How to Solve the PDE 2Uxxy + 3Uxyy - Uxy = 0?

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The discussion focuses on solving the partial differential equation 2Uxxy + 3Uxyy - Uxy = 0, with the substitution W = Uxy leading to a change of coordinates. This transformation simplifies the equation to Uxy = f(3x-2y)exp((2x+3y)/3). A general solution is proposed as U(x,y) = F1(x) + F2(y) + exp(x/2)F3(3x-2y), where F1, F2, and F3 are arbitrary functions. Another participant explains that the solution can also be derived through double integration, emphasizing the flexibility in methods for solving PDEs. The conversation highlights the complexity and multiple approaches available for tackling this type of equation.
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2Uxxy+3Uxyy-Uxy=0 where U=U(x,y)

I made the substitution W=Uxy and then used a change of coordinates (n= 2x+3y, and r=3x-2y) which reduced the problem to solving Uxy=f(3x-2y)exp((2x+3y)/3) because W=f(r)exp(n/3). Now I have no idea where to go from there. Any help would be much appreciated.

Thanks
 
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The general solution to your PDE is as follows

U(x,y) = F1(x)+F2(y)+exp(x/2)F3(3x-2y),

where F1, F2, F3 are arbitrary functions.
 
kosovtsov,

Would you mind explaining to me how you arrived at that solution? It would be greatly appreciated.

Thanks
 
There is a general principle. If a problem can be solved, it as a rule, can be solved by countless number of methods.

For example, from your

Uxy=f(3x-2y)exp((2x+3y)/3)

it is follows immediately by double integration that the general solution can be in form

U(x,y)=F1(x)+F2(y)+\int \int (f(3x-2y)exp((2x+3y)/3)) dx dy .

My solution only looks simpler, but is equivalent the one above.
 

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