What is the solution for y^{''}-4y^'-5y=e^x+2x ?

The solution for y^{''}-4y^'-5y=e^x+2x is y=c_{1}e^{5x}+c_{2}e^{-x}-1/8 e^x-(2x)/5+8/25

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y^{''}-4y^'-5y=e^x+2x

y^{^{''}} - 4 y^{^{'}} - 5 y = e^{x} + 2 x

y = c_{1} e^{5 x} + c_{2} e^{- x} - \frac{1}{8} e^{x} - \frac{2 x}{5} + \frac{8}{25}

Solution steps

y^{''} (x) - 4 y^{'} (x) - 5 y = e^{x} + 2 x

Solve linear ODE:

y = c_{1} e^{5 x} + c_{2} e^{- x} - \frac{1}{8} e^{x} - \frac{2 x}{5} + \frac{8}{25}

y = c_{1} e^{5 x} + c_{2} e^{- x} - \frac{1}{8} e^{x} - \frac{2 x}{5} + \frac{8}{25}

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

y^{^{''}} - 2 y^{^{'}} + 2 y = cos (t), y (0) = 1, y^{^{'}} (0) = 0

y^{^{'''}} + 2 y^{^{''}} = 24 x^{2} + 24 x + 4

y^{^{''}} + 5 y^{^{'}} + 6 y = 8 e^{5 t}

y^{^{'''}} + y^{^{''}} + y^{^{'}} + y = e^{- t} + 6 t

What is the solution for y^{''}-4y^'-5y=e^x+2x ?
The solution for y^{''}-4y^'-5y=e^x+2x is y=c_{1}e^{5x}+c_{2}e^{-x}-1/8 e^x-(2x)/5+8/25