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

Find the solution of the given initial value problem.
${y}^{\prime }-2y={e}^{2t},y\left(0\right)=2$
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Killaninl2
Given equation is ${y}^{1}-2y={e}^{2t}$
This is a linear equation of first order in y
where P(t)=-2 and $Q\left(t\right)={e}^{2t}$
$\int P\left(t\right)dt=-2\int 1\ast dt=-2t$
Integrating Factor $={e}^{\int P\left(t\right)dt}={e}^{-2t}$
Multiplying (1) with ${e}^{-2t}$, we get
${e}^{2t}{y}^{1}-2{e}^{-2t}y=1\to \left(2\right)$
(2) can be written as
$\frac{d}{dt}\left[y\ast {e}^{-2t}\right]dt=\int 1\ast dt$
$⇒y\ast {e}^{-2t}=t+C$
$⇒y\left(t\right)=\frac{t}{{e}^{-2t}}+\frac{C}{{e}^{-2t}}$
we know that y(0)=2
so $\frac{0}{{e}^{-0}}+\frac{C}{{e}^{-0}}=2⇒C=2$
Therefore the general solution of (1) is $y=\frac{t+2}{{e}^{-2t}}$