Respuesta :

This ODE is separable:

[tex]yy'-\cot t=0\implies y\,\mathrm dy=\cot t\,\mathrm dt[/tex]

Integrate both sides to get

[tex]\dfrac12y^2=\ln|\sin t|+C[/tex]

Given that [tex]y\left(\frac\pi2\right)=-1[/tex], we get

[tex]\dfrac12(-1)^2=\ln|\sin\dfrac\pi2\right|+C\implies C=\dfrac12[/tex]

Then

[tex]\dfrac12y^2=\ln|\sin t|+\dfrac12[/tex]

[tex]y^2=2\ln|\sin t|+1[/tex]

[tex]y^2=\ln\sin^2t+1[/tex]

[tex]\implies\boxed{y(t)=\pm\sqrt{\ln\sin^2t+1}[/tex]