# Solve separable DE with integrating factor and homogeneous substitution

I just came out of test which asked to solve $$\frac{dy}{dx}=\frac{y}{x}$$ with $x,y>0$ in three ways: by separating the variables, using the substitution $y=vx$ and using an integrating factor.

So what I did was the following \begin{align*}&\frac{dy}{y}=\frac{dx}{x}\\\Rightarrow&\ln y=\ln x +C\\\Rightarrow&y=e^c x=kx.\end{align*}

Then using the substitution $y=vx$, I had $$\frac{d}{dx}(vx)=v$$which by the product rule becomes $$x\frac{dv}{dx}+v=v\Rightarrow x\frac{dv}{dx}=0$$from which follows that $v$ is a constant and hence $y=vx=kx$.

Now with the integrating factor I said that the equation is equivalent to $$\frac{dy}{dx}+yP(x)=Q(x),$$with $P(x)=-1/x$ and $Q(x)=0$. Thus I rewrote the equation as $$\frac{d}{dx}(e^{-\ln x+C}y)=0\Rightarrow \frac{y}{x}e^C=0.$$ But now how do I proceed?

Or I am not allowed to have $Q(x)=0$ anyway?

• First your answer is wrong, $P(x)$ should be $-1/x$, second, you just integrate the right hand side also, you have left hand side without the derivative = a constant. – Lost1 Mar 26 '13 at 14:09

As I see your OE, it is $$y'=\frac{y}{x}$$ so $y'-\frac{1}{x}y=0$ so $P(x)=-\frac{1}{x}$ and $Q(x)=0$. Here we have $\mu(x)=\exp\int(-1/x)dx=x^{-1}$. Now applying this integrating factor you'll get: $$x^{-1}y'-x^{-2}y=0$$ so $$d(x^{-1}y)=0\to x^{-1}y=C$$