Solve the initial value problem
$$y' = \frac{1 + y^2}{x};\ y(1)=1;\ x>0$$
by separation of variables.
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$\begingroup$ What have you tried, and where did you get stuck? We will be able to help you more if you post your work. $\endgroup$– apnortonMar 21, 2013 at 13:33
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$\begingroup$ @anorton, I fail at integration part. $\endgroup$– Denys S.Mar 21, 2013 at 13:50
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1 Answer
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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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$\begingroup$ So we integrate both sides from 1 to something? And can you please give an explanation of integrating 1/(1+y^2)? $\endgroup$– Denys S.Mar 21, 2013 at 13:51
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$\begingroup$ No no ... just plain indefinite integration. $\int \frac{dy}{1+y^2} = \int \frac{dx}{x}$ $\endgroup$– S LMar 21, 2013 at 13:51
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$\begingroup$ Are you sure we're not supposed to integrate from 1 to y or from 1 to x respectfully? $\endgroup$– Denys S.Mar 21, 2013 at 14:01
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$\begingroup$ No ... this is indefinite integration that results in integral constant $C$, and we use that initial condition to find the value of this $C$ $\endgroup$– S LMar 21, 2013 at 14:02
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$\begingroup$ ok, so then I should end up with y=tan(ln(x)+1). Can you confirm? $\endgroup$– Denys S.Mar 21, 2013 at 14:05