Solve the initial value problem $ \ y'(t) +(\sin (t)) y=\sin t $

Question

Solve the initial value problem $ \ y'(t) +(\sin (t)) y=\sin t $ , $ \ y(0)=7 \ $ , $ 0 \leq t \leq \pi $ $$ $$ How can I solve this initial value problem . Any hints please

Answer 1

Hint:

The form of the equation is: $y'(t) + f(t)y = g(t)$, well suited for Integrating factor method.

We have

$f(t) = \sin(t)$, consider the integrating factor $\displaystyle e^{\int f(t)dt} = e^{-\cos(t)}$

Answer 2

Try using integrating factor method with integrating factor $e^{\int \sin(t)dt}$. Then after multiplying through you can rewrite the equation as $$\frac{d}{dt} ( e^{-\cos(t)}y(t))=e^{-\cos(t)}\sin(t)$$ Integrating with respect to $t$ (with a $u$-substitution, $u=-\cos(t)$), we get $$e^{-\cos(t)}y(t)=e^{-\cos(t)}+c$$ and so our general solution for $y$ is $$y(t)=1+c e^{\cos(t)}$$

Now using the initial value condition we can solve for $c$: $$y(0)=1 +ce=7$$ and solving for $c$ we get $$c=\frac{6}{e}$$ and the particular solution is:

$$y(t)=1 +\frac{6}{e}e^{\cos(t)}=1+6e^{\cos(t)-1}$$