MA 108 - Ordinary Differential Equationsdey/diffeqn_spring14/lecture2_D2.pdf · Santanu Dey Lecture 2. Outline of the lecture 1 Initial value problems 2 Separable DE 3 Equations reducible

MA 108 - Ordinary Differential Equations

Santanu Dey

Department of Mathematics,Indian Institute of Technology Bombay,

Powai, Mumbai [email protected]

February 25, 2014

Santanu Dey Lecture 2

Outline of the lecture

1 Initial value problems

2 Separable DE

3 Equations reducible to separable form


First order ODE & Initial Value Problem for first orderODE

We now consider first order ODE of the form F (x , y , y ′) = 0 or

y ′ = f (x , y) .

Consider a linear first order ODE of the form y ′ + a(x)y = b(x) .

If b(x) = 0, then we say that the equation is homogeneous.

Note that the solutions of a homogeneous differential equationform a vector space under usual addition and scalar multiplication

Definition

Initial value problem (IVP) : A DE along with an initial condition isan IVP.

y ′ = f (x , y), y(x0) = y0.




y ′ = f (x , y) .




Definition


y ′ = f (x , y), y(x0) = y0.




y ′ = f (x , y) .




Definition


y ′ = f (x , y), y(x0) = y0.




y ′ = f (x , y) .




Definition


y ′ = f (x , y), y(x0) = y0.




y ′ = f (x , y) .




Definition


y ′ = f (x , y), y(x0) = y0.


Examples

Given an amount of a radioactive substance, say 1 gm, find theamount present at any later time.

The relevant ODE isdy

dt= −k · y .

Initial amount given is 1 gm at time t = 0. i.e.,

y(0) = 1.

By inspection, y = ce−kt , for an arbitrary constant c , is a solutionof the above ODE. The initial condition determines c = 1. Hence

y = e−kt

is a particular solution to the above ODE with the given initialcondition.


Examples

Given an amount of a radioactive substance, say 1 gm, find theamount present at any later time.The relevant ODE is

dy

dt= −k · y .


y(0) = 1.


y = e−kt



Examples


dy

dt= −k · y .

Initial amount given is 1 gm at time t = 0.

i.e.,

y(0) = 1.


y = e−kt



Examples


dy

dt= −k · y .


y(0) = 1.


y = e−kt



Examples


dy

dt= −k · y .


y(0) = 1.

By inspection, y = ce−kt , for an arbitrary constant c , is a solutionof the above ODE.

The initial condition determines c = 1. Hence

y = e−kt



Examples


dy

dt= −k · y .


y(0) = 1.

By inspection, y = ce−kt , for an arbitrary constant c , is a solutionof the above ODE. The initial condition determines c =

1. Hence

y = e−kt



Examples


dy

dt= −k · y .


y(0) = 1.


y = e−kt



Geometrical meaning :dy

dt= −2 · y

1 Suppose that y has certain value.

From the RHS of the DE, we

obtaindy

dt. For instance, if y = 1.5,

dy

dt= −3. This means that the

slope of a solution y = y(t) has the value −3 at any point wherey = 1.5.

2 Display this information graphically in ty -plane by drawing short linesegments or arrows at several points on y = 1.5.

3 Similarly proceed for other values of y .

4 The figures given in next two slides are examples of direction fieldsor slope fields.

5 An isocline (a series of lines with the same slope) is often used tosupplement the slope field. In an equation of the formdy

dt= f (t, y), the isocline is a line in the ty -plane obtained by

setting f (t, y) equal to a constant.



dt= −2 · y

1 Suppose that y has certain value. From the RHS of the DE, we

obtaindy

dt.

For instance, if y = 1.5,dy











dt= −2 · y


obtaindy


dy











dt= −2 · y


obtaindy


dy











dt= −2 · y


obtaindy


dy











dt= −2 · y


obtaindy


dy











dt= −2 · y


obtaindy


dy











dt= −2 · y


obtaindy


dy










Slope field


Slope field


Examples

Find the curve through the point (1, 1) in the xy -plane having at

each of its points, the slope −y

x.

The relevant ODE isy ′ = −y

x.

By inspection,

y =c

xis its general solution for an arbitrary constant c ; that is, a familyof hyperbolas.The initial condition given is

y(1) = 1,

which implies c = 1. Hence the particular solution for the aboveproblem is

y =1

x.


Examples



x.


x.

By inspection,

y =c


y(1) = 1,


y =1

x.


Examples



x.


x.

By inspection,

y =c

xis its general solution for an arbitrary constant c ; that is, a familyof hyperbolas.

The initial condition given is

y(1) = 1,


y =1

x.


Examples



x.


x.

By inspection,

y =c


y(1) = 1,

which implies c = 1.

Hence the particular solution for the aboveproblem is

y =1

x.


Examples



x.


x.

By inspection,

y =c


y(1) = 1,


y =1

x.


Slope field


Slope field


Examples

A first order IVP can have

1 NO solution :

|y ′|+ |y | = 0, y(0) = 3.

2 Precisely one solution : y ′ = x , y(0) = 1. What is thesolution?

3 Infinitely many solutions: xy ′ = y − 1, y(0) = 1 The solutionsare y = 1 + cx .

Motivation to study conditions under which the solution wouldexist and the conditions under which it will be unique!

We first start with a few methods for finding out the solution offirst order ODEs, discuss the geometric meaning of solutions andthen proceed to study existence-uniqueness results.


Examples


1 NO solution : |y ′|+ |y | = 0, y(0) = 3.






Examples


1 NO solution : |y ′|+ |y | = 0, y(0) = 3.

2 Precisely one solution :

y ′ = x , y(0) = 1. What is thesolution?





Examples


1 NO solution : |y ′|+ |y | = 0, y(0) = 3.

2 Precisely one solution : y ′ = x , y(0) = 1.

What is thesolution?





Examples


1 NO solution : |y ′|+ |y | = 0, y(0) = 3.


3 Infinitely many solutions:

xy ′ = y − 1, y(0) = 1 The solutionsare y = 1 + cx .




Examples


1 NO solution : |y ′|+ |y | = 0, y(0) = 3.


3 Infinitely many solutions: xy ′ = y − 1, y(0) = 1

The solutionsare y = 1 + cx .




Examples


1 NO solution : |y ′|+ |y | = 0, y(0) = 3.






Examples


1 NO solution : |y ′|+ |y | = 0, y(0) = 3.






Examples


1 NO solution : |y ′|+ |y | = 0, y(0) = 3.






Separable ODE’s

An ODE of the form

M(x) + N(y)y ′ = 0

is called a separable ODE.

Let H1(x) and H2(y) be any functions such that H ′1(x) = M(x)and H ′2(y) = N(y).Substituting in the DE, we obtain

H ′1(x) + H ′2(y)y ′ = 0.

Using chain rule,d

dxH2(y) = H ′2(y)

dy

dx.

Hence,d

dx(H1(x) + H2(y)) = 0.

Integrating, H1(x) + H2(y) = c, where c is an arbirtaryconstant.Note: This method many times gives us an implicit solution andnot necessarily an explicit one!


Separable ODE’s

An ODE of the form

M(x) + N(y)y ′ = 0

is called a separable ODE.Let H1(x) and H2(y) be any functions such that H ′1(x) = M(x)and H ′2(y) = N(y).

Substituting in the DE, we obtain

H ′1(x) + H ′2(y)y ′ = 0.

Using chain rule,d

dxH2(y) = H ′2(y)

dy

dx.

Hence,d

dx(H1(x) + H2(y)) = 0.



Separable ODE’s

An ODE of the form

M(x) + N(y)y ′ = 0

is called a separable ODE.Let H1(x) and H2(y) be any functions such that H ′1(x) = M(x)and H ′2(y) = N(y).Substituting in the DE, we obtain

H ′1(x) + H ′2(y)y ′ = 0.

Using chain rule,d

dxH2(y) = H ′2(y)

dy

dx.

Hence,d

dx(H1(x) + H2(y)) = 0.



Separable ODE’s

An ODE of the form

M(x) + N(y)y ′ = 0


H ′1(x) + H ′2(y)y ′ = 0.

Using chain rule,d

dxH2(y) = H ′2(y)

dy

dx.

Hence,d

dx(H1(x) + H2(y)) = 0.



Separable ODE’s

An ODE of the form

M(x) + N(y)y ′ = 0


H ′1(x) + H ′2(y)y ′ = 0.

Using chain rule,d

dxH2(y) = H ′2(y)

dy

dx.

Hence,d

dx(H1(x) + H2(y)) = 0.



Separable ODE’s

An ODE of the form

M(x) + N(y)y ′ = 0


H ′1(x) + H ′2(y)y ′ = 0.

Using chain rule,d

dxH2(y) = H ′2(y)

dy

dx.

Hence,d

dx(H1(x) + H2(y)) = 0.

Integrating, H1(x) + H2(y) = c, where c is an arbirtaryconstant.Note:

This method many times gives us an implicit solution andnot necessarily an explicit one!


Separable ODE’s

An ODE of the form

M(x) + N(y)y ′ = 0


H ′1(x) + H ′2(y)y ′ = 0.

Using chain rule,d

dxH2(y) = H ′2(y)

dy

dx.

Hence,d

dx(H1(x) + H2(y)) = 0.



Separable ODE - Example 1

Solve the DE :y ′ = −2xy .

Separating the variables, we get :

dy

y= −2xdx .

Integrating both sides, we obtain :

ln |y | = −x2 + c1.

Thus, the solutions arey = ce−x

2.

How do they look?





dy

y= −2xdx .


ln |y | = −x2 + c1.


2.

How do they look?





dy

y= −2xdx .


ln |y | = −x2 + c1.


2.

How do they look?





dy

y= −2xdx .


ln |y | = −x2 + c1.


2.

How do they look?





dy

y= −2xdx .


ln |y | = −x2 + c1.


2.

How do they look?





dy

y= −2xdx .


ln |y | = −x2 + c1.

Thus, the solutions are

y = ce−x2.

How do they look?





dy

y= −2xdx .


ln |y | = −x2 + c1.


2.

How do they look?





dy

y= −2xdx .


ln |y | = −x2 + c1.


2.

How do they look?


If we are given an initial condition

y(x0) = y0,

then we get:c = y0e

x20

and y = y0ex20−x2 .

(y0 = e−x20 )



y(x0) = y0,

then we get:c = y0e

x20


(y0 = e−x20 )



y(x0) = y0,

then we get:c = y0e

x20


(y0 = e−x20 )



Find the solution to the initial value problem:

dy

dx=

y cos x

1 + 2y2; y(0) = 1.

Assume y 6= 0. Then,

1 + 2y2

ydy = cos x dx .

Integrating,ln |y |+ y2 = sin x + c .

As y(0) = 1, we get c = 1. Hence a particular solution to the IVPis

ln |y |+ y2 = sin x + 1.

Note: y ≡ 0 is a solution to the DE but it is not a solution to thegiven IVP.




dy

dx=

y cos x

1 + 2y2; y(0) = 1.

Assume y 6= 0.

Then,

1 + 2y2

ydy = cos x dx .



ln |y |+ y2 = sin x + 1.





dy

dx=

y cos x

1 + 2y2; y(0) = 1.


1 + 2y2

ydy = cos x dx .



ln |y |+ y2 = sin x + 1.





dy

dx=

y cos x

1 + 2y2; y(0) = 1.


1 + 2y2

ydy = cos x dx .

Integrating,

ln |y |+ y2 = sin x + c .


ln |y |+ y2 = sin x + 1.





dy

dx=

y cos x

1 + 2y2; y(0) = 1.


1 + 2y2

ydy = cos x dx .



ln |y |+ y2 = sin x + 1.





dy

dx=

y cos x

1 + 2y2; y(0) = 1.


1 + 2y2

ydy = cos x dx .


As y(0) = 1, we get c = 1.

Hence a particular solution to the IVPis

ln |y |+ y2 = sin x + 1.





dy

dx=

y cos x

1 + 2y2; y(0) = 1.


1 + 2y2

ydy = cos x dx .



ln |y |+ y2 = sin x + 1.





dy

dx=

y cos x

1 + 2y2; y(0) = 1.


1 + 2y2

ydy = cos x dx .



ln |y |+ y2 = sin x + 1.

Note: y ≡ 0 is a solution to the DE

but it is not a solution to thegiven IVP.




dy

dx=

y cos x

1 + 2y2; y(0) = 1.


1 + 2y2

ydy = cos x dx .



ln |y |+ y2 = sin x + 1.



Method of separation of variables doesn’t yield allsolutions!

Solve y ′ = 3y2/3, y(0) = 0.

y ≡ 0 is a solution.

If y 6= 0,dy

y2/3= 3dx =⇒ 3y1/3 = 3(x + c) =⇒ y = (x + c)3.

Initial condition yields c = 0.

Hence y = x3 and y = 0 are solutions which satisfy the initialconditions.Consider

φk(x) =

{0 −∞ < x ≤ k

(x − k)3 k < x <∞

Are these functions solutions of the DE? YES.There are infinitely many functions which are solutions of the DE.



Solve y ′ = 3y2/3, y(0) = 0.y ≡ 0 is a solution.

If y 6= 0,dy

y2/3= 3dx =⇒ 3y1/3 = 3(x + c) =⇒ y = (x + c)3.



φk(x) =

{0 −∞ < x ≤ k

(x − k)3 k < x <∞





If y 6= 0,dy

y2/3= 3dx

=⇒ 3y1/3 = 3(x + c) =⇒ y = (x + c)3.



φk(x) =

{0 −∞ < x ≤ k

(x − k)3 k < x <∞





If y 6= 0,dy

y2/3= 3dx =⇒ 3y1/3 = 3(x + c)

=⇒ y = (x + c)3.



φk(x) =

{0 −∞ < x ≤ k

(x − k)3 k < x <∞





If y 6= 0,dy

y2/3= 3dx =⇒ 3y1/3 = 3(x + c) =⇒ y = (x + c)3.



φk(x) =

{0 −∞ < x ≤ k

(x − k)3 k < x <∞





If y 6= 0,dy

y2/3= 3dx =⇒ 3y1/3 = 3(x + c) =⇒ y = (x + c)3.



φk(x) =

{0 −∞ < x ≤ k

(x − k)3 k < x <∞





If y 6= 0,dy

y2/3= 3dx =⇒ 3y1/3 = 3(x + c) =⇒ y = (x + c)3.


Hence y = x3 and y = 0 are solutions which satisfy the initialconditions.

Consider

φk(x) =

{0 −∞ < x ≤ k

(x − k)3 k < x <∞





If y 6= 0,dy

y2/3= 3dx =⇒ 3y1/3 = 3(x + c) =⇒ y = (x + c)3.



φk(x) =

{0 −∞ < x ≤ k

(x − k)3 k < x <∞





If y 6= 0,dy

y2/3= 3dx =⇒ 3y1/3 = 3(x + c) =⇒ y = (x + c)3.



φk(x) =

{0 −∞ < x ≤ k

(x − k)3 k < x <∞

Are these functions solutions of the DE?

YES.There are infinitely many functions which are solutions of the DE.




If y 6= 0,dy

y2/3= 3dx =⇒ 3y1/3 = 3(x + c) =⇒ y = (x + c)3.



φk(x) =

{0 −∞ < x ≤ k

(x − k)3 k < x <∞





If y 6= 0,dy

y2/3= 3dx =⇒ 3y1/3 = 3(x + c) =⇒ y = (x + c)3.



φk(x) =

{0 −∞ < x ≤ k

(x − k)3 k < x <∞



Homogeneous functions

Definition

A function f (x1, . . . , xn) is called homogeneous if

f (tx1, . . . , txn) = td f (x1, . . . , xn)

for some d ∈ Z and for all t 6= 0.

The number d is called the degree of f (x1, . . . , xn).Examples :f (x , y) = x2 + xy + y2 is homogeneous of degree 2.

f (x , y) = y + x cos2(yx

)is homogeneous of degree 1.



Definition


f (tx1, . . . , txn) = td f (x1, . . . , xn)


The number d is called the degree of f (x1, . . . , xn).

Examples :f (x , y) = x2 + xy + y2 is homogeneous of degree 2.





Definition


f (tx1, . . . , txn) = td f (x1, . . . , xn)


The number d is called the degree of f (x1, . . . , xn).Examples :f (x , y) = x2 + xy + y2 is homogeneous of degree

2.





Definition


f (tx1, . . . , txn) = td f (x1, . . . , xn)







Definition


f (tx1, . . . , txn) = td f (x1, . . . , xn)




)is homogeneous of degree

1.



Definition


f (tx1, . . . , txn) = td f (x1, . . . , xn)







Definition


f (tx1, . . . , txn) = td f (x1, . . . , xn)






Homogeneous Equations

Definition

The first order ODE

M(x , y) + N(x , y)dy

dx= 0

is called homogeneous if M and N are homogeneous of equaldegree.

Example :

(y2 − x2)dy

dx+ 2xy = 0.


Homogeneous ODE’s - Reduction to variable separableform

Let

M(x , y) + N(x , y)dy

dx= 0

where M and N are homogeneous of degree d .

Put

y

x= v .

Then,dy

dx= x

dv

dx+ v .

Substituting this in the given ODE, we get:

M(x , xv) + N(x , xv)

(xdv

dx+ v

)= 0.

Thus,

xdM(1, v) + xdN(1, v)

(xdv

dx+ v

)= 0.



Let

M(x , y) + N(x , y)dy

dx= 0

where M and N are homogeneous of degree d . Put

y

x= v .

Then,dy

dx= x

dv

dx+ v .


M(x , xv) + N(x , xv)

(xdv

dx+ v

)= 0.

Thus,

xdM(1, v) + xdN(1, v)

(xdv

dx+ v

)= 0.



Let

M(x , y) + N(x , y)dy

dx= 0


y

x= v .

Then,dy

dx= x

dv

dx+ v .


M(x , xv) + N(x , xv)

(xdv

dx+ v

)= 0.

Thus,

xdM(1, v) + xdN(1, v)

(xdv

dx+ v

)= 0.



Let

M(x , y) + N(x , y)dy

dx= 0


y

x= v .

Then,dy

dx= x

dv

dx+ v .


M(x , xv) + N(x , xv)

(xdv

dx+ v

)= 0.

Thus,

xdM(1, v) + xdN(1, v)

(xdv

dx+ v

)= 0.



Let

M(x , y) + N(x , y)dy

dx= 0


y

x= v .

Then,dy

dx= x

dv

dx+ v .


M(x , xv) + N(x , xv)

(xdv

dx+ v

)= 0.

Thus,

xdM(1, v) + xdN(1, v)

(xdv

dx+ v

)= 0.



Let

M(x , y) + N(x , y)dy

dx= 0


y

x= v .

Then,dy

dx= x

dv

dx+ v .


M(x , xv) + N(x , xv)

(xdv

dx+ v

)= 0.

Thus,

xdM(1, v) + xdN(1, v)

(xdv

dx+ v

)= 0.


Homogeneous ODE’s

Let x 6= 0.

Then,

M(1, v) + N(1, v) · v + N(1, v) · x dvdx

= 0.

Thus,dx

x+

N(1, v)

M(1, v) + N(1, v) · vdv = 0.

This is a separable equation.


Homogeneous ODE’s

Let x 6= 0. Then,

M(1, v) + N(1, v) · v + N(1, v) · x dvdx

= 0.

Thus,dx

x+

N(1, v)

M(1, v) + N(1, v) · vdv = 0.



Homogeneous ODE’s

Let x 6= 0. Then,

M(1, v) + N(1, v) · v + N(1, v) · x dvdx

= 0.

Thus,dx

x+

N(1, v)

M(1, v) + N(1, v) · vdv = 0.



Homogeneous ODE’s

Let x 6= 0. Then,

M(1, v) + N(1, v) · v + N(1, v) · x dvdx

= 0.

Thus,dx

x+

N(1, v)

M(1, v) + N(1, v) · vdv = 0.



Documents

MA 108 - Ordinary Differential Equationsdey/diffeqn_spring14/lecture2_D2.pdf · Santanu Dey Lecture 2. Outline of the lecture 1 Initial value problems 2 Separable DE 3 Equations reducible