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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{'''}} + 2 y^{^{''}} + y^{^{'}} = 10

y^{^{''''}} + 49 y^{^{''}} = sin (7 t) + t e^{8 t} + 4

y^{^{''}} = y^{^{'}} + x

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

y^{^{''}} + y = 3 sec (t)

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