Question

y'+2y=te^-2t, y(1)=0

Answer 1

The equation is linear, so find the integrating factor:
\[\mu(t)=\exp\left(\int 2~dt\right)=e^{2t}\]
Distribute the IF and solve:
\[\begin{align*}e^{2t}y'+2e^{2t}y&=te^{-2t}e^{2t}\\
e^{2t}y'+2e^{2t}y&=t\\
\frac{d}{dt}\left[e^{2t}y\right]&=t\\
e^{2t}y&=\int t~dt\\
y&=e^{-2t}\int t~dt
\end{align*}\]
and so on. Find the particular solution givne that initial value.

Answer 2

okay but what happened with the 2 coefficient after the second step?

Answer 3

Work backwards, using the chain rule and implicit differentiation:
\[\begin{align*}\frac{d}{dt}\left[e^{2t}y\right]&=\frac{d}{dt}[e^{2t}]y+e^{2t}\frac{d}{dt}[y]\\
&=2e^{2t}y+e^{2t}\frac{dy}{dt}\\
&=e^{2t}\frac{dy}{dt}+2e^{2t}y
\end{align*}\]
The coefficient appears a result of differentiation, and so it disappears as a result of antidifferentiation.

Answer 4

Thank you! You rock