Checking the solution to $2yy'=x(y')^2+4x,\ y(1)=-2$

I have to find a particular solution to the following differential equation: $$2yy'=x(y')^2+4x,\ y(1)=-2$$

I decided to substitute the values $$x=1$$ and $$y=-2$$ into the given equations. So, I got: \begin{aligned} &-4y'=(y')^2+4\iff(y'+2)^2=0\iff y'=-2\iff y=-2x+C\Rightarrow\\ &\Rightarrow [\text{Substitute y into the initial diff. eq.}]\Rightarrow (-4x+2C)(-2)=4x+4x\Rightarrow C=0\\ &\text{Answer: } y=-2x \end{aligned} But I'm not sure of my solution. Could anyone check it please?

• This isn't quite right - you've only shown that $y'(1) =-2$, which does not imply $y'(x) = -2x + C$ Feb 21, 2020 at 19:24
• Could you provide the correct solution then, please? Feb 21, 2020 at 19:32

Hint.

Solve for $$y'$$ giving

$$y'=\frac 1x\left(y\pm\sqrt{y^2-4x^2}\right)$$

or

$$y' = \frac yx\pm\sqrt{\left(\frac yx\right)^2-4}$$

and now making

$$z = \frac yx$$

we have

$$z'=\pm \frac 1x\sqrt{z^2-4}$$

which is separable.

NOTE

$$\frac{dx}{x}=\pm\frac{dz}{\sqrt{z^2-4}}$$

• that's .... awesome.+1 Feb 21, 2020 at 19:51
• Thank you for the solution! May I ask you to check the final result? It is either $x=C\left(\frac{y}{x}+\sqrt{\frac{y^2}{x^2}-4}\right)\ \ \left(C=-\frac{1}{2}\right)$ or $x=C/\left(\frac{y}{x}+\sqrt{\frac{y^2}{x^2}-4}\right)\ \ (C=-2)$. Feb 21, 2020 at 21:44
• According to my calculations, one of the solutions gives $$y = \frac{1}{C_0}+C_0 x^2$$ Feb 21, 2020 at 22:22
• I got it. Thank you! Feb 22, 2020 at 7:58