What is the solution for y^{''}+y=2e^{-x}+8x ?

The solution for y^{''}+y=2e^{-x}+8x is y=c_{1}cos(x)+c_{2}sin(x)+e^{-x}+8x

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y^{''}+y=2e^{-x}+8x

y^{^{''}} + y = 2 e^{- x} + 8 x

y = c_{1} cos (x) + c_{2} sin (x) + e^{- x} + 8 x

Solution steps

y^{''} (x) + y = 2 e^{- x} + 8 x

Solve linear ODE:

y = c_{1} cos (x) + c_{2} sin (x) + e^{- x} + 8 x

y = c_{1} cos (x) + c_{2} sin (x) + e^{- x} + 8 x

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

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

y^{^{''}} + 17 y^{^{'}} + 72 y = 4 x e^{2 x}

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

y^{^{''}} + 3 y^{^{'}} - 10 y = 3 x e^{- 2 x}

What is the solution for y^{''}+y=2e^{-x}+8x ?
The solution for y^{''}+y=2e^{-x}+8x is y=c_{1}cos(x)+c_{2}sin(x)+e^{-x}+8x