Question

Solve (y ^ 2 * z)/x * p + xzq = y ^ 2 .

Answer 1

Given $(y^2+z^2)p - xyq + xz = 0$

$\implies (y^2+z^2)p - xyq =- xz$

So, by Lagrange's method, we have

$\frac{dx}{y^2+z^2} = \frac{dy}{-xy} = \frac{dz}{-xz}$ ______________(1)

$\implies \frac{x dx + ydy + zdz}{xy^2 + xz^2 - xy^2 - xz^2} = \frac{dz}{-xz}$

$\implies x dx + ydy + zdz=0$

So by integration, we get $x^2 + y^2 + z^2 = c_1$ _____________(2)

Also from last two parts of equation (1), we have

$\frac{dy}{-xy} = \frac{dz}{-xz}$

$\implies \frac{dy}{y} = \frac{dz}{z}$

So by integration, we get

$\ln(y) = \ln(z) + \ln(c_2)$

$\implies \frac{y}{z} = c_2$ __________(3)

Now, from (2) and (3), solution of given differential equation is

$c_2 = f(c_1) \\ \implies \frac{y}{z} = f(x^2 + y^2 + z^2)$ is the solution.