What is the solution for y^{''}+3y^'-4y=6e^x ?

The solution for y^{''}+3y^'-4y=6e^x is y=c_{1}e^x+c_{2}e^{-4x}+(6e^xx)/5

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y^{''}+3y^'-4y=6e^x

y^{^{''}} + 3 y^{^{'}} - 4 y = 6 e^{x}

y = c_{1} e^{x} + c_{2} e^{- 4 x} + \frac{6 e ^{x} x}{5}

Solution steps

y^{''} (x) + 3 y^{'} (x) - 4 y = 6 e^{x}

Solve linear ODE:

y = c_{1} e^{x} + c_{2} e^{- 4 x} + \frac{6 e ^{x} x}{5}

y = c_{1} e^{x} + c_{2} e^{- 4 x} + \frac{6 e ^{x} x}{5}

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

y^{^{''}} - 3 y^{^{'}} - 4 y = 10 e^{t} - 250 cos (3 t)

y^{^{''}} + 4 y = H (x - π) - H (x - 2 π)

y^{^{''}} + 25 y = 60 cos (5 t)

y^{^{''}} + 2 y^{^{'}} + y = 1 - H (t - 1)

What is the solution for y^{''}+3y^'-4y=6e^x ?
The solution for y^{''}+3y^'-4y=6e^x is y=c_{1}e^x+c_{2}e^{-4x}+(6e^xx)/5