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Chapter 2

First-order linear equations

2.1 Solved Problems

Exercise 25. Find the general solution of 3u − 2u + 4u = x − 5y.
y
x
The characteristic change of coordinates for this equation is given by

ξ = 3x − 2y
η = −2x − 3y

From these we can also find the expressions of x and y in terms of (ξ, η). In

particular notice that
x − 5y = ξ + η.

In the characteristic coordinates the equation reduces to

13u + 4u = ξ + η.
ξ
The general solution of the homogeneous equation associated with the above equa-
tion is
4
u = e − 13 ξ f(η),
h
and the particular solution will be

Z ξ + η 13 169
4
4
4
4
4
u = e − 13 ξ e 13 ξ dξ = e − 13 ξ · (ξ + η)e 13 ξ − e 13 ξ =
p
13 4 16
13 169
= (ξ + η) −
4 16
Adding these two will give the general solution to the nonhomogeneous equation
u(ξ, η) = e − 13 ξ f(η) + 13 (ξ + η) − 169 .
4
4 16
Finally, substituting the expressions for ξ and η in terms of (x, y), we will obtain
the solution

u(x, y) = e − 13 (3x−2y) f(−2x − 3y) + 13 (x − 5y) − 169 .
4
4 16

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