Solve the initial value problem $4\left(\sin(t)\frac{dy}{dt}+\cos(t)y\right)=\cos(t)\sin^3(t)$ 
Solve the initial value problem
$$4\bigg(\sin(t)\frac{dy}{dt}+\cos(t)y\bigg)=\cos(t)\sin^3(t)$$
for $0<t<\pi$ and $y(\pi/2)=20$.
Put the problem in standard form.

I've calculated the integrating factor as being $\sin(t)$
I just can't seem to solve $y(t)$.

Edit (After reading your answers):
Here is the work that I have so far;
I multiplied every term by the integrating factor $p(t)=\sin(t)$
$$\sin(t)\frac{dy}{dt}+\cos(t)y=\frac{1}{4}\cos(t)\sin^3(t)$$
Then took the total differential of the left side;
$$\frac{d}{dt}(\sin(t)y)=\frac{1}{4}\cos(t)\sin^3(t)$$
Then integrated from ($\pi/2\to t$);
$$\sin(t)y(t)-\sin(\pi/2)y(\pi/2)=\frac{1}{16}\sin^4(t)-\frac{1}{16}\sin^4(\pi/2)$$
$$\implies \sin(t)y(t)=20+\frac{1}{16}(\sin^4(t)-1)$$
$$\implies y(t)=\frac{321}{16}\csc(t)+\frac{1}{16}\sin^3(t)$$
 A: If we put $$z=\sin (t)y $$
the equation becomes
$$\frac {dz}{dt}=\frac {1}{4}\sin^3 (t)\cos (t) $$
$$=\frac {1}{16}\frac {d \sin^4 (t)}{dt} $$
hence by integration,
$$z=\frac {1}{16}\sin^4 (t)+C $$
You can finish.
A: Since you don't share any ideas with us, I don't want to give a full answer to that problem. The key idea is to use a technique called "variation of constants". Try to solve the homogeneous problem (the left hand side equals zero) and solve this using seperation of variables. Keep the constant $C$ variable in $t$. You should end up with something like $$y(t) = -C(t) \, \sin(t).$$ Then, you derive this expression with respect to $t$, which gives you $$\dot{y}(t) = -\dot{C}(t) \, \sin(t) - C(t) \, \cos(t).$$ Using your differential equation on the left hand side, you end up with the condition $$-\dot{C}(t) \, \sin(t) = \frac{1}{4} \, \sin^2(t) \, \cos(t).$$ You will end up with a solution $$y(t) \sim \sin^3(t).$$ Be careful with those calculations, the details of some moduli are left out, maybe you devide by zero and you have to put in the right initial data, but I am sure you can figure out the rest.
A: Dividing the ODE by $4$, one obtains:
$$\sin(t)\cdot y'+\cos(t)\cdot y=\frac{1}{4}\cos(t)\sin^3(t) \tag{1}$$
Now we recall the product rule, and apply it in reverse on the LHS. Remember that $(\sin(t))'=\cos(t)$, therefore, we have from $(1)$:
$$\sin(t)\cdot y'+(\sin(t))'\cdot y=\frac{1}{4}\cos(t)\sin^3(t)$$
$$(\sin(t)\cdot y)'=\frac{1}{4}\cos(t)\sin^3(t) \tag{2}$$
Now integrate both sides of $(2)$ with respect to $t$, and you will have your general solution. All that remains to do now is to apply the condition you were given to obtain the particular solution: $y(\pi/2)=20$.

Edit: You've shown us what you've tried after reading the above. The only mistake is your fraction which I highlighted in red when transitioning between the following step:
$$\sin(t)y(t)=20+\frac{1}{16}(\sin^4(t)-1) \not\Rightarrow y(t)=\color{red}{\frac{321}{16}}\csc(t)+\frac{1}{16}\sin^3(t)$$
Remember that $20-\dfrac{1}{16}\equiv \dfrac{320}{16}-\dfrac{1}{16}\equiv \dfrac{319}{16}$.
