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:
- $\dfrac{\partial f(x^2-y^2)}{\partial x}=\dfrac{\partial f(x^2-y^2)}{\partial y}$
$f'$ is understood as the differental with respect to its argument, rather than the differential with respect to $x$.
$$\begin{align}\dfrac{\partial u}{\partial x} &= \dfrac{\partial f(x^2-y^2)}{\partial x} \\[1ex]&= \left.\dfrac{\partial (x^2-y^2)}{\partial x\hspace{8ex}}\dfrac{\mathsf d f(z)}{\mathsf d z~~~~~}\right\vert_{z=x^2-y^2}\\[1ex]&= 2x\;f'(x^2-y^2) \end{align}$$