# What type of initial value problem is this?

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?

## 2 Answers

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

• Also, I do believe it's called an homogeneous differential equation. – Cody Rudisill Nov 6 '15 at 3:32
• 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. – david mah Nov 6 '15 at 6:57
• 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. – Cody Rudisill Nov 6 '15 at 11:54

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.