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

`\frac12y^2=\ln|\sin t|+C`

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

`\frac12(-1)^2=\ln|\sin\frac\pi2\right|+C\implies C=\frac12`

Then

`\frac12y^2=\ln|\sin t|+\frac12`

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

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