What is the solution for y^{''}+4y^'+3y=e^{5x} ?

The solution for y^{''}+4y^'+3y=e^{5x} is y=c_{1}e^{-x}+c_{2}e^{-3x}+1/48 e^{5x}

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y^{''}+4y^'+3y=e^{5x}

y^{^{''}} + 4 y^{^{'}} + 3 y = e^{5 x}

y = c_{1} e^{- x} + c_{2} e^{- 3 x} + \frac{1}{48} e^{5 x}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{- x} + c_{2} e^{- 3 x} + \frac{1}{48} e^{5 x}

y = c_{1} e^{- x} + c_{2} e^{- 3 x} + \frac{1}{48} e^{5 x}

y^{^{'''}} - 6 y^{^{''}} + 11 y^{^{'}} - 6 y + x^{2} = - 1

x^{^{''}} + 5 x^{^{'}} + 4 x = 12.5 cos (4 t)

x^{^{''}} - 4 x^{^{'}} + 4 x = 4 e^{2 t}

y^{^{''}} - 4 y = 2 e^{4 t}

y^{^{''}} - 6 y^{^{'}} + 9 y = 25 e^{t} sin (t)

What is the solution for y^{''}+4y^'+3y=e^{5x} ?
The solution for y^{''}+4y^'+3y=e^{5x} is y=c_{1}e^{-x}+c_{2}e^{-3x}+1/48 e^{5x}