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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} + 10 y^{^{'}} + 25 y = - 8 - 5 x + 25 x^{2}

10 x^{^{''}} + 200 x = 600 sin (125 t)

5 y^{^{''}} + 6 y^{^{'}} + y = x^{3} + 18 x^{2} + 32 x + 13

y^{^{''}} - 6 y^{^{'}} + 8 y = 3 + e^{- 2 t}

\frac{d ^{2} y}{d x ^{2}} - y = cosh (x) + x

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