# Differential equation, substitution

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

• hint: separate the variables. use partial fraction etc. – abel Dec 8 '14 at 12:05
• @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. – Claude Leibovici Dec 8 '14 at 12:08
• 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. – abel Dec 8 '14 at 12:14

$$Y(x)=x$$

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

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

Which is in Bernoulli's Form.

• @Amzoti: Thanks for pointing out the mistake. It's now edited. – user 170039 Dec 8 '14 at 14:56

$$y' = v'Y + vY' = -\frac{vY}{x} + \frac{v^2Y^2}{x}$$ 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 $$y' +\frac{y}{x} = \frac{-C C_1}{\left(C+C_1x\right)^2} + \frac{C}{\left(C+C_1x\right)x} = \\ \frac{-C}{\left(C+C_1x\right)^2x}\left[C_1 x -\left(C+C_1x\right)\right] = \frac{C^2}{\left(C+C_1x\right)^2x} = \left(\frac{C}{\left(C+C_1x\right)}\right)^2\frac{1}{x} = \frac{y^2}{x}$$

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