# Solve and look for restrictions on x and y can those be returned as solutions. y^4 (x-4) -...

## Question:

Solve and look for restrictions on x and y can those be returned as solutions.

{eq}y^4 (x-4) - x^3(y^2-3) \ dy/dx = 0 {/eq}

## Differential Equation:

A differential equation is an equation that relates various functions to their derivatives.

A differential equation is said to be separable if it can be written separately as functions of different variables.

{eq}y^4 (x-4) - x^3(y^2-3)\frac{\mathrm{d} y}{\mathrm{d} x} = 0 {/eq}

This equation can be written as {eq}\frac{\mathrm{d} y}{\mathrm{d} x}=\frac{y^4 (x-4) }{x^3(y^2-3)} {/eq}

Integrate both sides.

{eq}\int \frac{y^2-3}{y^4}\,dy=\int \frac{x-4}{x^3}\,dx\\ \int \frac{y^2}{y^4}-\frac{3}{y^4}=\int \frac{x}{x^3}-\frac{4}{x^3}\,dx\\ \int \frac{1}{y^2}-\frac{3}{y^4}=\int \frac{1}{x^2}-\frac{4}{x^3}\,dx {/eq}

As {eq}\int t^n\,dt=\frac{t^{n+1}}{n+1} {/eq},

{eq}\int \frac{y^2-3}{y^4}\,dy=\int \frac{x-4}{x^3}\,dx\\ \int \frac{y^2}{y^4}-\frac{3}{y^4}=\int \frac{x}{x^3}-\frac{4}{x^3}\,dx\\ \int \frac{1}{y^2}-\frac{3}{y^4}=\int \frac{1}{x^2}-\frac{4}{x^3}\,dx\\ \frac{y^{-2+1}}{-2+1}-3\left ( \frac{y^{-4+1}}{-4+1} \right )=\frac{x^{-2+1}}{-2+1}-4\left ( \frac{x^{-3+1}}{-3+1} \right )+c\\ \frac{-1}{y}+\frac{1}{y^3}=\frac{-1}{x}+\frac{2}{x^2}+c {/eq}

Here, {eq}c {/eq} is constant. 