What is the solution for y^{''}-4y=(x+1)*e^x ?

The solution for y^{''}-4y=(x+1)*e^x is y=c_{1}e^{2x}+c_{2}e^{-2x}-(e^xx)/3-5/9 e^x

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y^{''}-4y=(x+1)*e^x

y^{^{''}} - 4 y = (x + 1) \cdot e^{x}

y = c_{1} e^{2 x} + c_{2} e^{- 2 x} - \frac{e ^{x} x}{3} - \frac{5}{9} e^{x}

Solution steps

y^{''} (x) - 4 y = (x + 1) \cdot e^{x}

Solve linear ODE:

y = c_{1} e^{2 x} + c_{2} e^{- 2 x} - \frac{e ^{x} x}{3} - \frac{5}{9} e^{x}

y = c_{1} e^{2 x} + c_{2} e^{- 2 x} - \frac{e ^{x} x}{3} - \frac{5}{9} e^{x}

y^{^{''}} - 4 y = 16 x^{3}, y (0) = 2, y^{^{'}} (0) = - 14

\frac{d ^{2} y}{d x ^{2}} + \frac{d y}{d x} = tan (2 x)

49 y^{^{''}} + 140 y^{^{'}} + 136 = 0

x^{^{''}} + \frac{1}{98} x = 8 cos (6 t)

33 y^{^{''}} - 22 y^{^{'}} - 3 y = 8 - t

What is the solution for y^{''}-4y=(x+1)*e^x ?
The solution for y^{''}-4y=(x+1)*e^x is y=c_{1}e^{2x}+c_{2}e^{-2x}-(e^xx)/3-5/9 e^x