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

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

• Use the Integrating Factor method, perhaps? – David Quinn Apr 7 '17 at 18:58
• Check out linear differential equations. There is a direct closed form solution for first order case. – Shraddheya Shendre Apr 7 '17 at 18:58
• But this is nono-linear equation , is not it? – M. A. SARKAR Apr 7 '17 at 19:00
• No, this is linear. Any equation of the form $y'(t) + a(t)y(t) = p(t)$ is linear. Nonlinear equations are those where the unknown $y$ shows up in some nonlinear way; for example, $y'(t) + \sin(t) y = \sin(y)$ is nonlinear as is $y'(t) = 1+ y(t)^2$. – User8128 Apr 7 '17 at 19:02

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:

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

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 $\displaystyle e^{\int f(t)dt} = e^{-\cos(t)}$