What is the solution for 2y^{''}-y^'-y=25e^{-x} ?

The solution for 2y^{''}-y^'-y=25e^{-x} is y=c_{1}e^x+c_{2}e^{-x/2}+25/2 e^{-x}

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2y^{''}-y^'-y=25e^{-x}

2 y^{^{''}} - y^{^{'}} - y = 25 e^{- x}

y = c_{1} e^{x} + c_{2} e^{- \frac{x}{2}} + \frac{25}{2} e^{- x}

Solution steps

2 y^{''} (x) - y^{'} (x) - y = 25 e^{- x}

Solve linear ODE:

y = c_{1} e^{x} + c_{2} e^{- \frac{x}{2}} + \frac{25}{2} e^{- x}

y = c_{1} e^{x} + c_{2} e^{- \frac{x}{2}} + \frac{25}{2} e^{- x}

\frac{d ^{2} y}{d x ^{2}} - 2 \frac{d y}{d x} + 5 y = e^{x} sin (2 x)

y^{^{''}} - 4 y = 12 sin (2 t)

y^{^{''}} + 6 y^{^{'}} + 45 y = 6

2 y^{^{''}} + 3 y^{^{'}} + 5 y = 3 cos (5 t)

y^{^{''}} + 25 y = 4 cos (5 x) + 6 sin (5 x)

What is the solution for 2y^{''}-y^'-y=25e^{-x} ?
The solution for 2y^{''}-y^'-y=25e^{-x} is y=c_{1}e^x+c_{2}e^{-x/2}+25/2 e^{-x}