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

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

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

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

y^{^{''}} - 2 y^{^{'}} + y = e^{x} (x^{2} - x + 1)

y^{^{''}} - 2 y^{^{'}} + y = e^{x} \frac{1}{x}

x^{^{''}} + 4 x^{^{'}} + 40 x = 4 e^{- 2 t} cos (6 t)

9 y^{^{''}} + 30 y^{^{'}} + 25 y = 2 x + 5

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