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

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

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

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

y = c_{1} e^{x} + c_{2} e^{- 4 x} + \frac{3 e ^{x} x}{5}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{x} + c_{2} e^{- 4 x} + \frac{3 e ^{x} x}{5}

y = c_{1} e^{x} + c_{2} e^{- 4 x} + \frac{3 e ^{x} x}{5}

y^{^{''}} - y^{^{'}} + y = 12 sin (t) - 4 e^{2 t}

y^{^{''}} - y^{^{'}} - 2 y = 1 + t

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

y^{^{''}} + y^{^{'}} - 2 y = e^{- 2 t} + t

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

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