What is the solution for y^{''}-4y=8x^3,y(0)=0,y^'(0)=1 ?

The solution for y^{''}-4y=8x^3,y(0)=0,y^'(0)=1 is y=e^{2x}-e^{-2x}-2x^3-3x

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y^{''}-4y=8x^3,y(0)=0,y^'(0)=1

y^{^{''}} - 4 y = 8 x^{3}, y (0) = 0, y^{^{'}} (0) = 1

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

Solution steps

y^{''} (x) - 4 y = 8 x^{3}

Solve linear ODE:

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

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

y^{^{''}} + a y^{^{'}} + b y = sin (x) + 1

y^{^{''}} - y^{^{'}} - 6 y = 5 cos (3 x)

y^{^{''}} - 18 y^{^{'}} + 81 y = t^{9} e^{9 t}

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

y^{^{''}} - 4 y^{^{'}} + 4 y = e^{2 x} \cdot ln (x)

What is the solution for y^{''}-4y=8x^3,y(0)=0,y^'(0)=1 ?
The solution for y^{''}-4y=8x^3,y(0)=0,y^'(0)=1 is y=e^{2x}-e^{-2x}-2x^3-3x