EXAMPLE: LINEAR DIFFERENTIAL EQUATIONS
110.302 DIFFERENTIAL EQUATIONS
PROFESSOR RICHARD BROWN
Problem. Solve the Ordinary Differential Equation t(y 0 − y) = (1 + t2 )et .
Strategy. Solving the ODE means finding the general solution (the 1-parameter
family of solutions). We place this linear, first order ODE into its standard
form and compute the integrating factor. Then we multiply the ODE by its
integrating factor and then integrate to find the general solution.
Solution. This ODE is almost in the standard form y 0 + p(t)y = q(t). To
get there, simply divide the entire equation by the coefficient of y 0 which in
this case is t. We get
1
[t(y 0 − y) = (1 + t2 )et
t
1 + t2 t
0
y −y =
e.
t
Here then, p(t) = −1, and the integrating factor is
R
e
p(t) dt
R
=e
− dt
= e−t .
Multiply this through the ODE to get
1 + t2 t
e
e−t y 0 − y =
t
1 + t2
e−t y 0 − e−t y =
.
t
Note. While this new expression certainly looks harder to play with, it winds
up making the problem easier to solve.
1
2
110.302 DIFFERENTIAL EQUATIONS PROFESSOR RICHARD BROWN
Continuing on, we get
1 + t2
e y −e y =
t
d −t 1 + t2
e y =
dx
Z
Z t
Z d −t 1 + t2
1
e y dt =
dt =
+ t dt
dt
t
t
t2
−t
e y = ln |t| + + C
2
t2
t
y(t) = e ln |t| + + C .
2
This is the general solution.
−t 0
−t
Is it correct? We check: Given y(t) as above, we calculate, using the Product Rule,
t2
0
t
t 1
y (t) = e ln |t| + + C + e
+t .
2
t
Substitute both y(t) and y 0 (t) into the original linear ODE, and we get
t(y 0 − y) = (1 + t2 )et
t2
t2
t
t 1
t
t
e ln |t| + + C + e
+t
− e ln |t| + + C
= (1 + t2 )et
2
t
2
t3
t3
tet ln |t| + et + Ctet + et + t2 et − tet ln |t| − et − Ctet = et + t2 et = (1 + t2 )et .
2
2
It all works.