What is the solution for y^{''}+y^'=e^{2x} ?

The solution for y^{''}+y^'=e^{2x} is y=c_{1}+c_{2}e^{-x}+1/6 e^{2x}

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y^{''}+y^'=e^{2x}

y^{^{''}} + y^{^{'}} = e^{2 x}

y = c_{1} + c_{2} e^{- x} + \frac{1}{6} e^{2 x}

Solution steps

y^{''} (x) + y^{'} (x) = e^{2 x}

Solve linear ODE:

y = c_{1} + c_{2} e^{- x} + \frac{1}{6} e^{2 x}

y = c_{1} + c_{2} e^{- x} + \frac{1}{6} e^{2 x}

y^{^{''}} - 4 y^{^{'}} + 4 y = 3 t^{2} + 4 + e^{2 t} + t sin (2 t)

\frac{d ^{2} y}{d x ^{2}} + 9 y = 2 sec (3 x)

y^{^{''}} - 5 y^{^{'}} - 6 y = 5 x e^{- x}

x^{^{''}} + x = 8 cos (5 t)

y^{^{''}} + 2 y^{^{'}} + 1 = 4 sin (2 x)

What is the solution for y^{''}+y^'=e^{2x} ?
The solution for y^{''}+y^'=e^{2x} is y=c_{1}+c_{2}e^{-x}+1/6 e^{2x}