Calculus problem (electric circuit)

[Cyannaca]

A electric circuit is made of a resistance, a solenoid and a power source.
Resistance, V= Ri
Solenoid, V = L (di/dt)
Power source, tension E(t)

Hence, L(di/dt)+ Ri = E(t). If E(t)= 20sin(5t), if L= 4 henrys and R= 20 ohms. If the intensity equals 0 at the beginning, find the amperage of i(t). Show that after a certain amount of time, the electric current becomes stationary.

I would really need help to start this problem, I'm completly lost :confused: . i was stuck at 20sin(5t)= 4(di/dt) + 20i... The answer is supposed to be
i(t)= 1/2( sin(5t)-cos(5t) +e^-5t)

[HallsofIvy]

A electric circuit is made of a resistance, a solenoid and a power source.
Resistance, V= Ri
Solenoid, V = L (di/dt)
Power source, tension E(t)

Hence, L(di/dt)+ Ri = E(t). If E(t)= 20sin(5t), if L= 4 henrys and R= 20 ohms. If the intensity equals 0 at the beginning, find the amperage of i(t). Show that after a certain amount of time, the electric current becomes stationary.

I would really need help to start this problem, I'm completly lost :confused: . i was stuck at 20sin(5t)= 4(di/dt) + 20i... The answer is supposed to be
i(t)= 1/2( sin(5t)-cos(5t) +e^-5t)

If 4 di/dt+ 20i= 20 sin(5t), then di/dt+ 5i= 5sin(5t). Since this is a "linear differential equation", the first thing I would do is look at the related homogeneous equation, di/dt+ 5i=0 which is the same as di/dt= -5i or
(1/i) di= -5dt. Integrating both sides, ln(i)= -5t+ C or i= C' e-5t.

We look for a solution to the entire equation of the form
i(t)= Asin(5t)+ Bcos(5t). Then i'(t)= 5Acos(5t)- Bsin(5t). Putting those into the equation i'+ 5i= 5 sin(5t), we have 5Acos(5t)- 5Bsin(5t)+ 5Asin(5t)+5Bcos(5t)= 5sin(5t) so we must have 5A+ 5B= 0 and -5B+ 5A= 5. Adding those two equations, 10A= 5 so A= (1/2) and B= -(1/2).

Putting those together, the general solution is i(t)= C'e-5t+ (1/2)sin(5t)- (1/2)cos(5t). i(0)= C'+ 1/2= 0 so C'= -1/2.
That is, i(t)= (1/2)[sin(5t)- cos(5t)- e-5t].