# 1stOrderLinearEquations(integratingfactormethod)

```Differential Equations Practice: 1st Order: Linear Equations (integrating factor method)
Page 1
Questions
Example (2.1.7) Draw a direction field for the differential equation:
2
dy
+ 2ty = 2te−t .
dt
Based on the direction field, describe how the solutions behave for large values of t. Find the general solution of the given
differential equation, and use it to determine how solutions behave as → ∞.
Example (2.1.18) Find the solution to the initial value problem ty 0 + 2y = sin t, y(π/2) = 1.
Example (2.1.28) Consider the initial value problem y 0 + 2y/3 = 1 − t/2, y(0) = y0 . Find the value of y0 for which the
solution solution touches, but does not cross, the t-axis.
Solutions
Example (2.1.7) Draw a direction field for the differential equation:
2
dy
+ 2ty = 2te−t .
dt
Based on the direction field, describe how the solutions behave for large values of t. Find the general solution of the given
differential equation, and use it to determine how solutions behave as → ∞.
The direction field analysis is contained in the Mathematica file.
To solve
2
dy
+ 2ty = 2te−t
dt
we can use the integrating factor method. Multiply by a function µ = µ(t):
µ
2
dy
+ 2µty = µ2te−t .
dt
Now, we want the following to be true:
d
[µy]
dt
=
µy 0 + µ0 y
=
µy 0 + 2µty
(by the product rule)
(the left hand side of our equation)
Comparing Eqs. (1) and (2), we arrive at the differential equation that the integrating factor must solve:
2µt = µ0 .
This differential equation is separable, so the solution is
2µt =
2t dt
=
2t dt
=
t2
=
dµ
µ
ln |µ| + c1
−c1
=
|µ|
Z
t
2
dµ
dt
dµ
µ
Z
e e
µ =
2
c2 et ,
where c2 = +e−c1
(1)
(2)
Differential Equations Practice: 1st Order: Linear Equations (integrating factor method)
Page 2
Therefore, the original differential equation becomes
µ
c2 et
dy
+ 2µty
dt
2
=
µ2te−t
=
c2 et 2te−t
=
2t
=
2t
=
=
2t dt
Z
2t dt
et y
=
t2 + c3
y(t)
=
t2 e−t + c3 e−t
2
dy
+ 2c2 et ty
dt
2
2 dy
+ 2et ty
et
dt
d h t2 i
e y
dt h
i
2
2
d et y
Z h
i
2
d et y
2
2
2
2
2
Notice how important it is that we insert the constant of integration properly into our solution!
Now, for the large t limit, we have
2
2
lim y(t) = lim t2 e−t + c3 e−t = 0
t→∞
t→∞
since the exponential decay term will dominate the t2 part.
Example (2.1.18) Find the solution to the initial value problem ty 0 + 2y = sin t, y(π/2) = 1.
To solve
ty 0 + 2y = sin t
we can use the integrating factor method. We want the coefficient in front of the y 0 to be 1, so divide through by t before
multiplying by a function µ = µ(t):
µ
2µ
y = sin t.
t
t
Now, we want the following to be true:
µy 0 +
d
[µy]
dt
=
µy 0 + µ0 y
=
µy 0 +
2µ
y
t
(by the product rule)
(the left hand side of our equation)
Comparing Eqs. (3) and (4), we arrive at the differential equation that the integrating factor must solve:
2µ
= µ0 .
t
This differential equation is separable, so the solution is
2µ
t
2
dt
t
Z
2
dt
t
2 ln |t|
ln |t2 |
e
−c1
dµ
dt
dµ
=
µ
Z
dµ
=
µ
= ln |µ| + c1
=
=
ln |µ| + c1
2
|t | = |µ|
µ = c2 t2 ,
where c2 = +e−c1
(3)
(4)
Differential Equations Practice: 1st Order: Linear Equations (integrating factor method)
Therefore, the original differential equation becomes
2µ
y
t
2c2 t2
y
c2 t2 y 0 +
t
t2 y 0 + 2ty
d 2 t y
dt d t2 y
Z
d t2 y
µy 0 +
t2 y
µ
sin t
t
c2 t2
=
sin t
t
= t sin t
=
= t sin t
= t sin t dt
Z
=
t sin t dt
Z
=
t sin t dt
This remaining integral can be done using parts:
Let u = t, dv = sin t dt, so du = dt, v = − cos t.
Z
Z
t sin t dt
=
u dv
Z
= uv − v du
Z
= t(− cos t) − (− cos t) dt
= −t cos t + sin t + c3
Substituting back, we find
Z
t2 y
=
t2 y
=
y
=
t sin t dt
−t cos t + sin t + c3
cos t sin t c3
+ 2 + 2
−
t
t
t
Now we can use the initial condition to determine the constant c3 :
y(π/2) = 1
cos π/2 sin π/2
c3
+
+
2
(π/2)
(π/2)
(π/2)2
1
c3
0+
+
(π/2)2
(π/2)2
= −
1
=
(π/2)2
=
c3
=
1 + c3
π2
−1
4
The solution to the initial value problem is
y(t) = −
cos t sin t
π2
1
+ 2 + 2 − 2.
t
t
4t
t
The solution is valid for t > 0.
Page 3
Differential Equations Practice: 1st Order: Linear Equations (integrating factor method)
Page 4
Example (2.1.28) Consider the initial value problem y 0 + 2y/3 = 1 − t/2, y(0) = y0 . Find the value of y0 for which the
solution solution touches, but does not cross, the t-axis.
To solve
2
t
y0 + y = 1 −
3
2
we can use the integrating factor method. Multiply by a function µ = µ(t):
µy 0 +
2µ
µt
y =µ− .
3
2
Now, we want the following to be true:
d
[µy]
dt
=
µy 0 + µ0 y
=
µy 0 +
(by the product rule)
2µ
y
3
(the left hand side of our equation)
Comparing Eqs. (5) and (6), we arrive at the differential equation that the integrating factor must solve:
2µ
= µ0 .
3
This differential equation is separable, so the solution is
dµ
dt
dµ
=
µ
Z
dµ
=
µ
2µ
3
=
2
dt
3
Z
2
dt
3
2t
3
−c1 2t/3
e e
=
ln |µ| + c1
= |µ|
µ = c2 e2t/3 ,
where c2 = +e−c1
Therefore, the original differential equation becomes
µy 0 +
2µ
y
3
2c2 e2t/3
y
3
2e2t/3
e2t/3 y 0 +
y
3
h
d 2t/3 i
e
y
dt
h
i
d e2t/3 y
Z h
i
d e2t/3 y
c2 e2t/3 y 0 +
e2t/3 y
= µ−
µt
2
c2 e2t/3 t
2
2t/3
e
t
e2t/3 −
2 t
e2t/3 1 −
2
t
2t/3
dt
e
1−
2
Z
t
2t/3
e
1−
dt
2
Z
t
e2t/3 1 −
dt
2
Z
3 2t/3 1
e
−
e2t/3 t dt
2
2
= c2 e2t/3 −
=
=
=
=
=
=
(5)
(6)
Differential Equations Practice: 1st Order: Linear Equations (integrating factor method)
Page 5
This remaining integral can be done using parts:
Let u = t, dv = e2t/3 dt, so du = dt, v = 32 e2t/3 .
Z
e2t/3 t dt
Z
=
=
=
=
u dv
Z
uv − v du
Z
3
3
t( e2t/3 ) − ( e2t/3 ) dt
2
2
3t 2t/3 9 2t/3
e
− e
+ c3
2
4
Substituting back, we find
e
2t/3
y
=
e2t/3 y
=
y
=
y
=
Z
3 2t/3 1
e
−
e2t/3 t dt
2
2
3 2t/3 1 3t 2t/3 9 2t/3
e
−
e
− e
+ c3
2
2 2
4
3 3t 9 c3 −2t/3
−
+ − e
2
4
8
2
21 3t
−2t/3
−
− c4 e
, where c4 = c3 /2
8
4
Now we can use the initial condition to determine the constant c3 :
y(0) = y0
=
c4
=
21
− c4
8
21
− y0
8
The solution to the initial value problem is
21 3t
21
y(t) =
−
−
− y0 e−2t/3 .
8
4
8
If this touches, but does not cross the t-axis, then the tangent line must be horizontal. Assume this happens at a point t̄.
Then we have:
y(t̄) = 0 and y 0 (t̄) = 0.
We can solve these two equations for the two unknowns t̄ and y0 .
21 3t̄
21
−
−
− y0 e−2t̄/3
y(t̄) = 0 =
8
4
8
3 2
21
y 0 (t̄) = 0 = + ·
− y0 e−2t̄/3
4 3
8
These equations must be solved numerically. Using Mathematica, you can solve the second equation for t̄ in terms of y0 ,
then substitute into the first equation and solve for y0 . You should find:
y0 = −
3
−7 + 3e4/3 ∼ −1.64288.
8
See the Mathematica file for the computational details.
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