Question

Solve the initial value problem (explicit solution). y y' - cot t = 0 y(pi/2) = -1

Answer 1

This ODE is separable:

$yy'-\cot t=0\implies y\,\mathrm dy=\cot t\,\mathrm dt$

Integrate both sides to get

$\dfrac12y^2=\ln|\sin t|+C$

Given that $y\left(\frac\pi2\right)=-1$, we get

$\dfrac12(-1)^2=\ln|\sin\dfrac\pi2\right|+C\implies C=\dfrac12$

Then

$\dfrac12y^2=\ln|\sin t|+\dfrac12$

$y^2=2\ln|\sin t|+1$

$y^2=\ln\sin^2t+1$

$\implies\boxed{y(t)=\pm\sqrt{\ln\sin^2t+1}$