Using the substitution $z=y^2$ solve the differential equation ... given that when $x=4$ and $y=-5$ in the form $y=f(x)$

Question

Differential equation is $2y\frac{dy}{dx}+\frac{y^2}{x}=x^2$

Solving this we get $y^2=\frac{x^3}{4}+\frac{c}{x}$

And by the initial points given c = 36

Therefore, $y=\pm\sqrt{\frac{x^3}{4}+\frac{36}{x}}$

But in the answer it is given as $y=-\sqrt{\frac{x^3}{4}+\frac{36}{x}}$

Where the $\pm$ is omitted and replaced with a $-$. Is this an error in the text or is it something I am missing?

Answer 1

Note that your initial point is $(4,-5)$, so we should take the negative since the positive square root would instead go through the point $(4,5)$.