Solve the equation $y''\sin t-2y'\cos t-y\sin t=0$

Question

Solve the differential equation $y''\sin t -2y'\cos t-y\sin t=0$

My try:

I assumed $z=y \sin t$

Then we have:

$$z'=y\cos t+y'\sin t$$

$$z''=-y\sin t+2y'\cos t+y''\sin t$$

Is there a way out here?

Answer 1

Hint :

Alternativelly to LutzL's proposal, you can also use the "simpler"

$$y = v(t)\sin(t) \implies \frac{\mathrm{d}y(t)}{\mathrm{d}t} =\cos(t)\frac{\mathrm{d}v(t)}{\mathrm{d}t}-\sin(t)v(t)$$

Answer 2

Make the following substitutions: $$y(t)=v(t)cos(t)$$ then we get $$\cos(t)v''(t)\sin(t)-2v'(t)=0$$ now let $$v'(t)=u(t)$$ and we get $$\int\frac{\frac{du(t)}{dt}}{u(t)}dt=\int2\csc(t)\sec(t)dt$$ I hope you can finish!