Question: Find the solution of the given initial value problem.
$$y'-2y=e^{2t}, y(0)=2$$
Solution:
Notice that the given DE is a first order linear ODE.
Let $\mu(t)$ denote an integrating factor for the given DE.
$$\mu(t) = e^{\int -2 dt} = e^{-2t} $$
Multiply the given DE by $\mu(t)$: note that $\mu(t) ≠ 0$ for all $t$
$$e^{-2t}y'- 2e^{-2t}y = e^{-2t} e^{2t}$$
Simplify the equation:
$$\frac{d}{dt} (e^{-2t}y)= 1$$
Integrate both sides with respect to $t$:
$$e^{-2t}y = t + C$$
Isolate $y$:
$$y = (t+C)e^{2t}$$
Since $y(0)=2$, then $2 = (0+C)\cdot e^{0} = C\cdot 1 = C$
Thus, solution to the given IVP is
$$y = (t+2)e^{2t}$$
