I was given a problem, it has been worded as follows: Use the substitution $z=\frac{y}{x}$ to transform the differential equation $\frac{dy}{dx}=\frac{-3xy}{{y}^{2}-3{x}^{2}}$, into a linear equation. Hence obtain the general solution of the original equation.

Workings:

$\frac{dy}{dx}=\frac{-3(\frac{y}{x})}{(\frac{y}{x}{)}^{2}-3}$

Multiplied by $\frac{\frac{1}{{x}^{2}}}{\frac{1}{{x}^{2}}}$

$\frac{dy}{dx}=\frac{-3z}{{z}^{2}-3}$

${z}^{\prime}=\frac{{y}^{\prime}}{x}-\frac{y}{{x}^{2}}$

$({z}^{\prime}+\frac{y}{{x}^{2}})x=\frac{-3z}{{z}^{2}-3}$

After a whole page of workings, I arrive at this

${z}^{\prime}+\frac{z}{x}=\frac{3}{z}$

Technically, this is a linear equation since it takes the form y'+p(x)y=f(x) . Since z=f(y,x). However, I'm unable to compute this, since I won't know how to integrate z with respect to x. I do know this can easily be done by separating variables, however, we are basically told to solve it this way.

Workings:

$\frac{dy}{dx}=\frac{-3(\frac{y}{x})}{(\frac{y}{x}{)}^{2}-3}$

Multiplied by $\frac{\frac{1}{{x}^{2}}}{\frac{1}{{x}^{2}}}$

$\frac{dy}{dx}=\frac{-3z}{{z}^{2}-3}$

${z}^{\prime}=\frac{{y}^{\prime}}{x}-\frac{y}{{x}^{2}}$

$({z}^{\prime}+\frac{y}{{x}^{2}})x=\frac{-3z}{{z}^{2}-3}$

After a whole page of workings, I arrive at this

${z}^{\prime}+\frac{z}{x}=\frac{3}{z}$

Technically, this is a linear equation since it takes the form y'+p(x)y=f(x) . Since z=f(y,x). However, I'm unable to compute this, since I won't know how to integrate z with respect to x. I do know this can easily be done by separating variables, however, we are basically told to solve it this way.