dy/dt + 3y = t + e^(-2t)
Find the general solution of the given differential equation and use it to determine how solutions behave at t --> inf
So we did this problem in class and I'm a little confused about how the TA integrated it.
(e^(3t)y)' = te^(3t) + e^t
e^(3t)y = INT(te^(3t) + e^t)dt
u = t
du = dt
dv = e^(3t)dt
v = (1/3)e^(3t)
e^(3t)y = (1/3)te^(3t) - (1/9)e^(3t) + e^t + C
y = (1/3)t - (1/9) + e^(-2t) + Ce^(-3t)
If someone could please explain how they got that after integration. Thank you.