What is the solution for y^{''}-y=e^{3x} ?

The solution for y^{''}-y=e^{3x} is y=c_{1}e^x+c_{2}e^{-x}+1/8 e^{3x}

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y^{''}-y=e^{3x}

y^{^{''}} - y = e^{3 x}

y = c_{1} e^{x} + c_{2} e^{- x} + \frac{1}{8} e^{3 x}

Solution steps

y^{''} (x) - y = e^{3 x}

Solve linear ODE:

y = c_{1} e^{x} + c_{2} e^{- x} + \frac{1}{8} e^{3 x}

y = c_{1} e^{x} + c_{2} e^{- x} + \frac{1}{8} e^{3 x}

1 q^{^{''}} + 4 q^{^{'}} + 5 q = 5 cos (40 t)

y^{^{''}} - 2 y^{^{'}} - 3 y = 12 e x - 3

(d^{3} + 10 d^{2} + 25 d) y = e^{x}

y^{^{''''}} + 2 y^{^{''}} + y = 11 + cos (4 t)

2 y^{^{''}} + 3 y^{^{'}} - 2 y = 10 x^{2} - 4 x - 13

What is the solution for y^{''}-y=e^{3x} ?
The solution for y^{''}-y=e^{3x} is y=c_{1}e^x+c_{2}e^{-x}+1/8 e^{3x}