Question

Put the homogeneous differential equation $\left(5x^2-2y^2\right)\mathit{d}\mathit{x}\mathbf{-}\mathit{x}\mathit{y}\mathit{d}\mathit{y}\mathbf{=}\mathbf{0}$into the form given in Problem 31.

Short answer

As a result,$\frac{\mathrm{d}y}{\mathrm{d}x}=F\left(\frac{y}{x}\right),$ where $F\left(u\right)=\frac{5-2u^2}{u},u=\frac{y}{x}$.

Step-by-step solution

Given Information:

The given equation is$\left(5x^2-2y^2\right)dx-xydy=0$

Solve the given homogeneous differential equation:

The given differential equation can be written as follows:

$\begin{array}{c}\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{5x^2-2y^2}{xy}\\ =\frac{x^2\left(5-2\frac{y^2}{x^2}\right)}{x^2\left(\frac{y}{x}\right)}\\ =\frac{5-2{\left(\frac{y}{x}\right)}^2}{\frac{y}{x}}\end{array}$

The above equation is in the form $\frac{\mathrm{d}y}{\mathrm{d}x}=F\left(\frac{y}{x}\right),$where $F\left(u\right)=\frac{5-2u^2}{u},u=\frac{y}{x}$.