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

The solution for y^{''}+y^'-2y=e^{-2x} is y=c_{1}e^x+c_{2}e^{-2x}-(e^{-2x}x)/3

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

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

y = c_{1} e^{x} + c_{2} e^{- 2 x} - \frac{e ^{- 2 x} x}{3}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{x} + c_{2} e^{- 2 x} - \frac{e ^{- 2 x} x}{3}

y = c_{1} e^{x} + c_{2} e^{- 2 x} - \frac{e ^{- 2 x} x}{3}

\int \frac{( 2 + x )}{9 + x ^{2}} d x

\int_{0}^{1} x (\frac{2 ( x + 3 )}{7}) d x

f (x) = - 3 cos (x)

\int \frac{x ^{2}}{4 x ^{3} + 5} d x

f (t) = 1 + cos (t)

What is the solution for y^{''}+y^'-2y=e^{-2x} ?
The solution for y^{''}+y^'-2y=e^{-2x} is y=c_{1}e^x+c_{2}e^{-2x}-(e^{-2x}x)/3