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I have been asked to find the "Most general solution" for $u(x,y)$ of the PDE

$$\frac{\partial u}{\partial x} = \frac{x}{\sqrt{x^2+y^2}} + 3y\cos(3xy) + 3x^2y^2$$

I know you must take the integral of both sides to "undo" the partial derivative. Is it reasonable to then split the right hand side into three separate integrals as below?

$$u(x, y) = \int\frac{x}{\sqrt{x^2+y^2}}dx + \int3y\cos(3xy)dx + \int3x^2y^2dx.$$

From here I got: $$\int\frac{x}{\sqrt{x^2+y^2}}dx = \frac{1}{2}\int\frac{2x}{\sqrt{x^2+y^2}}dx = \sqrt{x^2+y^2} + C(y)$$ $$\int3y\cos(3xy)dx = \sin(3xy) + C(y)$$ $$\int3x^2y^2dx = y^2x^3+C(y)$$

The final answer is:

$$u(x, y) = \sqrt{x^2+y^2} + \sin(3xy) + y^2x^3+C(y)$$

Is this correct?

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Integration is a linear operation, so you can distribute it among addition/subtraction in general. That's perfectly legitimate.

This looks correct to me. Be sure to note that $C(y)$ may have a constant term, but I would even argue that the notation is sufficiently general as to make that clear. Nice work. If I were to nitpick, the symbol $C$ is so canonically associated with constants of integration that you might want to use a different symbol, but it is fine the way it is. And from the point view of the integrals, it is constant in $x$ anyway.

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    $\begingroup$ Thank you for the help The Count, one more question if I may. If I were given the initial condition $u(0, 0) = 3$ would $C(y) = 3$? $\endgroup$
    – Asyu7
    May 17, 2019 at 4:02
  • $\begingroup$ You're very welcome. $C(y)$ is just some function of $y$. $C(0)=3$, yes. But only because for $(x,y)=(0,0)$, the other terms are all zero. $\endgroup$
    – The Count
    May 17, 2019 at 15:40

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