Solve the equation y′ + 3y = t + e^(−2t).

Hello,

I am going to remember:

y'+3y=0==>y=C*e^(-3t)

y'=C'*e^(-3t)-3C*e^(-3t)

y'+3y=C'*e^(-3t)-3Ce^(-3t)+3C*e^(-3t)=C'*e^(-3t) = t+e^(-2t)
==>C'=(t+e^(-2t))/e^(-3t)=t*e^(3t)+e^t
==>C=e^t+t*e^(3t) /3-e^(3t)/9

==>y= (e^t+t*e^(3t)/3-e^(3t)/9)*e^(-3t)+D
==>y=e^(-2t)+t/3-1/9+D
==>y=e^(-2t)+t/3+k

Whose mistake? can you give more details to what the question is asking?