# Find the solution of the given initial value problem. y'−2y=e2t,y(0)=2

Find the solution of the given initial value problem. ${y}^{\prime }-2y=e2t,y\left(0\right)=2$
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Usamah Prosser
Write equation in the form of standard Linear differential equation
$\frac{dy}{dt}+P\left(t\right)\cdot y=Q\left(t\right)$
Compare with it and get P(t).
Now calculate integrating factor $\mu ={e}^{\int P\left(t\right)dt}$
Multiply each term of differential equation by integrating factor
$P\left(t\right)=-2$
Integrating factor $={e}^{\int }\left(-2\right)dt$
$\mu ={e}^{-2t}$
Multiply equation by integrating Factor we get,
${e}^{-2t}\cdot {y}^{\prime }-2{e}^{-2t}y={e}^{-2t}\cdot {e}^{2t}$
${e}^{-2t}\cdot {y}^{\prime }-2{e}^{-2t}y=1$
Consider $\left({e}^{-2t}\cdot y\right)$
Differentiate with respect to t we get,
${\left({e}^{-2t}\cdot y\right)}^{\prime }={e}^{-2t}\cdot {y}^{\prime }-2{e}^{-2t}\cdot y$
Plug this value in equation
we get,
${\left({e}^{-2t}\cdot y\right)}^{\prime }=1$
Integrate both sides,
$\int {\left({e}^{-2t}\cdot y\right)}^{\prime }=\int 1$
${e}^{-2t}\cdot y=t+C$
Where C is integration constant
Apply initial condition. Plug t=0 in equation
${e}^{0}\cdot y\left(0\right)=0+C$
$y\left(0\right)=C$
Given y(0)=2
C=2
Plug value of C=2 in equation
${e}^{-2t}\cdot y=t+2$
Multiply ${e}^{2t}$ both sides we get,
$y={e}^{2t}\cdot \left(t+2\right)$