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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} + 4 y^{^{'}} + 4 y = x^{2} - 2 x

y^{^{''}} + 25 y = 11 sec^{2} (5 t)

y^{^{''}} - 2 y^{^{'}} + 5 y = 3 e^{x} sin (x) + 24 x + 4

y^{^{'''}} - y^{^{'}} = 7 sin (t)

(D^{4} + D^{2}) y = 18 x - 4 sin (x)

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