Solve $\dot y+y\sin t=\sin^3 t,\ y(0)=1$

Question

I want to solve the IVP $$\dot y+y\sin t=\sin^3 t,\\ y(0)=1$$ and I'm pretty sure it can be done by separation of variables. How though? I have trouble putting this equation into the form $\dot y=f(t)g(y)$. Or do I need another approach there?

Answer 1

One way to simplify the equation is by making the substitution $x=\cos t$. Then

$$ \dot y = \frac{dy}{dx}\frac{dx}{dt} = -\sin t \frac{dy}{dx} $$

and the equation becomes

$$ \frac{dy}{dx} - y = x^2 - 1, \quad y(1) = 1 $$

The homogeneous solution for this is $y_h(x) = ce^x$. You can use the method you know to obtain the general solution.

Answer 2

HINT

This equation is not separable. Use an integrating factor $I(t)$ given by

$$ I(t) = e^{\int \sin(t) dt} = e^{-\cos(t)}. $$

Answer 3

Not all first order differentail equation are separable.

You can try the method of integrating factor

$$\dot y+y\sin t=\sin^3 t,\ y(0)=1$$ $$(ye^{-\cos t})'=\sin^3 t e^{-\cos t}$$

Answer 4

You cannot simply put it in the form of separation of variables because LHS cannot be written as $f(y)g(t)$ . This is a linear differential equation of form $$\dot{y}+A(t)y=B(t)$$

Multiplying $e^z$ both sides we get $$e^z\dot{y}+e^zA(t)y=e^zB(t)$$ Now we have to find z such that LHS is a perfect differential (So that you can integrate both sides).

Look a $d(e^{\int{A(t)dt}})$ it gives LHS . Taking $z=\int{A(t)dt}$ does our job.

Now,$$d(e^{\int{A(t)dt}})=e^{\int{A(t)dt}}B(t)$$

Now integrate both sides to get your answer.