Question

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

Answer 1

first you have to find your integrating factor:
\[u(t)=e^{\int\limits 3 dt}=e^{3t}\]
Now:
\[e^{3t}y=\int\limits e^{3t}(t=e^{-2t})dt\]
\[e^{3t}y=\int\limits (te^{3t}+e^{t})dt\]
\[e^{3t}y=\int\limits te^{3t}dt+\int\limits e^tdt\]
use integrations by parts on the first integral and the second integral is basic integration:
\[e^{3t}y= \frac{e^{3t}}{3}-\frac{e^{3t}}{9}+e^t+C\]
Divide both side by \[e^{3t}\]
\[y=\frac{t}{3}-\frac{1}{9}+e^{-2t}+Ce^{-3t}\]

Answer 2

made a typo the equal sign should be a +

Answer 3

thank you! It makes more sense now