# Obtaining the general solution

I was doing a problem in the book A Collection of Problems on MATHEMATICAL PHYSICS by B. M. BUDAK, A. A. SAMARSKII and A. N. TIKHONO of the form $$x^2 u_{xx} - y^2 u_{yy} = 0$$

In the answer section it only said $ε = \frac{y}{x} , η = xy$ and the general solution $u(x,y)=F(xy)+{x}~G\left(\dfrac{x}{y}\right)$

On my attempts i got the conical form to be $$u_{\eta \epsilon} - \frac{u_{\epsilon}}{2\eta}=0$$

and the general solution to be
$$u(x,y)=F(xy)+\sqrt{xy}~G\left(\dfrac{x}{y}\right)$$

Looking at posts on this question i have seen people also get $u(x,y)=F(xy)+\sqrt{xy}~G\left(\dfrac{x}{y}\right)$ as well as $u(x,y)=F(xy)+{xy}~G\left(\dfrac{x}{y}\right)$ I feel confused about this question as i have seen 3 different answer and no method. So i wish to ask if my answer is correct and if not how do i arrive at the correct answer?

Remember that $F$ and $G$ will be arbitrary functions of a single variable (well, almost arbitrary - there will be certain differentiability conditions). Your answer can be written as \eqalign{u(x,y) &=F(xy)+\sqrt{xy}G\Bigl(\frac xy\Bigr)\cr &=F(xy)+x\sqrt{\frac yx}G\Bigl(\frac xy\Bigr)\cr &=F(xy)+xH\Bigl(\frac xy\Bigr)\cr} where $$H(t)=\frac1{\sqrt t}G(t)\ ,$$ so it is really the same as the book answer. For the final answer $$u(x,y)=F(xy)+xyG\Bigl(\frac xy\Bigr)$$ you can carefully differentiate using the chain rule, product rule and quotient rule: if I have done this right you end up with $$x^2u_{xx}-y^2u_{yy}=2x^2G'\Bigl(\frac xy\Bigr)\ ;$$ this is not (usually) $0$, so the answer is wrong. You can use a similar method to check that your answer and the book answer are correct.
• Your answer is the same as the book answer for $x,y > 0$. You'll run into some trouble outside that quadrant. Note that if you change the signs of both $x$ and $y$ you change the sign of $x H(x/y)$, but not of $\sqrt{xy} G(x/y)$. – Robert Israel Nov 6 '17 at 6:07