , y )= y e∫ pdx−∫ ( f e∫ pdx ) dx+k ( y )

F y=e∫ pdx=e∫ pdx+k ' ( y ) from k ' ( y )=0 , k is constant , let0General solution: F(x,y)=c, y=e−∫ pdx [∫ ( f e∫ pdx ) dx+c ] , c is arbitrary real number.

Examples:

10

1. y’+y tanx=sin2xa) Give the general solution.b) Give the particular solution satisfying the initial condition y(0)=1a) solution of homogeneous differential equationy’+ytanx=0, separate and integrate, y=0 is a solution, if y ≠0 dyy

=−sinxcosx

dx, ln|y|=ln|cosx|+c1 , |y|=ec1|cosx| from we get (using c2=± ec 1)

y=c2 cosx, we know that y=0 is a solution, we get yh=c ∙ cosxuse the method variation of parameter to find yp , yp =c(x)cosx. Substitute into the nonhomogeneous differential equation

c ' ( x ) cosx+c ( x ) (−sinx )+c ( x )cosx sinxcosx

=2 sinxcosx , c ' ( x )=2 sinx , c ( x )=−2cosx

y p=−2cos2 x , ynh=c ∙ cosx−2cos2 xb) y(0)=1, 1=c-2, c=3 the particular solution satisfying the initial condition:

y=3 cosx−2 cos2 x2. y '− y=e2 x

solution of y’-y=0, y=0, if y≠0 we get dy=ydx, dyy

=dx , ln|y|=x+c 1, y=c2ex , yh=c ex

use variation of parameter, y p=c ( x ) ex , c ' ( x ) ex+c (x ) ex−c ( x ) ex=e2 x ,c ' (x )=ex,one function for c ( x )=ex , y p=e2 x , ynh=c ex+e2x

3. x y '+2 y=4ex2

, y '+ 2x

y=4x

ex2

integrating factor: x2 , multiply the equation with this function we get x2 y '+2 xy=( x2 y )'=4 x ex2

, after integration x2 y=2 ex2

+c,

y=cx2 +

2x2 ex2

4. xy’+2y=9xUse the integrating factor: x2 y '+2 xy=9 x2 , ( x2 y ) '

=9 x2 , , x2 y=3 x3+c

y=cx2 +3 x

5. y’+ay=0Find the values of a for which the limit of all solutions at plus infinity is equal to 0.The general solution (separable differential equation) y=ce−ax. lim

x→+∞c e−ax

=0 if the limit of the exponent is negative infinity, that is a is positive real number.

6. y'+ y

x2 =e1/ x

y'+ y

x2 =0 , y=0 is a solution. If y≠0, then separate and integrate: dyy

=−1x2 dx ,

ln|y|=1x+c1 , y=c2e1 / x ,c 2≠ 0 , yh=ce

1x , c∈R

Particular solution: y p=c (x)e1 / x

11

y p' =c' ( x )e

1x+c (x )e

1x (−1

x2 ) Substitute into the nonhomogeneous diff. eq.

we get c ' ( x ) e1x+c ( x )(−1

x2 e1x+

1x2 e

1x )=e1 / x , c’(x)=1, c(x)=x (it is enough to give one function),

y p=x e1/ x , ynh=(x+c)e1/ x

7. y'+ 4 cosx

sinxy= x

sin3 x y=0 is a solution of the homogeneous differential equation.

y≠0 , separate the homogeneous differential equation dyy

=−4 cosxsinx

dx

ln|y|=−4 ln|sinx|+c1 , y=c2 ∙ 1sin4 x

, yh=c 1sin4 x

y p=c (x) 1sin4 x

, y p' =c' ( x )(sinx)−4+c ( x ) (−4 ) (sinx )−5 cosx .After substitution we get

c' ( x ) (sinx )−4+c ( x ) (−4 (sinx )−5 cosx+4 ( sinx )−5cosx )=c ' (x ) (sinx )−4= x

(sinx)3

c ' ( x )=xsinx , c ( x )=∫ xsinxdx=x (−cosx )−∫1 (−cosx ) dx=−xcosx+sinx+0Integration by parts : u=x, v’=sinx. The integration constant is 0, because we have to give one c(x) function.

ynh=c

(sinx)4 +−xcosx+sinx(sinx )4

8. Solve the following differential equations

a) x y '−2 y=x4 ex b) y '−2x

y=x2 ex

c) y '+ay=ebx , a ,b∈ R d) y'= 2 y

6 x− y2 , it is linear for x(y)

e) y '− 2sin (2x )

y= 1sinx

, find the solution which is bounded at π /2

f) y '= y2

y2+2 xy−x , it is linear for x(y)

Reduction of nonlinear differential equation to linear form.

Bernoulli differential equation: The differential equation is called the Bernoulli differential equation if it is of the form: y '+p ( x ) y=f (x ) ya , a∈RIf a=0 or a=1, the equation is linear, otherwise it is nonlinear.

Divide by ya , we get y 'ya + p 1

ya−1= f ( x ) , let u= 1ya−1 , u

'=(1−a ) y−a y '=(1−a) y 'ya

y 'ya =u' 1

1−a,a ≠ 1

u '

1−a+ pu=f ( x ) , it is linear for u

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Examples:1. y’+y=x/y ,Multiply by y we get yy’+y2 =x. Use u=y2 , u’=2yy’. After the substitution we get u’/2+u=x, u’+2u=2x, the integrating factor is e2x. e2x u '+2 e2x u=( e2x u )'=2 x e2 x, integrate both sides

e2 x u=∫2 xe2 x dx=( integration by parts )=xe2 x−∫e2 x dx=x e2 x−12

e2 x+c

u= y2=x−12+ce−2x, from you can get y.

2. 2 x y '=10 x3 y5+ y , 2 x y 'y5 − 1

y 4 =10x3 , u=y-4 , u’= - 4y-5 y’ from

y 'y5 =−u '

4

The differential equation for u is -xu’-2u=20x3 . Multiply by -x we get

x2u'+2 xu=( x2 u )'=−20 x4, x2u=−4 x5+c , u= 1

y4 =−4 x3+ cx2 .

3. ( x2 cosy ) y '−2 xsiny=−1Use the new variable z=siny.

z’=(cosy)y’, after the substitution we get x2 z '−2 xz=−1 , integrating factor is m (x )= 1x2 .

multiply the normal equation z'−2

xz=−1

x2 by m(x), we get z 'x2−

2x3 z=( z

x2 )'

=−1x4 .

The left side is the derivative of zx2 .

zx2 =

13 x3 +c , the solution z=siny= 1

3 x+c x2.

Solve the following examples1. 3 y '+ y=(1−2 x ) y4 2. 2 xy y '+( x−1 ) y2=x2e x

3. y '= 2x+3

+x e− y (z=e y ) 4. y '= yx(xsinx−lny )

SOME APPLICATIONS of DIFFERENTIAL EQUATIONS.

A. Local property of particular solution.

Some example for this. 1. y’=x+1. The particular solution may have a local extr. value if x=-1. At this point y”=1 ˃0 therefore the solutions have a min.2. y’=y-x2 . Give the points at which the part. sol. have inflection point. y”=y’-2x=0 y’=2x, y-x2=2x, y= x2 +2xy’’’=y”-2= -2 ≠0 The solution has infl. point.3.Does the solution of y '=e y+ 2−x through the point P(e,-1) have a relative extreme value at P? y ' ( e ,−1 )=0 , y=y' {e} ^ {y+2} -1, y''(e,-1)=0-1=- At P the solution has a rel. max.

Decide whether the part. solution through the given point has a relative extreme value or inflection point.

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1. y’=(y2 -4)x+x-1, P(1,2)

2. y’=y3-x3 , P(1,1)3. y’= xy3 -y2 +2 , P(1,-1)

B. Orthogonal trajectories.

Given the family of curves F(x,y,c)=0 . At first we will find the differential equation of this

curves: We have to equations F (x.y.c)=0 , dFdx

=0 . Eliminate c , we get the differential

equation. Let us assume that this differential equation is y’=f(x,y) (differential equation of family of curves)

Differential equation of orthogonal trajectories : y '= −1f ( x , y) ( the product of slopes is equal

to -1). We have to solve this differential equation.Examples:1. Find the orthogonal trajectories. of y=cx2 . (parabolas)y’=2cx and c=y/x2 . y’=2y/x. The differential equation of orthogonal trajectories is: y’=-x/(2y), it is separable differential equation.2

ydy= - (x/2)dx, integrate, y2

2=−x2

4+c , y2+ x2

2=c∗¿(ellipses)

2. Find the orthogonal trajectories of x2+ y2=r2 (circles)Differential equation: x+yy’=0, y’= - x/yThe differential equation of orthogonal trajectories is: y’= y/x. It is a separable differential equation whose solution y= cx (lines) Solve the following examples: Find the orthogonal trajectories.1) x2− y2=c2 2) y=cx3 3) x2+4 y2=c 4) xy=c5) yx2 =c 6) y=ce8 x

Examples from the homepage:Differential equation: First order diff. eq.: 1,2,3,4,5,6,7

higher order diff, eq.: 1,2,3,4,5,6,8Exact diff. equations

9. a,b,c,g,h, system of linear diff. eq.Differential equation 1

1,2,3,4,5,6,7,14??10-13,15-17

From homepage of Ilona Nagy : Differential equationsHigher order differential equation: p: 1-11

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HIGHER ORDER DIFFERENTIAL EQUATIONS

F ( x , y , y ' , …, y ( n) )=0 it is called implicit nth order differential equation. y(n )=f (x , y , y ' , …, y (n−1)) it is called explicit nth order differential equation.Initial value problem: y(n )=f (x , y , y ' , …, y (n−1)) initial conditions: y ( x0 )= y0 , y ' ( x0 )= y ' 0 , …, y ( n−1 ) ( x0 )= y0

n−1

General solution: y= y (x , c1 ,…,cn)

Particular solution: Specify the values of the constantsy p= y (x , c10 ,…,cn 0)

Some second order differential equations which can be reduced to first order differential equation. F ( x , y , y ' , y ' ' )=0or y ' '=f ( x , y , y ' )

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General solution: y= y (x , c1 ,c2)Particular solution: c1=a , c2=b , y=y(x,a,b)Initial value problem: y ' '=f ( x , y , y ' ) , y ( x0 )= y0, y ' ( x0 )= y0

' (this is the slope of the tangent line to the particular solution through the point (x0 , y0) )

Reduction to first order differential equation.1. y’’=f(x), y '=∫ f (x ) dx , y=∫ (∫ f (x ) dx )dx2. F(x,y’,y’’)=0 y’=p(x), y’’=p’(x) we get F (x,p,p’)=0 from this equation we get p=p(x,c)=y’

3. F(y, y’,y’’)=0 , y’=p(y) , y ' '=dp( y )

dy∙ dydx

=p ' p , F(y,p,p’p)=0 From this equation we get

p(y), y’=p(y) , ∫ dyp( y )

=∫ dx

In this case the new independent variable is y.

Examples: 1. y’’=sinx+cosxy’=-cosx+sinx+c1 , y=-sinx-cosx+c1x+c2

2. xy’’=y’y’=p(x), xp’=p, separate and integrate p= c1xy’=c1x , y=c1x2/2+c2 3. y’’=2y3 , y(0)=1, y’(0)=1 y’=p(y), y”=p’p , p’p=2y3 , p(1)=1 after integration we get p2 /2=y4/2+c1 , from the given condition c1=0,p2=y4 from this using the initial condition p=y2 =y’.

Separate and integrate we get -1/y=x+c2 , -1=c2 , -1/y=x-1, y=−1x−1

Exercises:

1. y' '=lnx

x2 2.x y ' '= (1+2 x2 ) y ' 3. ( xlnx ) y' '= y '

4. y’’=2yy’, y(0)=1, y’(0)=15. y3y’’=-1, y(1)=1, y’(1)=0

SECOND ORDER LINEAR DIFFERNTAL EQUATION

A second order differential equation is called linear differential equation if it can be written in the form

y ' '+ p( x) y'+q (x ) y= f (x) and nonlinear if it cannot be written in this form. It is linear in the unknown function y and its derivatives, p, q f are continuous functions of x on the interval I.Standard form (1) is the form if coefficient of y’’ is equal to 1. If f(x)≡ 0, then the equation (2) y ' '+ p ( x ) y '+q ( x ) y=0 is called homogeneous differential equation. If f (x)≠ 0 , then (1) y ' '+ p ( x ) y '+q ( x ) y=f (x) is called nonhomogeneous differential equation.

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Some examples: y ' '+4 y=e−x , (1−x2) y ' '−2 x y '+6 y=0 are linear differential equations. x¿ is nonlinear differential equation.

A solution of differential equation on (a,b) is a function h(x) that has derivatives and satisfies the differential equation for all x in (a,b). Example: y’’-y=0, y1=ex , y2=e− x are solutions. The second order derivatives of these functions are the functions, therefore after substitution we get 0. y=3 ex−2 e−xis also solution. We can show that c1 y1+c2 y2isalso solution of differential equation. It is the linearity principle.

Theorem: Fundamental theorem for the homogeneous differential equation (2). For a homogeneous differential equation (2) any linear combination of two solutions on an open interval I is again solution of (2) on I. Sums and constant multiples of solutions are again solution.Proof: y1∧ y2 are solutions. Substitute c1 y1+c2 y2into the differential equation (2) (c1 y1+c2 y2 )' '+ p (c1 y1+c2 y2 )'+q (c1 y1+c2 y2 )=¿

¿c1 ( y1' '+ p y1

' +q y1 )+c2 ( y2' '+ p y2

' +q y2 )=0+0=0

This one does not hold for nonhomogeneous differential equation:Example: y’’+y=1 , solutions y1=1+sinx, y2=1+cosx. y1+y2 is not solution.-sinx-cosx+2+sinx+cosx=2≠1

Existence and uniqueness theorem for initial value problem:

If y’’+p(x)y’+q(x)y=0, y ( x0 )=k0 , y '(x0)=k 1 p ang q are continuous on an open interval I , x0 is in I, then the initial value problem has a unique solution y(x) on I.

Linear independence of solution, Wronskian (Wronski determinant)Definitions:a) y1 and y2 are called linearly independent on the interval I if k1 y1+k2 y2=0 implies k1=k2=0 b) y1 and y2 are called linearly dependent on I if k1 y1+k2 y2=0 holds for k1 , k2 not both zero.In this case one function is scalar multiple of the other one (y1=ky2 or y2=ly1). c) Wronski determinant of y1 and y2 defined by

W ( y1 , y2 )=| y1 y2

y '1 y '2|= y1 y '2− y ' 1 y2

Theorem: Given y’’+p(x)y’+q(x)y=0 (2) homogeneous linear differential equation. p and q are continuous on an open interval I , y1 and y2 are solutions of (2) on I. y1 and y2 are linearly dependent on I if and only if W(y1 ,y2 )=0 at x0 in I. If W(y1(x0), y2(x0))=0 , then W=0 on I.If there is an x1 in I at which W≠0, then y1 and y2 are linearly independent on I.

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(Very important that y1 and y2 are solutions of homogeneous linear differential equation. Without this the following statement is true: If W≠ 0 on the interval I, then y1 and y2 are linearly independent on I. Example: y1 and y2 are linearly independent on (−∞, ∞) and W=0

y1=t 3 , y2=|t|3 , W =|t 3 3 t2

t 3 3 t2|=0 for pos .numbers , W =| t 3 3 t 2

−t 3 −3 t2|=0 for negative numbers.

)

Proof of the theorem:

a) y1 and y2 are linearly dependent , y1=ky2, W =|y1 y2

y '1 y '

2|=k y2 y '

2−k y '2 y2=0

b) If W(y1,y2)=0 for x0 in I, then y1 and y2 are linearly dependent. The system k1 y1 ( x0 )+k2 y2 ( x0 )=0

k1 y '1 ( x0 )+k 2 y '

2 ( x0 )=0 The value of the determinant of the coefficient matrix is 0, therefore the system has a nontrivial solution, that is not both unknowns are equal to 0. Using this solution y(x)=k1y1(x)+k2y2(x) is a solution satisfying the initial conditions y(x0)=y’(x0)=0. Other solution y*=0 also satisfies the same initial conditions. p and q are continuous. From the uniqueness follows y=y*. k1y1+k2y2 =0, |k1|+|k2|>0 from follows that y1 and y2 are linearly dependent. c) W(y1(x0),y2(x0))=0 implies y1 and y2 are linearly dependent. From a) W=0 on I.If W≠0 at x1 in I, then y1 and y2 are linearly independent.

Theorem: If p and q are continuous on I, then (2) homogeneous linear differential equation has two linearly independent solutions (y1, y2).General solution: y(x)= c1y1+c2y2 where c1 and c2 are arbitrary real numbers.Proof: we will construct these solutions. x0 is in I.

y1(x0)=1, y1’(x0)=0y2(x0)=0, y2’(x0)=1

The Wronskian at x0 is W =|1 00 1|=1≠ 0 . The given solutions are linearly independent. These

functions form a basis for the solution space.

Theorem: If p, q are continuous on I , then every solution of homogeneous linear differential equation is of the form

Y= C1y1+C2y2 where y1 and y2 form a basis of solutions of differential equation, C1

and C2 are suitable constants. The homogeneous differential equation does not have singular solutions (solutions not obtainable from general solution). Proof: General solution y= c1y1+c2y2 on I Initial condition: y(x0)=Y(x0), y’(x0)=Y’(x0)The system: c1y1(x0)+c2y2(x0)=Y(x0) , c1y’1(x0)+c2y’2(x0)=Y’(x0)W≠0, From follows that the system has a unique solution, c1=C1 , c2=C2

y*(x)= C1y1+C2y2 , y*(x0)=Y(x0), y*’(x0)=Y’(x0)from the uniqueness theorem y*=Y on I. We have to give a basis. We get one solution by guessing or by other method. The method of reduction of order applicable to any equation as follows:y1 is a solution on I.

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Substitute y2= uy1 , multiply by qy’2= u’y1+uy’1 , multiply by py2’’=u”y1+2u’y1’+uy1’’ , multiply by 1

Substitute into (2) u”y1+u’( 2y1’+py1)+u(y1’’+py1’+qy1)=0, the coefficient of u is 0 .u’’y1+u’(2y1’+py1)=0 , this equation can be reduced to first order linear differential equation by the substitution z=u’z’y1+z(2y1’+py1)=0, solve for z, after by integration you will get u and y1 , y2=uy1 will give a basis for the solution space. (y1 and y2 are linearly independent)

Example: Reduction the order, given one solution y1, find y2 such that y1 and y2 form a basis. 1. (x+1)2 y ' '−2 ( x+1 ) y '+2 y=0, y1=x+1 is a solution, after substitution into the differential equation we get this.y2=u(x+1) , multiply by 2y2’=u’(x+1)+u , multiply by -2(x+1)y2’’= u’’(x+1)+2u’ , multiply by (x+1)2

u’’(x+1)3+u’(2(x+1)2-2(x+1)2)+u( 2(x+1)-2(x+1))=0u’’(x+1)3=0, u’’=0, u’=c, u= cx+k (k=0, c=1), it is enough to give one function. u=x, y2=x(x+1). y1 and y2 is a basis, general solution y= c1(x+1)+c2(x2+x)2. (x-1)y’’-xy’+y=0, y1=ex

3. (1-x)2y’’-4(1-x)y’+2y=0, y1=1/(1-x) , x≠1

Nonhomogeneous linear differential equations.

y’’+p(x)y’+q(x)y=f(x) (1)y’’+p(x)y’+q(x)y=0 (2)

We have taken the theorems for linear first order differential equations. I will repeat these theorems, but the proof is the same. Theorem: a) The difference of two solutions of (1) on an open interval I is a solution of (2) on Ib) The sum of a solution of (1) on I and a solution of (2) on I is a solution of (1) on I. Proof:a) y1 and y2 are solutions of (1). Substitute the difference into the (2)(y1-y2)’’+p(x)(y1-y2)’+q(x)(y1 -y2)= (y1’’+py1’+qy1)-(y2’’+py2’+qy2)=f-f=0b) y1 is a solution of (1), y2 is a solution of (2)(y1+y2)’’+p(y1+y2)’+q(y1+y2)=(y1”+py1’+qy1)+(y2’’+py2’+qy2)=f+0=f

Definition: General solution of (1) on I is of the form y=yh+yp (3)where yh = c1y1+c2y2 is the general solution of (2) on I and yp is any solution of (1) on I.

Theorem: If p, q, f are continuous on I, then every solution of (1) on I is obtained by assigning suitable values to arbitrary constants in a general solution (3) of (1) on I. Proof: ~y is a solution of (1), y is general solution of (1) Y=~y− y p is a solution of (2), Y is obtained from yh by assigning suitable values to the arbitrary constants c1 , c2 . From this ~y=Y + y p .

Finding particular solution by Variation of Parameters.

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yh=c1y1+c2y2 , y1 and y2 are linearly independent solutions of (2) .Variation of parameters: Replace the constants by the functions u(x), v(x) y p=u ( x ) y1+v (x) y2 which satisfies the (1).

y p' =u ' ( x ) y1+u ( x ) y1

' +v ' ( x ) y2+v (x) y2 ' It is enough to give one u(x) and v(x) for which we get a solution of (1). We can take the following condition:

u' (x ) y1+v ' (x ) y2=0 It is one condition for u(x) and v(x).This reduces the expression yp’ to the form

yp’=u(x)y1’+v(x)y2’ By differentiating

yp’’=u’y1’+uy1’’+v’y2’+vy2’’Substitute into (1) (the differential equation is normal form) u’y1’+v’y2’+u(y1’’+py1’+qy1)+v(y2’’+py2’+qy2)=f , the coefficients of u and v are 0 because y1 and y2 are solutions of (2).The system for u’ and v’ is

u’y1+v’y2=0u’y1’+v’y2’=f .

This system is a nonhomogeneous linear system for u’ and v’. The determinant of coefficients is the Wronski determinant which is not 0. There exist a unique solution, after integrate u’ and v’ we get a particular solution.

Example: (x-1)y’’-xy’+y= (x−1)2e x , the basis for the solution set of the corresponding homogeneous differential equation y1=ex , y2=x .

y p=u ( x ) ex+v ( x ) xThe system for derivatives is: u' ex+v ' x=0

u' ex+v ' 1=(x−1)ex (the coefficient of y’’ is equal to 1)Solution of this system: u’=x, v'=−ex

One function for u and v: u= x2

2, v=−ex , y p=

x2

2ex−x ex

General solution: y=c1 ex+c2 x+( x2

2−x)ex .

To find a basis for solution set of homogeneous linear differential equation is not difficult if the coefficients are constants. Second order linear differential equation with constant coefficients.

The differential equation is: y”+py’+qy=f(x), p and q are real numbers.The homogeneous linear differential equation : (2) y’’+py’+qy=0There exists one function whose derivatives are proportional to the function, therefore we are looking for the solution in this form: y=eλx , y '= λ eλx , y ' '=λ2e λx, substitute these function into the differential equation we get e λx ( λ2+ pλ+q )=0.The function e λx is solution if λ2+ pλ+q=0 . This quadratic equation is called characteristic equation.Let us denote the roots by λ1, λ2. For the roots are three cases:

I. The roots are different real numbers.

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The basis for the solution set is: e λ1 x , eλ2 x (the functions are linearly independent)The general solution: y=c1 eλ1 x+c2 eλ2 x

II. The equation has a double root: λ1,2=−p

2=λ0

The basis is: y1=eλ0 x , y2=x eλ0 x .The general solution: y=(c1+c2 x )e λ0 x

III. The equation has the complex roots (the coefficients of the equation are real numbers, the roots are complex conjugate of each other)

λ1,2=α ± iβ The real basis are: eαx cosβx, eαx sinβxThe general solution: y=eαx (c1 cosβx+c2 sinβx)(We have used the complex exponential function ez=ex+ iy=ex (cosy+isiny ) . In the complex algebra we have used Euler’s formula: e iφ=cosφ+isinφ. )

Some examples1. y’’-3y’+2y=0, the characteristic equation: λ2−3 λ+2=0 ,the roots are1∧2. The general solution: y=c1 ex+c2 e2 x.2. y’’-6y’+9y=0, the characteristic equation : λ2−6 λ+9=0 , λ1,2=3.The general solution: y=(c1+c2 x )e3 x .3. y’’+y=0: the characteristic equation: λ2+1=0 , λ1,2=±i.The general solution: y=c1 sinx+c2 cosx

Nonhomogeneous second order linear differential equation with constant coefficients.

(1) y’’+py’+qy=f(x), p and q are real numbers. We have yh=c1 y1+c2 y2 and we know that ynh= yh+ y p where yp is a particular solution of (1).Finding a particular solution:1. Variation of parameters2. Method of undetermined coefficients. It can be used if f(x) is a linear combination of exponential function, polynomial, sines, cosines. These functions have derivatives of the form similar to the function itself. The cases: 1. f ( x )=P (x)ewx, P(x) is a polynomial having degree n. . y p=xr Pn 1(x)ewx, Pn 1 (x )=an xn+…+a1 x+a0 , ai are unknowns, r=0,1,2 which is the multiplicity of w in characteristic equation.2. f ( x )=ewx ( Pn ( x ) cosax+Qm (x ) sinax )n=max(n,m) y p=xr ewx(Pn1 ( x )cosax+Qn1 ( x ) sinax) Pn 1 (x )=an xn+…+a1 x+a0 ,Q n1 ( x )=bn xn+…+b1 x+b0 , ai and bk are unknowns.r=0,1 which is the multiplicity of w+ia in the characteristic equation.

Superposition principle: If f=f1 +f2 , then we are finding particular solution corresponding to the terms of f and add these functions.

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Examples: 1. y’’-3y’+2y=f(x)We know that the general solution of homogeneous linear differential equation yh=c1 ex+c2 e2 x .a) f(x)=4x+3, yp=ax+b, y’=a, y’’=0Substitute into the differential equation, combine like terms

2ax+2b-3a=4x+3. Equating the corresponding coefficients, we get a system:

2a=4, 2b-3a=3Solution: a=2, b=9/2yp=2x+9/2, ynh=c1 ex+c2 e2 x+2 x+9/2 b) f(x)= sinxyp=asinx+bcosx , multiply by 2yp’=acosx-bsinx , multiply by -3yp’’=-asinx-bcosx , multiply by 1After multiplication and combining like terms we get :sinx( a+3b)+cosx(b-3a)=sinx. The system foe a, b : a+3b=1, b-3a=0, solution a=1/30, b=1/10

ynh=c1 ex+c2 e2 x+ 130

sinx+ 110

cosx

c) f ( x )=2e− x

y p=a e−x , yp' =−a e− x , y p

' '=a e−x , After the substitution and equating the corresponding coefficients we get 6a=2, a=1/3

ynh=c1 ex+c2 e2 x+ 13

e− x

d) f ( x )=5 ex

If we are finding the particular solution in the form y p=a ex, the we get (a−3a+2a )ex=5 ex, this is a contradiction .(exis a solution of the homogeneous differential equation)The correct form for is y p=xaex. There is an outer resonance. y p

' =aex+ax ex , y p' '=2a ex+axe x .

After substitution we get x ex (2 a−3a+a )+ex (−3 a+2a )=5ex. The first term is 0. We get a=-5, ynh= yh−5 x ex .

2. y’’-6y’+9y=f(x) yh=(c1+c2 x )e3 x

a) f(x)=9x2 yp=ax2+bx+c , multiply by 9yp’=2ax+b , multiply by -6yp’’=2a , multiply by 1After multiplication and combining like term we get9ax2+x(9b-12a)+9c-6b+2a=9x2 , the system: 9a=9, 9b-12a=0, 9c-6b+2a=0 The solution: a=1, b=4/3, c= 2/3

ynh=( c1+c2 x ) e3 x+x2+ 4 x3

+ 23

b) f ( x )=x ex

y p=(ax+b)ex , multiply by 9 y p

' =aex+(ax+b)ex , multiply by -6 y p

' '=2 a ex+(ax+b)ex , multiply by 1

22

After multiplication and combine like term we get x ex ( 4a )+ex (4 b−4a )=x ex , the system : 4a=1, b=a=1/4

yh=( c1+c2 x ) e3 x+ 14

(x+1)ex

c) f ( x )=3 e3 x (3 is double root of the characteristic equation) y p=x2 ae3x , multiply by 9 y p

' =a 2 x e3 x+a x2 3e3 x , multiply by -6 y p

' '=a2 e3 x+a12 x e3 x+a9 x2 e3x , multiply by 1After multiplication and combining like terms we get x2e3x (9 a−18 a+9a )+x e3x (−12a+12a )+2a e3 x=3e3 x, 2a=3, a=3/2

ynh=( c1+c2 x ) e3 x+ 32

x2 e3 x

d) f ( x )= e3x

x3 . We can use the variation of parameters.

y p=c1 ( x ) e3 x+c2 ( x ) x e3 x ,The system for derivatives: c1

' (x ) e3x+c2' ( x ) xe3 x=0

c1' (x ) 3 e3 x+c2

' ( x ) ( e3 x+x 3e3 x )= e3 x

x3

c2' (x )= 1

x3 , c1' (x )=−1

x2 , c1 (x )=1x

, c2 (x )=−12 x2 , y p=

12 x

e3 x

y p=( c1+c2 x ) e3 x+ 12 x

e3 x

3. y’’+y=f(x) yh=c1 sinx+c2 cosxa) f(x)=sin2xyp=asin2x+bcos2x , multiply by 1yp’=2acos2x-2bsin2x , multiply by 0yp’’=-4asin2x-4bcos2x , multiply by 1After substitution we get: -3asin2x-3bcos2x=sin2x

a= -1/3 , b= 0 , yp=(-1/3)sin2xynh=c1sinx+c2cosx-(sinx)/3b) f ( x )=(x+2)ex y p=(ax+b)ex , multiply by 1 y p

' =a ex+(ax+b)ex , multiply by 0 y p

' '=2 aex+(ax+b)ex , multiply by 1After substitution we get x ex (2 a )+ex (2 b+2 a )=(x+2)ex

The system for a, b: 2a=1, 2a+2b=2, the solution: a=b=1/2

The particular solution is: y p=12(x+1)ex

ynh=c1 sinx+c2cosx+ 12(x+1)ex

c) f(x)=6x+1yp=ax+byp’=a

23

yp’’=0Substitute into the differential equation we get : ax+b=6x+1From the solution for a, b is : a=6, b=1ynh=c1sinx+c2cosx+6x+1d) f ( x )= (x+2 ) ex+6 x+1We have a particular solution for the terms of f(x) , from the superposition principle follows that we have to add these particular solutions.

y p=12

( x+1 ) ex+6 x+1

ynh=c1 sinx+c2cosx+ 12

( x+1 ) ex+6 x+1

e) f(x)=2sinxyp=x ( asinx+bcosx) , multiply by 1yp’= asinx+bcosx+x(acosx-bsinx) , multiply by 0yp’’=(acosx-bsinx)2 + x(-asinx-bcosx) , multiply by 1After substitution we get xsinx(a-a)+xcosx(b-b)+sinx(-b-b)+cosx(a+a)=2 sinxThe system for a,b is -2b=2, 2a=0The solution is : a=0, b= -1, yp=-xcosxynh=c1sinx+c2cosx -xcosx

Reduction to linear differential equation with constant coefficients.

Euler-Cauchy differential equation.

General form of this differential equation is: x2 y ' '+ax y '+by=f ( x )I will show one transformation which leads to linear differential equation with constant coefficient. If x is positive (x is different from 0), then x=e t ¿ ) and for the new function y(t) we will get a differential equation with constant coefficients.

We will solve the homogeneous linear differential differential equation. x2 y ' '+ax y '+by=0 Let x=e t , t=lnx.

y '=dydx

=dydt

∙ dtdx

= y 1x

y ' '= ddx ( dy

dx )= y dtdx

1x+ y (−1

x2 )= y 1x2 − y 1

x2

Substitute: y− y+a y+by=0 , y=eλt , the characteristic equation λ2+(a−1 ) λ+b=0Some examples:

1. x2 y ' '−2 x y '+2 y=1x , x=e t

y−3 y+2 y=e−t , the characteristic equation: λ2−3 λ+2=0 ,the roots are1∧2.The basis for the solution set is: e t , e2t .

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y p=ae−t ,substitute the derivatives we get e−t (2 a+3 a+a )=e−t , a=1/6

y (t )=c1e t+c2 e2 t+ e−t

6

y ( x )=c1 x+c2 x2+ 16 x

2. x2 y ' '−x y '+ y=0 , x=e t

y−2 y+ y=0 , the characteristic equation is: λ2−2 λ+1=0 , roots λ1,2=1The basis is: e t , t e t.General solution is: y (t )=c1e t+c2 t e t , y ( x )=c1 x+c2 xlnx .3. x2 y ' '−x y '+2 y=0 , x=e t

y−2 y+2 y=0 , the characteristic equation is: λ2−2 λ+2=0 , λ1,2=1±iThe basis is: e t cost , et sintGeneral solution is: y (t )=c1e t cost+c2 et sint , y ( x )=c1 xcos ( lnx )+c2 xsin(lnx) .

HIGHER ORDER LINEAR DIFFERENTIAL EQUATIONS

The differential equation which are of the form are called(1) y(n )+ pn−1 ( x ) y(n−1)+…+ p1 ( x ) y '+p0 (x ) y=f (x) nonhomogeneous differential equation, standard form that is the coefficient of highest order derivative is 1.(2) y(n )+ pn−1 ( x ) y (n−1)+…+ p1 ( x ) y '+ p0 (x ) y=0 homogeneous differential equation .

Theorem: For (2) the sum and scalar multiples of solutions on some open interval I are again solutions of (2) on I. Definitions:

1. y1 , y2 , …, yn are called linearly independent on an open interval I where they are defined if

∑i=1

n

k i y i=0 implies that k1=…=kn=0

2. These functions are called linearly dependent on an open interval I if the equation

∑i=1

n

k i y i=0 holds for some k1 , …,kn not all zero.

3. Wronskian or Wronski determinant of y1 , …, yn is

W ( y1 , …, yn )=det [ y1 ⋯ y n

⋮ ⋱ ⋮y1

(n−1) ⋯ yn(n−1)]

4. General solution of (2)

y=∑i=1

n

c i y i(x ), c í∈ R , y1 , …, yn is a basis for the solution set, these functions are linearly

independent on an open interval I. Particular solution: we assigns specific values to the n constants c1,…,cn .

Theorem: If in (2) pi ( x ) , i=0,1 ,…,n−1are continuous on an open interval I, x0∈ I ,

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y (l ) ( x0 )=k l , l=0,…,n-1 , then the initial value problem has a unique solution on I.

Examples:

1. Show that x, 3x, x2 are linearly dependent on any interval. 3x +(-1)3x+0x2=0

2. Show that x, x2, x3 are linearly independent on any interval, for example (-4,4).

Let us consider the equation ∑i=1

3

k i xi=0 . taking x = -1, 1, 2 we get a system for the

unknowns. −k 1+k2−k3=0 k1+k2+k3=0 2k1+4 k2+8 k3=0 , solution: k1=k2=k3=0. This means the functions are linearly independent.

3. Solve the initial value problem x3 y ' ' '−3x2 y+6xy'-6y=0. y(1)=2, y'(1)=1, y''(1)=-

yh=∑i=1

3

c i xi. In example two we have shown that x, x2, x3 are linearly independent. You can

show that these functions are solutions. We have to find the particular solution satisfying the initial conditions.The system for ci i=1,2,3 is: c1+c2+c3=2

c1+2 c2+3 c3=1 2c2+6 c3=−4

The solution is: c1=2 , c2=1 , c3=−1 . The particular solution is: y=2 x+x2−x3 .

Theorem: The coefficients in (2) are continuous functions on an open interval I. y1 , …, yn are solutions of (2). These solutions are linearly dependent on I if and only if their W (Wronskian) is zero for some x0 in I.If W is zero for x0 in I , then W=0 on I. If there is x1 in I at which W≠0, then y1 , …, yn are linearly independent on I.

Theorem: Existence of general solution.If p0 ( x ) , …, pn−1 ( x ) are continuous on an open interval I, then (2) homogeneous linear differential equation has a general solution.For this one we have to show that there are n linearly independent solution.

Theorem: (general solution of (2) includes all solution)If p0 ( x ) , …, pn−1(x) are continuous on an open interval I and y1 , …, yn is a basis of solution of homogeneous linear differential equation (2), then every solution Y(x) of (2) on I is of the

form Y ( x )=∑i=1

n

C i y i , Ci are suitable constants.

Proof: the initial value problem: Y=∑i=1

n

c i y i , the initial condition is given.

x=x0 , c1 y1 ( x0 )+…+cn yn ( x0 )=Y (x0)…..

26

c1 y1(n−1) ( x0 )+…+cn yn

(n−1 ) ( x0 )=Y ❑(n−1)(x0)

This system has unique solution, C1, …, Cn .

y∗( x )=∑i=1

n

Ci y i(x) , for this function y∗¿❑(k ) ( x0 )=Y (k )(x0)¿ for k=0, 1,…, n-1.

from the uniqueness theorem follows y∗( x )=Y (x) .

Important general properties of linear differential equation(1) y(n )+ pn−1 ( x ) y (n−1)+…+ p1 ( x ) y '+ p0 (x ) y=f (x )(2) y(n )+ pn−1 ( x ) y (n−1)+…+ p1 ( x ) y '+ p0 (x ) y=0

1. The trivial solution y=0 is a solution of (2).2. The sum of a solution of (1) and a solution of (2) is a solution of (1).3. The difference of two solutions of (1) is a solution of (2).4. The sum of two solutions of (1) is not solution of (1).5. A multiple cy of a solution of y of (1) with c≠1 is not solution of (1).

Examples:1. Show that the given function form a basis of solutions of the given differential equation on an open interval.a) y1=e−x , y2=x e−x , y3=x2e− x is a basis of the solution set of y ' ' '+3 y ' '+3 y '+ y=0, y1 is a solution: substitute into the differential equation, −e− x+3 e− x−3 e−x+e− x=0You can substitute y2 and y3.These functions form a basis:

W =| e− x x e−x x2e− x

−e− x (1−x)e−x (2 x−x2)e−x

e− x (x−2)e−x (x2−4 x+2)e−x|=2e−3 x ≠ 0 , the functions are linearly independent.

b) y1=cosx , y2=sinx , y3=e− x is a basis of solutions of y ' ' '+ y +y'+y=.yi (i=1,2,3) is a solution:y1 : sinx-cosx-sinx+cosx=0y2 : -cosx-sinx+cosx+sinx=0y3 : −e− x+e−x−e− x+e−x=0The given function are linearly independent:

W =| cosx sinx e−x

−sinx cosx −e−x

−cosx −sinx e−x |=2 e− x ≠ 0

The functions form a basis.2. The general solution of y’’’-y’=0 is y=c1+c2 e−x+c3 ex. Find the particular solution satisfying the initial condition.y(0)=6, y’(0)= -4, y’’(0)=2.The system for the coefficients is: c1+c2+c3=6 ,−c2+c3=−4 , c2+c3=2.The solution is c1=4, c2=3, c3= -1. The particular solution is: y=4+3 e−x−ex .

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HIGHER ORDER LINEAR DIFFERENTIAL EQUATION with CONSTANT COEFFICIENTS

The homogeneous linear differential equation with constant coefficients.

y(n )+an−1 y(n−1)+…+a1 y '+a0 y=0 , ai∈ RThe solution can be looked for in the form y=eλx .For λ we get the equation so called the characteristic equation: λn+an−1 λn−1+…+a1 λ+a0=0

The basis depends on roots:1. The roots are different real numbers. e λ1 x , eλ2 x ,…,eλn x are linearly independent, they form a basis.2. The equation has a simple complex roots. λ1,2=α ± iβ, y1=eαx cosβx , y2=eαx sinβx are real linearly independent solutionsThe λ1,2=α ± iβ are double roots: y1=eαx cosβx , y2=eαx sinβx , y3=xeαx cosβx , y4=x eαx sinβx are linearly independent solutions.3. The equation has real multiple roots. λ is a root of multiplicity m. The other roots are different from λ.There are m corresponding linearly independent solutions: e λx , xe λx , …, xm−1e λx

THE NONHOMOGENEOUS LINEAR DIFFERENTIAL EQUATIONS

(1) y ( n)+pn−1 ( x ) y (n−1)+…+ p1 ( x ) y '+ p0 ( x ) y=f (x )(2) y(n )+ pn−1 ( x ) y (n−1)+…+ p1 ( x ) y '+ p0 ( x ) y=0

Theorem:a) The difference of two solutions of (1) on an open interval I is a solution of (2) on I.b) The sum of a solution of (1) and a solution of (2) on an open interval I is a solution of (1) on I. (the proof is the same which is for second order linear differential equation)

Theorem: The general solution of the nonhomogeneous linear differential equation: (this is the same which is for second order linear differential equation) ynh= yh+ y p where yhis the general solution of the corresponding homogeneous linear

differential equation yh(x )=∑i=1

n

c i y i(x) and y pis any solution of nonhomogeneous linear

differential equation.

Finding particular solution: (the methods are the same which are for second order differential equation)Variation of parameters:

28

y p=∑i=1

n

c i(x) y i

We can write a system for c i' (x), solve this and integrate.

NONHOMOGENEOUS LINEAR DIFFERENTIAL EQUATION with CONSTANT COEFFICIENTS y(n )+an−1 y(n−1)+…+a1 y '+a0 y= f ( x ) , a i∈R

Finding the particular solution:

1. Variation of parameters:yp=c1(x)y1+…+cn(x)yn

2. The method of undetermined coefficients:for example f ( x )=eax , y p=c xk eαx where k is the multiplicity of α in the characteristic equation.

Examples:

1. y’’’-y’=0, the characteristic equation: λ3−λ=λ ( λ2−1 )=0The roots are: 0, 1, -1The general solution: y=c1+c2 ex+c3 e−x.2. y ' ' '−2 y ' '− y '+2 y=0 The characteristic equation: λ3−2 λ2−λ+2=( λ−2 ) ( λ2−1 )=0 . The roots are : 1, -1, 2The general solution: y=c1 ex+c2 e−x+c3 e2 x .3. y(5 )+ y ' ' '=0 ,the characteristic equation: λ5+ λ3=λ3 ( λ2+1 )=0The roots are: 0 (multiplicity is 3), ± i.The general solution: y=c1+c2 x+c3 x2+c4 cosx+c5 sinx .4. y(4)−5 y ' '+4 y=sinx , the characteristic equation: λ4−5 λ2+4=0The roots are: ± 1, ± 2, yh=c1 ex+c2 e−x+c3 e2 x+c4 e−2 x

A particular solution can be looked for in the form: yp=asinx+bcosx. yp’=acosx-bsinx, yp’’=-asinx-bcosx, yp’’’=-acosx+bsinx, y p

(4)=asinx+bcosxThe system for a and b is: 10a=1, 10b=0. a= 1/10, b=0yp=sinx/10, ynh=yh+yp .5. y(4)−2 y ' '+ y=64 e3 x, the characteristic equation: λ4−2 λ2+1=( λ2−1 )2=0The roots are : 1, -1 (multiplicity is 2)The general solution: yh=( c1+c2 x ) ex+ (c3+c4 x ) e− x .A particular solution is of the form: y p=a e3 x, y p

' =3 a e3 x , y p' '=9 a e3 x ,

y p' ' '=27 ae3 x , y p

(4)=81a e3 x

The equation for a: 64a=64, a=1 , y p=e3 x , ynh=yh +yp .

SYSTEM OF DIFFERENTIAL EQUATION

29

The system of first order differential equations are of the form:

y i'=f i (t , y1 , …, yn ) , i=1 ,… ,n

Special case: the functions fi are linear. We will consider only system of first order linear differential equations with constant coefficients. A=[aik ] , i,k=1,2,…,n, aik are real numbers. y1 (t ) ,…, yn ( t ) are the unknown function, fi(t) are coordinates of f(t).The system: y ' (t )=A y (t)+ f (t).

y i' ( t )=aí 1 y1 ( t )+…+a¿ yn ( t )+ f i(t ) , i=1,2,…,n

If f(t)=0, the system is called homogeneous system.

The general solution: ynh= yh+ y p where yh is the general solution of the corresponding homogeneous system. yp is any particular solution of the nonhomogeneous system.

Definition: If φ1( t) ,… ,φn(t) are linearly independent on I , then they form a basis for the solution set. (the determinant of coordinates of solutions is different from 0) The general solution of the homogeneous system:

yh=∑i=1

n

c iφ i ( t )=ϕ( t)c , Φ(t):= basic matrix, whose column vectors are the elements of

the linearly independent set.

Finding a particular solution: variation of parameter. Write p(t) instead of c . y p=ϕ p( t)

Substitute into the nonhomogeneous system we get ϕ ' p (t )+ϕ p ' (t)=Aϕ p( t )+f ( t) , ϕ p' (t)=f ( t)

p ' (t )=ϕ−1(t ) f (t) , from with integration we get p(t) and the particular solution.

The next problem: How we can get the basic matrix? (there is a matrix function, using this we can get the basic matrix y ' ( t )=A y ( t) , y (t )=e At c . The method to find the matrix e At depends on eigenvectors of the matrix A. ) There is a special case in which you can use linear algebra to find the basic matrix.

This case is the following: The matrix A is a matrix of simple structure, that is A has a complete set of eigenvectors. (there are n linearly independent eigenvectors)

Theorem: The eigenvalues of the matrix are λ1 , …, λn and the corresponding eigenvectors are u1 , …,un (these vectors are linearly independent, complete set of eigenvectors) . The column vectors of basic matrix are φ i( t)=e λi t ui , i=1,2,…,n

You can show this by usage the spectral decomposition of the matrix.Let the matrix U is the matrix whose column vectors are the eigenvectors

30

U=[u1 u2 ….un ] , the spectral decomposition of the matrix A is A=U ΛU−1 .Using this the system can be written in the form y '=U Λ U−1 y , U−1 y '=Λ U−1 y , the new variable is x=U−1 y . For this new variable the system is: x '=Λ x or x i

'=λ i x i , i=1,…,nThe solution is x i=e λi t c i , i=1,…,n x= ⟨e λ1 t …eλ n t ⟩ c , y=U x=U ⟨e λ1 t …. eλ nt ⟩ c=[eλ 1 t u1…. e λn t un ] c=ϕc

Each nth order linear differential equation with constant coefficient is equivalent to system of first order linear differential equation with constant coefficient.I will show this for second order differential equation. Given y”+py’+qy=0, y=y1

y’=y1’=y2

y’’=y2’=-py2-qy1,

The system: [ y ' 1

y2 ' ]=[ 0 1−q −p][ y1

y2]The characteristic equation det ( λ I−A )=|λ −1

q λ+ p|=λ2+ pλ+q=0

It is the same equation which we have got for the second order differential equation.Example:

1. [ y1 'y2 ' ]=[ 0 1

−3 −4] [ y1

y2] , det [ λ −13 λ+4]=λ2+4 λ+3=0

The eigenvalues are: λ1=−1 , λ2=−3

The eigenvectors are: λ1=−1 , x+ y=0 ,u1=[ 1−1]

λ2=−3 , 3 x+ y=0 ,u2=[ 1−3 ]

Solution: y=c1[ 1−1]e−t+c2[ 1

−3]e−3 t

2. [ y1 'y2 ' ]=[0 1

1 0 ][ y1

y2]+[ 6−3 ]e2 t

The characteristic equation: λ2−1=0

λ1=1 ,u1=[11] , λ2=−1 ,u2=[−11 ]

yh=c1[11]e t+c2[−11 ]e−t

The coordinates of particular solution:y p 1=c1 (t ) et−c2( t)e−t

y p 2=c1 ( t ) e t+c2(t )e−t

The system for derivatives of c i '(t) isc1

' ( t ) e t−c2' (t )e−t=6 e2 t

31

c1' (t ) e t+c2

' (t ) e−t=−3e2 t

The solution is: c1' ( t )=3

2et , c2

' (t )=−92

e3 t

One solution is : c1 (t )=32

et , c2 ( t )=−32

e3 t

Particular solution is: y p 1=3 e2 t , y p 2=0The general solution of nonhomogeneous system is:

y1=c1 e t−c2 e−t+3e2 t , y2=c1 et+c2e−t

BOUNDARY VALUE PROBLEM

When the values of a solution to a differential equation are specified at two different points , these conditions are called boundary conditions.

The purpose of this exercise is to be show that for boundary value problems there no existence -uniqueness theorem that is analogous to Existence and Uniqueness Theorem for initial value problem.

Given that every solution to(1) y’’+y=0 is of the form y ( x )=c1 cosx+c2 sinx where c1 , c2 are arbitrary constants , show that (a) There is a unique solution to (1) satisfies the boundary conditions y(0)=2 and y(π/2)=0. The system 2=c1 ,0=c2 has a unique solution.(b) There is no solution to (1) that satisfies the boundary conditions y(0)=2 and y(π)=0.The system 2=c1 , 0=−2+0 , contradiction.(c) There are infinitely many solutions to (1) that satisfy y(0)=2 and y(π)=-2.The system 2=c1 ,−2=−c1+0, c2∈ R The solutions are: y ( x )=2 cosx+c2 sinx .

There are examples in which the boundary conditions contains derivative.

32

Example:Find the non-trivial solution of the given differential equation satisfying the boundary conditions y ”+ λy=0 , y ' (0 )=0 , y ' ( π )=0 The characteristic equation is: μ2+λ=0There are 3 cases:a) λ=0, μ=0 , y=c1+c2 x y '=c2 , from the boundary conditions we get c1≠ 0 , c2=0one non -trivial solution y=1b) λ<0 , μ=±√− λ, y=c1 e√−λ x+c2 e−√−λ x , y '=√−λ c1 e√−λ x−√−λ c2 e−√− λx

the system for the unknowns ís : c1=c2 , c1 (e√− λ π−e−√−λ π )=0 , solution: c1=c2=0There is only the trivial solution: y=0.c) λ>0 , λ=ν2, μ=±iυ , y=c1 cosνx+c2 sinνx, y '=−c1 νsinνx+c2 νcosνxfrom the boundary conditions we get: 0=c2 ν ,¿c2=0 , 0=−c1 νsinν π ,We know that c1≠ 0 , sinνπ=0 , ν=1,2, …The solutions are called the eigenfunctions: 1 , cosx ,cos 2 x ,…,coskx ,…The numbers: λ= 0, 1, 4,, are called eigenvalues.

33