Upload
philippa-barrett
View
261
Download
1
Embed Size (px)
Citation preview
First-Order Differential Equations
Part 1
First-Order Differential Equations
Types:
• Variable Separable
• Linear Equations
• Exact Equations
• Solvable by Substitutions
Variable Separable
• The simplest of all differential equations are those of the first order with separable variables.
• A first-order differential equation of the form
is said to be separable or to have separable variables.
)y(h)x(gdx
dy
Variable Separable
• To solve variable separable first-order differential equations, proceed as follows:
c)x(G)y(H
dx)x(gdy)y(p
dx)x(gdy)y(p
dx)x(g)y(h
dy
)y(h)x(gdx
dy
Let 1/h(y) = p(y)
H(y) and G(x) are antiderivatives of
p(y) and g(x), respectively.
ExampleSolve:
0dy)x1(dx)y1( 22 Solution:
c)xarctan()yarctan(
x1
dx
y1
dy
x1
dx
y1
dy
)x1)(y1(
10dy)x1(dx)y1(
0dy)x1(dx)y1(
22
22
2222
22
Alternate Solution:
xy1
yxc
)xy1(cxy
xccxyycx1
xcy
)xarctan(tancarctantan1
)xarctan(tancarctantany
carctan)xarctan(tan)yarctan(tan
carctan)xarctan()yarctan(
x1
dx
y1
dy
x1
dx
y1
dy
)x1)(y1(
10dy)x1(dx)y1(
0dy)x1(dx)y1(
22
22
2222
22
ExampleSolve:
2
3)0(y,0dyx1dxy1 22
Solution:
Cxarcsinyarcsin
dxx1
1dy
y1
1
0dxx1
1dy
y1
1
x1y1
10dyx1dxy1
22
22
22
22
Example
Initial Boundary Condition:2
3)0(y
Solving for C:
3C
C03
C)0arcsin(2
3arcsin
Cxarcsinyarcsin
Example
Initial Boundary Condition:2
3)0(y
Solving for C:
3xarcsinyarcsin
Cxarcsinyarcsin
Example
0dyx1dxy1 22
22
x12
3
2
xy)
2
3)(
1
x1()
2
1(xy
)3
sin()xcos(arcsin)3
cos()xsin(arcsiny
)3
xsin(arcsin)ysin(arcsin
3xarcsinyarcsin
Alternate Form of Final Answer:
Linear Equations
A first-order differential equation of the form
is said to be a linear equation in the dependent variable y.
When g(x) = 0, the linear equation is said to be homogeneous; otherwise, it is nonhomogeneous.
)x(gy)x(adx
dy)x(a 01
Linear Equations
We can divide both sides of the equation by the lead coefficient a1(x):
)x(fy)x(Pdx
dy
)x(a
)x(gy)x(a
)x(a
dx
dy
)x(gy)x(adx
dy)x(a
11
0
01
Linear Equations
Standard form of a linear 1st-order DE:
This differential equation has the property that its solution is the sum of two solutions:
y = yc + yp
)x(fy)x(Pdx
dy
Linear Equations
Now, yc is a solution of the associated homogeneous equation
and yp is a particular solution of the nonhomogeneous equation:
0y)x(Pdx
dy
)x(fy)x(Pdx
dy
Linear EquationsProof:
)x(f)x(f0
)x(fy)x(Pdx
dyy)x(P
dx
dy
)x(fy)x(Pdx
dyy)x(P
dx
dy
)x(fyy)x(Pyydx
d
)x(fy)x(Pydx
d
?
pp
cc
?
pp
cc
?
pcpc
Now, the previous homogeneous equation is also separable:
dx)x(P)ykln(
dx)x(Pklnyln
dx)x(Py
dy
0dx)x(Py
dy
0y)x(Pdx
dy
dx)x(P1
1c
dx)x(Pc
dx)x(P
dx)x(P
dx)x(P
eywhere
cyy
ceyyLet
cey
ek
1y
eyk
Now, let yp = u(x)y1:
)x(f)x(udx
dy
)x(f)x(udx
dy0)x(u
)x(f)x(udx
dyy)x(P
dx
dy)x(u
)x(fy)x(u)x(P)x(udx
dy
dx
dy)x(u
)x(fy)x(u)x(Py)x(udx
d
)x(fy)x(Pydx
d
1
1
111
111
11
pp
Separating variables and integrating gives
dx)x(feu
dxe
)x(fu
dx)x(y
)x(fu
dx)x(y
)x(fdu
)x(f)x(udx
dy
dx)x(P
dx)x(P
1
1
1
Now, going back to yp = uy1:
Now, going back to y = yc + yp:
dx)x(Pdx)x(P
p
1p
edx)x(fey
uyy
dx)x(feecey
yyy
dx)x(Pdx)x(Pdx)x(P
pc
Remember this special term called the “integrating factor”:
We can use the integrating factor as follows:
dx)x(Pe
dx)x(fecye
edx)x(feecey
dx)x(Pdx)x(P
dx)x(Pdx)x(Pdx)x(Pdx)x(P
General Solution
RECALL AGAIN
Standard form of a linear 1st-order DE:
)x(fy)x(Pdx
dy
Left-hand side of the standard form, to be used for deriving the solution
(see next slide)
A Simpler Derivation
This derivation hinges on the fact that the left hand side of the 1st-order differential equation (in standard form) can be recast into the form of the exact derivative of a product by multiplying both sides of the equation by a special function (x).
Left side of standard form of 1st order, linear DE
multiplied by .
dx
dyy
dx
dy)x(P
dx
dy
Left-hand side of the standard form of a 1st
order linear D.E.
Derivative of a product of two variables
ydx
dy)x(P
dx
dy
?
A Simpler Derivation
The derivation then involves solving the encircled elements as follows:
1cdx)x(P
1
e)x(
cdx)x(P||ln
Pdxd
Pdx
d
A Simpler Derivation
Continuing, we have:
dx)x(P2
dx)x(Pc
cdx)x(P
ec)x(
ee)x(
e)x(
1
1
A Simpler Derivation
Even though there are infinite choices of (x), all produce the same result. Hence, to simply, we let c2 = 1 and obtain the integrating factor.
dx)x(P
dx)x(P
dx)x(P2
e)x(
e)1()x(
ec)x(
A Simpler Derivation
• This is what we have derived so far. We multiply both sides of the standard form of the 1st-order equation by the integrating factor (x).
• We can then integrate both sides of the resulting equation and solve for y, resulting in a one-parameter family of solutions.
A Simpler Derivation
dx)x(Pdx)x(Pdx)x(P
dx)x(Pdx)x(P
dx)x(Pdx)x(P
dx)x(Pdx)x(P
dx)x(Pdx)x(Pdx)x(P
ecdx)x(feey
cdx)x(feye
dx)x(feyed
)x(feyedx
d
)x(fey)x(Pedx
dye
)x(fy)x(Pdx
dy
Solving a Linear, 1st-Order DE
1. Put the differential equation in standard form.
2. From the standard from, identify P(x) and then find the integrating factor
3. Multiply the standard form equation by the integrating factor.
dx)x(Pe
Solving a Linear, 1st-Order DE
4. The left hand side of the resulting equation is automatically the derivative of the integrating factor and y:
5. Integrate both sides of this last equation.
)x(feyedx
d dx)x(Pdx)x(P
ExampleFind the general solution of:
Solution:
Step 1:
0xdydx)y3x( 5
4
5
5
xyx
3
dx
dy
0x
y3x
dx
dy
xdx
10xdydx)y3x(
ExampleSolution:
Step 2:
Integrating Factor:
4xyx
3
dx
dy
P(x): include the negative
sign if present
3xln3dx)x
3(dx)x(P
xeee
ExampleSolution:
Step 3:
Step 4:
Recall:
4333 xxyx
3x
dx
dyx
dx
duv
dx
dvuuv
dx
d
xyx3dx
dyx 43
Example
Solution:
Step 3:
Step 4:
Thus:
xyx3dx
dyx 43
xyxdx
d 3
Derivative of y
Derivative of x –3
ExampleSolution:
Step 5:
35
35
32
3
23
3
3
3
Cxxy2
x'c2xy2
x2'c2
xyx
'c2
xyx
xdxyxd
xdxyxd
xyxdx
d
Solve the following variable separable differential equations.
1.
2.
3.
4.
5.
Exercises
ar,0when;dsinrdr)ra2( 322
0y,0xwhen);xyexp(x'y 2
0dye)1y(dxxy x3
ycosxcos'y 2
xdtsec)t1(tdx 22
Answers:
c)1t(x2sinx2)5
cx2sinx2|ytanysec|ln4)4
cy2
1y2)1x(e)3
a
rlnracosr)2
)e1ln(2lny)1
22
2x
222
x2
Solve the following 1st-order, linear differential equations.
1.
2.
3.
4.
5.
0xdydx)y3x( 5
xcotyxcsc'y 0xdydx)xcotxyxy(
)dydx)(1x(ydx2 2 3x2y'y)3x2(
Exercises
|3x2|ln3x2y2)5
)c|1x|ln2x)(1x(y)1x()4
cxsinxcosxxsinxy)3
xcosxsincy)2
cxxy2)1 35
Answers: