[note: Ux=∂U/∂x, Uy=∂U/∂y](adsbygoogle = window.adsbygoogle || []).push({});

Example: Solve the partial differential equation 2Ux + 3Uy + U = 0 by using the change of variables V(x,y)=ln[U(x,y)]

Solution:

Vx = Ux/U

Vy = Uy/U

2Ux + 3Uy + U = 0

Dividing both sides by U, we have

2Ux/U + 3Uy/U + 1 = 0

=> 2Vx + 3Vy +1 = 0

=> 2Vx + 3Vy = -1

The corresponding homogeneous equation has the general solution V = f(3x-2y) where f is arbitrary function.

2Vx + 3Vy = -1

SetV(x,y)=f(x)

=> 2f ' + 0 = -1

=> f ' = -1/2

=> f= -x/2 + C

=> f= -x/2 (take C=0)

Therefore, a particular solution to 2Vx + 3Vy = -1 isV = -x/2

So the general solution to 2Vx + 3Vy = -1 is V = -x/2 + f(3x-2y)

=> the general solution to the original PDE is U = exp(-x/2) exp[f(3x-2y)] =exp(-x/2) g(3x-2y)

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Now, I don't understand the parts in red:

1) In the solution, they divided both sides by U. Why is this always allowed? How do we know that U is not 0?

2) For the part of finding a particular solution to 2Vx + 3Vy = -1, they first setV(x,y)=f(x). What is the logic behind this step? Why would this lead us to a particular solution? I just don't get the idea.

3) At the end, they claimed that the general solution to the original PDE is

U = exp(-x/2) exp[f(3x-2y)]

i.e. U= exp(-x/2) g(3x-2y) (final answer)

I don't understand why exp[f(3x-2y)] can be replaced by g(3x-2y). Why do we have to do that? Is g here anarbitraryfunction?

Could someone please kindly explain? Any help is greatly appreciated!

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# PDE 2Ux + 3Uy + U = 0 with change of variables V(x,y)=ln[U(x,y)]

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