What is the solution for y^{''}+3y^'-4y=14e^x ?

The solution for y^{''}+3y^'-4y=14e^x is y=c_{1}e^x+c_{2}e^{-4x}+(14e^xx)/5

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y^{''}+3y^'-4y=14e^x

y^{^{''}} + 3 y^{^{'}} - 4 y = 14 e^{x}

y = c_{1} e^{x} + c_{2} e^{- 4 x} + \frac{14 e ^{x} x}{5}

Solution steps

y^{''} (x) + 3 y^{'} (x) - 4 y = 14 e^{x}

Solve linear ODE:

y = c_{1} e^{x} + c_{2} e^{- 4 x} + \frac{14 e ^{x} x}{5}

y = c_{1} e^{x} + c_{2} e^{- 4 x} + \frac{14 e ^{x} x}{5}

(D^{2} + 4 D + 4) y = x e^{- x}

y^{^{''}} - 2 y^{^{'}} - 3 y = 9 t^{2}

y^{^{''}} - 3 y^{^{'}} - 40 y = (t + 1) e^{t}

2 y^{^{''}} + 5 y^{^{'}} + 2 y = x e^{- x}

y^{^{''}} - 4 y^{^{'}} + 4 y = x^{2} \cdot e^{2 x} - e^{2 x}

What is the solution for y^{''}+3y^'-4y=14e^x ?
The solution for y^{''}+3y^'-4y=14e^x is y=c_{1}e^x+c_{2}e^{-4x}+(14e^xx)/5