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

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

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y^{''}+8y=8x

y^{^{''}} + 8 y = 8 x

y = c_{1} cos (2^{1 + \frac{1}{2}} x) + c_{2} sin (2^{1 + \frac{1}{2}} x) + x

Solution steps

y^{''} (x) + 8 y = 8 x

Solve linear ODE:

y = c_{1} cos (2^{1 + \frac{1}{2}} x) + c_{2} sin (2^{1 + \frac{1}{2}} x) + x

y = c_{1} cos (2^{1 + \frac{1}{2}} x) + c_{2} sin (2^{1 + \frac{1}{2}} x) + x

y^{^{'''}} + 12 y^{^{''}} + 48 y^{^{'}} + 64 y = 8 e^{- 4 x}

y^{^{''}} + y = 8 e^{- 6 x}, y (0) = 0, y^{^{'}} (0) = 0

y^{^{''}} - 6 y^{^{'}} - 6 y = 2 x e^{- x}

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

y^{^{''}} + y^{^{'}} - 1 = 0

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