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Consider the homogeneous differential equation

$$\cfrac{ds}{dt}=\cfrac{2s^2}{s^2+t^2}$$

Find the general solution $s(t)$ of this equation and the solution that satisfies the initial condition $s(0)=1$.

I don't seem to be able to separate these equations or rearrange them into a linear form

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    $\begingroup$ Try introducing a new variable, call it $v$, by $s=tv$. Then $ds/dt=v+(t)(dv/dt)$. $\endgroup$ May 18, 2014 at 12:05

1 Answer 1

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$\dfrac{ds}{dt}=\dfrac{2s^2}{s^2+t^2}$

$\dfrac{dt}{ds}=\dfrac{s^2+t^2}{2s^2}$

$\dfrac{dt}{ds}=\dfrac{1}{2}\left(1+\left(\dfrac{t}{s}\right)^2\right)$

Let $u=\dfrac{t}{s}$ ,

Then $t=su$

$\dfrac{dt}{ds}=s\dfrac{du}{ds}+u$

$\therefore s\dfrac{du}{ds}+u=\dfrac{1+u^2}{2}$

$s\dfrac{du}{ds}=\dfrac{u^2+1}{2}-u$

$s\dfrac{du}{ds}=\dfrac{u^2-2u+1}{2}$

$s\dfrac{du}{ds}=\dfrac{(u-1)^2}{2}$

$\dfrac{du}{(u-1)^2}=\dfrac{ds}{2s}$

$\int\dfrac{du}{(u-1)^2}=\int\dfrac{ds}{2s}$

$-\dfrac{1}{u-1}=\dfrac{\ln s}{2}+c$

$\dfrac{1}{u-1}=\dfrac{C-\ln s}{2}$

$u=\dfrac{2}{C-\ln s}+1$

$\dfrac{t}{s}=\dfrac{2}{C-\ln s}+1$

$t=\dfrac{2s}{C-\ln s}+s$

$s(0)=1$ :

$0=\dfrac{2}{C}+1$

$C=-2$

$\therefore t=s-\dfrac{2s}{\ln s+2}$

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