Problem 16
Solve the differential equation 2st^2s'=1+t^2
Solve the differential equation $2st^2s'=1+t^2$.
Explanation of Method
Divide by $t^2$ to separate variables into $s\,ds$ and a function of $t$, then integrate term by term.
Step 1: Separate variables
\[2st^2\frac{ds}{dt}=1+t^2 \quad\Longrightarrow\quad 2s\,ds=\left(\frac{1}{t^2}+1\right)dt,\qquad t\neq0.\]
Step 2: Integrate
\[\int 2s\,ds=\int\left(t^{-2}+1\right)dt \quad\Longrightarrow\quad s^2=-\frac1t+t+C.\]
Result
Final answer:
\[\boxed{\,s^2=t-\dfrac{1}{t}+C,\qquad t\neq0.}\]