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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} + 4 y = x^{2} + 1

\frac{d ^{2} y}{d x ^{2}} - 7 \frac{d y}{d x} + 6 y = e^{x}

y^{^{''}} + y = 12 cos^{2} (x)

y^{^{''}} + y^{^{'}} - 2 y = t^{2}

x^{^{''}} + 3 x^{^{'}} + 33 x = 10 cos (5 t)

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