Answer to Question #152776 in Differential Equations for Abdulrahman

Question

2yzdx-2xzdy-(x^2-y^2)(z-1)dz=0

Answer 1

$2yzdx-2xzdy-(x^2-y^2)(z-1)dz=0........................(1)$

First we take $z$ to be constant. Then the equation reduces to

$2yzdx-2xzdy=0$

Divide through by $2xyz$

$\frac{dy}{y}-\frac{dx}{x}=0$

Integrating, we obtain;

$\ln(\frac{y}{x})=c\\ y=xc_1$

We assume that the solution of Eqn(1) is

$y=x\Psi(z)$

Now, let $x=\alpha$ and, therefore,

$y=\alpha\Psi(z) \text{ is a solution of eqn(1).}$

Since $x=\alpha,dx=0$ . Put this into eqn(1)

$2\alpha zdy+(\alpha^2-y^2)(z-1)dz=0$

Divide through by $z(\alpha^2-y^2)$

$\frac{2\alpha}{\alpha^2-y^2}dy+\left(1-\frac{1}{z}\right)dz=0$

Integrate through

$\ln|\frac{\alpha+y}{\alpha-y}|-\ln|z|+z=\ln|c| \text{ where } c \text{ is a constant.}$

$\implies \frac{\alpha+y}{\alpha+y}=cze^{-z}$

But $y=\alpha \Psi,$ therefore,

$\frac{\alpha(1+\Psi)}{\alpha(1-\Psi)}=\frac{1+\frac{y}{x}}{1-\frac{y}{x}}=\frac{x+y}{x-y}=cze^{-z}$

Hence the solution is

$(x+y)e^z=c(x-y)z$