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Given the following

$\ y'= \frac{3y^2-x^2}{2xy}$

I need to tell if the equation is linear, which I think it is because:

$\ y'= \frac{3y^2-x^2}{2xy} = \frac{3y}{2x}-\frac{x}{2y}$

Now I need to solve the equation with separation of variables which is only possible with substitution. So I substitute with $\ u(x)= \frac{y}{x}$ and do the following

$\ y'= \frac{3y}{2x}-\frac{x}{2y} => y'=\frac{3y}{2x}-\frac{1}{2u(x)} = \frac{3}{2x}y-\frac{1}{2u(x)}$

so I can use separation of variables and get the homogeneous solution and the inhomogeneous. For the homogeneous I got

$\ y_h=e^cx^{3/2} = Cx^{3/2} $

but I really don't know how to get the inhomogeneous solution because I don't exactly know what to do with the substitution, I am glad for help.

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    $\begingroup$ The DE is not linear, so you can't extract "homogeneous" or "inhomogeneous" solutions with the usual method. Also, note that this DE is first order, so you don't really need to use homogeneous and inhomogeneous solutions, since you can solve it exactly. You have the correct substitution, but when you substitute, you want to make sure you get rid of every y in your equation. $\endgroup$ – Kaynex Mar 7 '16 at 16:42
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A start: The DE is not linear. But it is homogeneous. A standard way to begin is to let $y=ux$. Then $\frac{dy}{dx}=\frac{du}{dx}+u$. The right-hand side simplifies to $\frac{3u^2-1}{2u}$. After some further simplification the equation can be written as $2u\frac{du}{dx}=u^2-1$. This can be solved by separation of variables.

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$$y'(x)=\frac{3y(x)^2-x^2}{2xy(x)}\Longleftrightarrow$$ $$y'(x)-\frac{3y(x)}{2x}=-\frac{x}{2y(x)}\Longleftrightarrow$$ $$2y(x)y'(x)-\frac{3y(x)^2}{2x}=-x\Longleftrightarrow$$


Let $r(x)=y(x)^2$, which gives $r'(x)=2y(x)y'(x)$:


$$r'(x)-\frac{3r(x)}{x}=-x\Longleftrightarrow$$


Let $v(x)=\exp\left[\int-\frac{3}{x}\space\text{d}x\right]=\frac{1}{x^3}$:


$$\frac{r'(x)}{x^3}-\frac{3r(x)}{x^4}=-\frac{1}{x^2}\Longleftrightarrow$$


Substitute $-\frac{3}{x^4}=\frac{\text{d}}{\text{d}x}\left(\frac{1}{x^3}\right)$:


$$\frac{r'(x)}{x^3}-\frac{\text{d}}{\text{d}x}\left(\frac{1}{x^3}\right)r(x)=-\frac{1}{x^2}\Longleftrightarrow$$


Apply the reverse product rule:


$$\frac{\text{d}}{\text{d}x}\left(\frac{r(x)}{x^3}\right)=-\frac{1}{x^2}\Longleftrightarrow$$ $$\int\frac{\text{d}}{\text{d}x}\left(\frac{r(x)}{x^3}\right)\space\text{d}x=\int-\frac{1}{x^2}\space\text{d}x\Longleftrightarrow$$ $$\frac{r(x)}{x^3}=\frac{1}{x}+\text{C}\Longleftrightarrow$$ $$r(x)=x^2\left(\text{C}x+1\right)\Longleftrightarrow$$ $$y(x)^2=x^2\left(\text{C}x+1\right)\Longleftrightarrow$$ $$y(x)=\pm\sqrt{x^2\left(\text{C}x+1\right)}\Longleftrightarrow$$ $$y(x)=\pm x\sqrt{\text{C}x+1}$$

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    $\begingroup$ Thanks a lot! this helped me to verify my solution! $\endgroup$ – gammaALpha Mar 7 '16 at 18:10
  • $\begingroup$ @gammaALpha You're welcome! You can find $\text{C}$ when you got the right conditions for your question $\endgroup$ – Jan Eerland Mar 7 '16 at 18:10

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