# How to solve simple differential equation

Solve the initial value problem

$$y' = \frac{1 + y^2}{x};\ y(1)=1;\ x>0$$

by separation of variables.

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What have you tried, and where did you get stuck? We will be able to help you more if you post your work. – apnorton Mar 21 '13 at 13:33
@anorton, I fail at integration part. – Denys S. Mar 21 '13 at 13:50

write $y' = \frac{dy}{dx}$ and you get $\frac{dy}{dx} = \frac{1+y^2}{x} \implies \frac{dy}{1+y^2} = \frac{dx}{x}$ and integrate both sides. use $y(1) = 1$ to find the value of constant after integration.
No no ... just plain indefinite integration. $\int \frac{dy}{1+y^2} = \int \frac{dx}{x}$ – Santosh Linkha Mar 21 '13 at 13:51
No ... this is indefinite integration that results in integral constant $C$, and we use that initial condition to find the value of this $C$ – Santosh Linkha Mar 21 '13 at 14:02