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So I'm trying to solve this initial value problem:

$x^2 dy/dx + xy = 1$, $y(-1)=1$

Now I think that it's some sort of Linear Equation and I know how to solve linear equations like $dA/dt+1/100A=6$, $A(0)=50$ but for this equation, there's multiple variables and an $x^2$ in front of the dy/dx. So is this even a linear equation and if it is a linear equation then how could I solve it compared to equations like the one above?

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Find an integrating factor. Start by diving by $x^2$, so that $\frac{dy}{dx}+\frac{1}{x}y=\frac{1}{x^2}$

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  • $\begingroup$ Also, I do believe it's called an homogeneous differential equation. $\endgroup$ – Cody Rudisill Nov 6 '15 at 3:32
  • $\begingroup$ Ok so I tried to find the integrating factor and arrived at $d/dx[e^{1/x}y]=1/x^2e^{1/2}$ but I'm not sure where to go from there. $\endgroup$ – david mah Nov 6 '15 at 6:57
  • $\begingroup$ Your next step is to integrate after multiplying by the integrating factor, and then clean up with division. So, $e^{\int\frac{1}{dx}}=e^{ln(x)}=x:$ becomes $\frac{d}{dx}[xy]=\frac{1}{x}$. Integrating gets, $xy=ln(x)+C$ which becomes $y=\frac{ln(x)}{x}+\frac{C}{x}$ with division. $\endgroup$ – Cody Rudisill Nov 6 '15 at 11:54
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Hint

Start changing variable $x y=z$, $y=\frac z x$, $y'=\frac{x z'-z}{x^2}$ and replace in the initial equation. You will arrive to something simple and easy.

I am sure that you can take from here.

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