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Suppose we are given with the equation $u(x,y)=f(x^2-y^2)$ (here $f$ is any arbitrary function of $x$ and $y$) and we need to form a partial differential whose solution is $u(x,y)$. Proceeding to do that using elimination method. This method was used by in books. Partially differentiate $u(x,y)$ with respect to $x$ and $y$ respectively. Hence we get:

  • $\dfrac{\partial u}{\partial x} = 2x\ f^{'}(x^2-y^2)$
  • $\dfrac{\partial u}{\partial y} = -2y\ f^{'}(x^2-y^2)$

Dividing these two equations we get:

  • $\dfrac{\dfrac{\partial u}{\partial x}}{\dfrac{\partial u}{\partial y}}=\dfrac{2x\ f^{'}(x^2-y^2)}{-2y\ f^{'}(x^2-y^2)}$

Which can be simplified to:

  • $x\dfrac{\partial u}{\partial y}+y\dfrac{\partial u}{\partial x} = 0$

This is what the differential equation is what $u(x,y)$ satisfies as it is said in the book and it surely is the correct answer but the doubt that arises is when I differentiated $u(x,y)$ w.r.t $x$ and $y$ both the time we represented both $\dfrac{\partial u}{\partial x}$ & $\dfrac{\partial u}{\partial y}$ as $f^{'}(x^2-y^2)$ and not like:

  • $\dfrac{\partial u}{\partial y} = -2y\ \dfrac{\partial f(x^2-y^2)}{\partial y}$
  • $\dfrac{\partial u}{\partial x} = 2x\ \dfrac{\partial f(x^2-y^2)}{\partial x}$.

Which clearly will not lead to the solution we found above. The super-scripted dash notation implies that:

  • $\dfrac{\partial f(x^2-y^2)}{\partial x}=\dfrac{\partial f(x^2-y^2)}{\partial y}$

Am I missing something? Why does this method work?

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  • $\begingroup$ Shouldn't it be $x\dfrac{\partial u}{\partial y} + y\dfrac{\partial u}{\partial x} = 0$? $\endgroup$
    – Chee Han
    Mar 4, 2018 at 20:45
  • $\begingroup$ Correct. I'll edit it. $\endgroup$ Mar 4, 2018 at 20:45
  • $\begingroup$ That fixes everything then. $\endgroup$
    – Chee Han
    Mar 4, 2018 at 20:47

1 Answer 1

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$f'$ is understood as the differental with respect to its argument, rather than the differential with respect to $x$.

$$\begin{align}\dfrac{\partial u}{\partial x} &= \dfrac{\partial f(x^2-y^2)}{\partial x} \\[1ex]&= \left.\dfrac{\partial (x^2-y^2)}{\partial x\hspace{8ex}}\dfrac{\mathsf d f(z)}{\mathsf d z~~~~~}\right\vert_{z=x^2-y^2}\\[1ex]&= 2x\;f'(x^2-y^2) \end{align}$$

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