Question

Find the solution of the given initial value problem: y' - y= 6e^2t. y(0) = 1.

Answer 1

I figured you would use the product rule in reverse

Answer 2

\[y'-y=6e^{2t}\]
Find your integrating factor first:
\[\mu(t)=\exp\left(\int (-1)~dt\right)=e^{-t}\]
Multiply both sides of the eq. by \(\mu\):
\[e^{-t}y'-e^{-t}y=6e^{t}\]
As you mentioned, the left side is the result of differentiation via product rule:
\[\frac{d}{dt}\left[e^{-t}y\right]=6e^{t}\]
Anti-differentiate both sides:
\[e^{-t}y=6\int e^{t}~dt\]
And so on. You'll have a constant of integration on the right side; plug in the initial conditions to solve for it.

Answer 3

thanks @SithsAndGiggles !

Answer 4

yw