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which can be treated as an ODE in ξ. The solution to this ODE has the form
− c ξ
u h (ξ, η) = e a 2 +b 2 f(η)

with f again being an arbitrary single-variable function. Changing the coordinates back
to the original (x; y), we will obtain the general solution to the homogeneous equation

c(ax+by)
−
u h (x, y) = e a 2 +b 2 f(bx − ay).
You should verify that this indeed solves equation (2.11).
To find a particular solution of (2.10), we can use the characteristic coordinates to
reduce it to the nonhomogeneous ODE
c g(ξ, η)
2
2
(a + b )u ξ + cu = g(ξ, η) or u ξ + u = .
2
2
a + b 2 a + b 2
Having found the solution to the homogeneous ODE, we can find the solution to this
nonhomogeneous equation by e.g. variation of parameters. So the particular solution will
be
Z
− c ξ g(ξ, η) − c ξ
u p = e a 2 +b 2 e a 2 +b 2 dξ
a + b 2
2
The general solution of (2.10) is then
Z
− c ξ g(ξ, η) − c ξ
u(ξ, η) = u h + u p = e a 2 +b 2 f(η) + e a 2 +b 2 dξ
2
a + b 2
To find the solution in terms of (x, y), one needs to first carry out the integration in ξ in
the above formula, then replace ξ and η by their expressions in terms of x and y.
Example 2.2 Find the general solution of −2u x + 4u y + 5u = e x+3y .
The characteristic change of coordinates for this equation is given by

ξ = −2x + 4y
(2.12)
η = 4x + 2y
From these we can also find the expressions of x and y in terms of (ξ, η). In particular
notice that
ξ + η
x + 3y = .
2
In the characteristic coordinates the equation reduces to
ξ+η
20u ξ + 5u = e 2 .

The general solution of the homogeneous equation associated with the above equation is

1
4 f(η),
u h = e − ξ
and the particular solution will be

Z e ξ+η 1 1
2
3
1
1
1
1
η
ξ
ξ
u p = e − ξ e 4 dξ = e − ξ e 2 e 4 = e − ξ e 4 1 (3ξ+2η)
4 ·
4
4 ·
20 15 15
11