What is the solution for y^{''}-4y^'-5y=2e^{(3x)} ?

The solution for y^{''}-4y^'-5y=2e^{(3x)} is y=c_{1}e^{5x}+c_{2}e^{-x}-1/4 e^{3x}

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y^{''}-4y^'-5y=2e^{(3x)}

y^{^{''}} - 4 y^{^{'}} - 5 y = 2 e^{(3 x)}

y = c_{1} e^{5 x} + c_{2} e^{- x} - \frac{1}{4} e^{3 x}

Solution steps

y^{''} (x) - 4 y^{'} (x) - 5 y = 2 e^{(3 x)}

Solve linear ODE:

y = c_{1} e^{5 x} + c_{2} e^{- x} - \frac{1}{4} e^{3 x}

y = c_{1} e^{5 x} + c_{2} e^{- x} - \frac{1}{4} e^{3 x}

y^{^{''}} + 5 y^{^{'}} + 6 y = 19 t e^{5 t}

y^{^{''}} - 2 y^{^{'}} + y = 15 x e^{x}

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

y^{^{''}} - 12 y^{^{'}} + 36 y = 5 e^{2} x

a y^{^{''}} + b y^{^{'}} + y = c

What is the solution for y^{''}-4y^'-5y=2e^{(3x)} ?
The solution for y^{''}-4y^'-5y=2e^{(3x)} is y=c_{1}e^{5x}+c_{2}e^{-x}-1/4 e^{3x}