PDES: GENERAL AND PARTICULAR SOLUTIONS
is of the form
A(x, y)
∂u
∂x
+B(x, y)
∂u
∂y
+C(x, y)u=R(x, y), (20.9)
whereA(x, y),B(x, y),C(x, y)andR(x, y) are given functions. Clearly, if either
A(x, y)orB(x, y) is zero then the PDE may be solved straightforwardly as a
first-order linear ODE (as discussed in chapter 14), the only modification being
that the arbitrary constant of integration becomes anarbitrary functionofxory
respectively.
Find the general solutionu(x, y)of
x
∂u
∂x
+3u=x^2.
Dividing through byxwe obtain
∂u
∂x
+
3 u
x
=x,
which is a linear equation with integrating factor (see subsection 14.2.4)
exp
(∫
3
x
dx
)
=exp(3lnx)=x^3.
Multiplying through by this factor we find
∂
∂x
(x^3 u)=x^4 ,
which, on integrating with respect tox,gives
x^3 u=
x^5
5
+f(y),
wheref(y)isanarbitrary functionofy. Finally, dividing through byx^3 , we obtain the
solution
u(x, y)=
x^2
5
+
f(y)
x^3
.
When the PDE contains partial derivatives with respect to both independent
variables then, of course, we cannot employ the above procedure but must seek
an alternative method. Let us for the moment restrict our attention to the special
case in whichC(x, y)=R(x, y) = 0 and, following the discussion of the previous
section, look for solutions of the formu(x, y)=f(p)wherepis some, at present
unknown, combination ofxandy. We then have
∂u
∂x
=
df(p)
dp
∂p
∂x
,
∂u
∂y
=
df(p)
dp
∂p
∂y
,