Originally Posted by

**pakman** 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.