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Consider following differential equation in $$\frac{dx}{(3x^2+y^2)/x}=\frac{dy}{(3y^2+x^2)/y}=\frac{dz}{-(x^2+y^2)/z}$$ $\require{enclose}$ Taking multipliers $x, y, 4z$ we get $$xdx+ydy+4zdz=0 \\ \enclose{box}{x^2+y^2+4z^2=C_1}$$

I have this function correct.

Book writes: for taking first two fraction $xdx-ydy=0$ and hence $$\bbox[5px,border:2px solid red]{x^2-y^2=C_2}$$

I am sure this is the correct solution. We would have on subtracting numerator and denominator $$\frac{xdx-ydy}{2x^2-2y^2}$$ for which denominator isn't 0 so how can we assert $xdy-ydx=0$

I am not missing anything am I?

Possible Solution

$$\frac{xdx-ydy}{2x^2-2y^2}=\frac{xdx+ydy}{4x^2+4y^2} \\ \frac{d(x^2-y^2)}{4(x^2-y^2)}=\frac{d(x^2+y^2)}{8(x^2+y^2)} \\ \enclose{box}{\frac{(x^2-y^2)^2}{x^2+y^2}=C_2}$$

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    $\begingroup$ You can check via differentiation that the solution $$x^{2}+y^{2}+4z^{2} = f(x^{2}-y^{2})$$ doesn't satisfy the PDE. So the book is wrong. $\endgroup$ Sep 27, 2019 at 7:38
  • $\begingroup$ Thanks +1.............. $\endgroup$
    – Sonal_sqrt
    Sep 27, 2019 at 7:40

1 Answer 1

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You are right, two charecteristic equations are : $$x^2+y^2+4z^2=C_1$$ $$\frac{(x^2-y^2)^2}{x^2+y^2}=C_2$$ From that, the general solution of the PDE expressed on the form of implicit equation $C_1=F(C_2)$ with arbitrary function $F$ : $$x^2+y^2+4z^2=F\left(\frac{(x^2-y^2)^2}{x^2+y^2}\right)$$ On explicit form : $$z(x,y)=\pm\frac12\sqrt{-x^2-y^2+F\left(\frac{(x^2-y^2)^2}{x^2+y^2}\right)}$$

NOTE :

$xdx-ydy=0$ is not correct in the general case thus $x^2-y^2=C_3$ is not a correct characteristic equation.

NOTE :

The problem is simplified with the change of variables $\quad\begin{cases} X=x^2\\ Y=y^2\\ Z=z^2 \end{cases}$ : $$\frac{dX}{3X+Y}=\frac{dY}{3Y+X}=-\frac{dZ}{X+Y}$$

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