I need help solving $y'(t)+7\sin(t)y(t)=(te^{\cos(t)})^7$ (differential equation)

Question

I need help solving the following differential equation:

$$y'(t)+7\sin(t)y(t)=(te^{\cos(t)})^7$$

Can anyone push me in the right direction?

Answer 1

$$y'(t)+7\sin(t)y(t)=(te^{\cos(t)})^7$$ $$y'(t)e^{-7\cos(t)}+7\sin(t)y(t)e^{-7\cos(t)}=t^7$$

Rewrite it as $$\implies (y(t)e^{-7\cos(t)})'=t^7$$ And integrate...

Answer 2

Hint: Notice that you can rewrite the LHS as $$ y'(t) + 7\sin(t) y(t) = e^{7\cos(t)}\frac{\mathrm d}{\mathrm dt}\left(y(t)e^{-7\cos(t)}\right). $$ Can you continue?

Answer 3

you have a function of the form $y'(t)+a(t)y=b(t)$.

You could multiply both sides of the expression by the integrating factor $e^{A(t)}$ where $A(t)=\int a(t)dt$.

In your case $a(t)=7sin(t)$ and hence $A(t)=-7cos(t)$.

After multiplying by the integrating factor you'll end up with:

$y'(t)e^{-7cos(t)}+y(t)7sin(t)e^{-7cos(t)}=t^7e^{7cos(t)}e^{-7cos(t)} $

$y'(t)e^{-7cos(t)}+y(t)7sin(t)e^{-7cos(t)}=t^7$

$\frac{d}{dt} y(t)e^{-7cos(t)}=t^7$

And once you are here, integrate...