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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. –  anorton Mar 21 '13 at 13:33
    
@anorton, I fail at integration part. –  Denys S. Mar 21 '13 at 13:50
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1 Answer

up vote 4 down vote accepted

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.

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So we integrate both sides from 1 to something? And can you please give an explanation of integrating 1/(1+y^2)? –  Denys S. Mar 21 '13 at 13:51
    
No no ... just plain indefinite integration. $\int \frac{dy}{1+y^2} = \int \frac{dx}{x}$ –  Santosh Linkha Mar 21 '13 at 13:51
    
Are you sure we're not supposed to integrate from 1 to y or from 1 to x respectfully? –  Denys S. Mar 21 '13 at 14:01
    
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
    
ok, so then I should end up with y=tan(ln(x)+1). Can you confirm? –  Denys S. Mar 21 '13 at 14:05
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