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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?

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

1 Answer 1

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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

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

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    $\begingroup$ that's .... awesome.+1 $\endgroup$
    – Qurultay
    Feb 21, 2020 at 19:51
  • $\begingroup$ 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 $x=C/\left(\frac{y}{x}+\sqrt{\frac{y^2}{x^2}-4}\right)\ \ (C=-2)$. $\endgroup$
    – Bonrey
    Feb 21, 2020 at 21:44
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    $\begingroup$ According to my calculations, one of the solutions gives $$y = \frac{1}{C_0}+C_0 x^2$$ $\endgroup$
    – Cesareo
    Feb 21, 2020 at 22:22
  • $\begingroup$ I got it. Thank you! $\endgroup$
    – Bonrey
    Feb 22, 2020 at 7:58

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