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Unformatted text preview: 4.
uxy + uy = 0
Let v = uy then the equation becomes
vx + v = 0
For ﬁxed y , this is a separable ODE
ln v = − x + C (y )
v = K (y ) e− x
In terms of the original variable u we have
uy = K (y ) e− x
u = e− x q (y ) + p(x)
You can check your answer by substituting this solution back in the PDE. 3 ...
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- Fall '08