Consider the homogeneous differential equation


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

  • 1
    $\begingroup$ 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





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

Then $t=su$


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






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

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

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

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

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

$s(0)=1$ :



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


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