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

The solution for y^{''}-3y^'-4y=3e^2x is y=c_{1}e^{4x}+c_{2}e^{-x}-(3e^2x)/4+(9e^2)/(16)

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

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

y = c_{1} e^{4 x} + c_{2} e^{- x} - \frac{3 e ^{2} x}{4} + \frac{9 e ^{2}}{16}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{4 x} + c_{2} e^{- x} - \frac{3 e ^{2} x}{4} + \frac{9 e ^{2}}{16}

y = c_{1} e^{4 x} + c_{2} e^{- x} - \frac{3 e ^{2} x}{4} + \frac{9 e ^{2}}{16}

y^{^{''}} - 4 y^{^{'}} + 5 y = \frac{2 e ^{2 x}}{sin ( x )}

3 y^{^{''}} - 6 y^{^{'}} + 6 y = e^{x} \cdot sec (x)

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

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

(D^{2} - 3 D + 2) y = 72 x e^{- x}

What is the solution for y^{''}-3y^'-4y=3e^2x ?
The solution for y^{''}-3y^'-4y=3e^2x is y=c_{1}e^{4x}+c_{2}e^{-x}-(3e^2x)/4+(9e^2)/(16)