# I was given a problem, it has been worded as follows: Use the substitution z = y

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:
$\frac{dy}{dx}=\frac{-3\left(\frac{y}{x}\right)}{\left(\frac{y}{x}{\right)}^{2}-3}$
Multiplied by $\frac{\frac{1}{{x}^{2}}}{\frac{1}{{x}^{2}}}$
$\frac{dy}{dx}=\frac{-3z}{{z}^{2}-3}$
${z}^{\prime }=\frac{{y}^{\prime }}{x}-\frac{y}{{x}^{2}}$
$\left({z}^{\prime }+\frac{y}{{x}^{2}}\right)x=\frac{-3z}{{z}^{2}-3}$
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.
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bamenyab4mxn
From y=zx, draw
${y}^{\prime }={z}^{\prime }x+z.$
Then
${z}^{\prime }x+z=-\frac{3z}{{z}^{2}-3}$
which is separable:
${z}^{\prime }x=-z-\frac{3z}{{z}^{2}-3}=-\frac{{z}^{3}}{{z}^{2}-3}.$
The equation can be seen as linear (and homogeneous) by swapping the roles of the dependent and independent variables and rewriting
$\frac{dx}{dz}+\frac{{z}^{2}-3}{{z}^{3}}x=0.$
This doesn't ease the solution.
###### Not exactly what you’re looking for?
llunallenaipg5r
Setting
$y=ux$
then we have
${u}^{\prime }x+u=-\frac{3u}{{u}^{2}-3}$
simplifying we get
${u}^{\prime }x=-u\left(\frac{3}{{u}^{2}-3}+1\right)$
or
${u}^{\prime }x=-\frac{{u}^{3}}{{u}^{2}-3}$
this equation is separble and it follows
$-\frac{{u}^{2}-3}{{u}^{3}}du=\frac{dx}{x}$