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$x^2y'^2 + 3xyy' +2y^2 = 0 $

Usually, to solve an ODE with respect to $y'=p$, we first isolate the $y$, to get $y = f(x,p)$ and then differentiate with respect to $x$ to get an expression that only depends on $x$ and $p$. Then, we can write $x$ and $y$ in terms of $p$ and get to a solution.

But what do we do if, like in this particular example, we can't isolate the y? Is there another method for these kind of ODEs?

My manual lists the solutions for this equation as $xy=c$ or $yx^2 =c$, but I have no clue as to how they come to that conclusion.

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  • $\begingroup$ Either your manual is wrong or the second solution should be $x^2 y = c$. $\endgroup$ Feb 4, 2020 at 18:58
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    $\begingroup$ for $w=xy'$ you get a $w^2+3yw+2y^2 \implies (w+y)(w+2y)=0$ $\endgroup$ Feb 4, 2020 at 19:30

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You can factor your differential equation to get $$ (x y' + y)(x y' + 2 y) = 0 $$

so either $x y' + y = 0$ or $x y' + 2 y = 0$. One gives you $y = c/x$, the other $c/x^2$.

Actually we should be careful to check that you can't switch from one of these to the other while maintaining differentiability: it turns out that you can't.

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$$x^2y'^2 + 3xyy' +2y^2 = 0$$ I suppose $y \ne 0$ $$x^2 \frac {y'^2}{y^2} + 3x\frac {y'}{y} +2 = 0$$ $$x^2 \left ( \frac {y'}{y} \right )^2 + 3x \left ( \frac {y'}{y} \right ) +2 = 0$$ $$x^2 \left ( (\ln y)' \right )^2 + 3x(\ln y )'+2 = 0$$ It's a quadtratic equation. $$ \left ( x(\ln y)' \right )^2 + 3(x(\ln y )')+2 = 0$$

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