what is e'
??
Do you mean
y' + y = te^t ?
If so, use the integrating factor e^t. Then we have
e^t y' + e^t y = te^(2t)
(e^t y)' = te^(2t)
e^t y = 1/4 e^(2t) (2t-1) + c
y = 1/4 e^t (2t-1) + c*e^-t
Hello, I am needing help to find the general solution to:
dy/dt + y = te'
Thanks!
1 answer