Is f(xy)=e^(2y)sin(pix) a Solution to the Differential Equation Given?

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let f(xy)=e^(2y)sin(pix), is f a solution to (fxy)^2 -fxx(fyy)=4pi^2 e^(4x)


So I found fxx and fyy and fxy, which are -pi(e^(2y))sin(pix), 4sin(pix)e^(2y), 2pi(e^(2y))cos(pix) respectively,

When i reduced everything i got e^4y(sin^2(pix))+pie^4ycos^2(pix)=pie^(4x). I am assuming it is not a solution to 4pi^2 e^(4x) because first, it doesn't add up, and second, e^(4y) is never e^(4x). Am I right, I am not sure?
 
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You are right that it is not a solution, but I wonder whether it's because there is a typo in the question. Maybe it should say fxy2 -fxxfyy=4pi2 e4y
 
Yeah, me too. But the problem on the paper clearly states that. I'll bring it up to my professor. Thanks!
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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