Question

Differential equations: The general solution is y(t)=Ce^(-2t) +t+1 , what is p(t) and g(t)

Answer 1

just take a derivative. y'=-2Ce^(-2t)+1
y=-y'/2+3/2+t

Answer 2

you have a formula to find out y(t) from original one which is
\[\large y(t) = e^{-\int P(t)}*\int Q(t)* e^{\int P(t)} dt + C e^{-\int P(t)}= Ce^{-2t}+t+1\]
make agreement both sides, you have \(\large Ce^{-\int P(t)} = Ce^{-2t}\)
therefore, \(\int P(t) = 2t \rightarrow P(t) =2\)
2/ as @eashy guided, you have y' , replace that y' and your y into the standard form of the first order y' + P(t)y = Q(t) you have
\[-2Ce^{-2t}+1+2*(Ce^{-2t} +t+1)= Q(t)\] you have Q(t) = 2t+2
That's it

Answer 3

Thank you both!