What is the solution for x^{''}-2x^'-3x=e^{4t} ?

The solution for x^{''}-2x^'-3x=e^{4t} is x=c_{1}e^{3t}+c_{2}e^{-t}+1/5 e^{4t}

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x^{''}-2x^'-3x=e^{4t}

x^{^{''}} - 2 x^{^{'}} - 3 x = e^{4 t}

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

Solution steps

x^{''} (t) - 2 x^{'} (t) - 3 x = e^{4 t}

Solve linear ODE:

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

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

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

y^{^{'''}} - 6 y^{^{''}} = x^{2}

y^{^{'''}} - 6 y^{^{'}} = 3 - cos (x)

y^{^{'''}} - 3 y^{^{''}} = 9 e^{x}

y^{^{''}} + 2 y^{^{'}} + 2 y = e^{- t} sin (3 t)

What is the solution for x^{''}-2x^'-3x=e^{4t} ?
The solution for x^{''}-2x^'-3x=e^{4t} is x=c_{1}e^{3t}+c_{2}e^{-t}+1/5 e^{4t}