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By means of the substitute $y = v(x)Y (x)$, where $Y (x)$ is to be specified, solve the differential equation:

$$\dfrac{dy}{dx}+\dfrac{y}{x}=\dfrac{y^2}{x}$$

with $y=2$ at $x=1$

Anyone can solve it for me with explaining the steps please, I have no idea how to do it. Thank you very much

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  • $\begingroup$ hint: separate the variables. use partial fraction etc. $\endgroup$ – abel Dec 8 '14 at 12:05
  • $\begingroup$ @abel. Your idea is very good but the OP seems to be asked to use $y=v(x)Y(x)$. I don't see what to do with the requirement. Cheers. $\endgroup$ – Claude Leibovici Dec 8 '14 at 12:08
  • $\begingroup$ i don't get it. why would you want as a product? in fact $y = {2 \over 2 - e^{(x^2-1)/2}}$ is the solution and it does not seems to be the product of two simpler functions. $\endgroup$ – abel Dec 8 '14 at 12:14
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$$Y(x)=x$$

$\therefore y=vx\implies \dfrac{dy}{dx}=v+x\dfrac{dv}{dx}$

$\therefore\dfrac{dy}{dx}+\dfrac{y}{x}=\dfrac{y^2}{x}\\\implies 2v+x\dfrac{dv}{dx}=v^2x\\\implies \dfrac{dv}{dx}+2\left(\dfrac{1}{x}\right)v=v^2$

Which is in Bernoulli's Form.

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  • $\begingroup$ @Amzoti: Thanks for pointing out the mistake. It's now edited. $\endgroup$ – user 170039 Dec 8 '14 at 14:56
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$$ y' = v'Y + vY' = -\frac{vY}{x} + \frac{v^2Y^2}{x} $$ if we re-write $$ v'Y = -vY' -\frac{vY}{x} + \frac{v^2Y^2}{x} = -\left(Y' + \frac{Y}{x}\right)v + \frac{v^2Y^2}{x} $$ set the brackets term to zero i.e. $$ Y' = -\frac{Y}{x}\implies Y(x) = \frac{C}{x} $$ then we obtain $$ v' = \frac{1}{Y}\frac{v^2Y^2}{x} = \frac{Y}{x}v^2 = \frac{C}{x^2} v^2 $$ which is separable. Thus $$ v = \frac{x}{C+C_1x}\implies y(x) = \frac{C}{C+C_1x} $$ put it back into the equation we find $$ y' +\frac{y}{x} = \frac{-C C_1}{\left(C+C_1x\right)^2} + \frac{C}{\left(C+C_1x\right)x} = \\ \frac{-C}{\left(C+C_1x\right)^2x}\left[C_1 x -\left(C+C_1x\right)\right] = \frac{C^2}{\left(C+C_1x\right)^2x} = \left(\frac{C}{\left(C+C_1x\right)}\right)^2\frac{1}{x} = \frac{y^2}{x} $$

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    $\begingroup$ The unknown is $v$, $Y$ is a function of your choice. $\endgroup$ – Yves Daoust Dec 8 '14 at 14:01
  • $\begingroup$ then swap it round no? $\endgroup$ – Chinny84 Dec 8 '14 at 14:03
  • $\begingroup$ Yes, that's the fix. $\endgroup$ – Yves Daoust Dec 8 '14 at 14:04
  • $\begingroup$ I also noticed another problem (namely the brackets was a silly expression) I fixed that also. cheers $\endgroup$ – Chinny84 Dec 8 '14 at 14:05
  • $\begingroup$ But then you should specify $Y$. $\endgroup$ – Yves Daoust Dec 8 '14 at 14:07

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