What is the solution for y^{''}-4y=16x^3,y(0)=2,y^'(0)=-14 ?

The solution for y^{''}-4y=16x^3,y(0)=2,y^'(0)=-14 is y=-e^{2x}+3e^{-2x}-4x^3-6x

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

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

y = - e^{2 x} + 3 e^{- 2 x} - 4 x^{3} - 6 x

Solution steps

y^{''} (x) - 4 y = 16 x^{3}

Solve linear ODE:

y = - e^{2 x} + 3 e^{- 2 x} - 4 x^{3} - 6 x

y = - e^{2 x} + 3 e^{- 2 x} - 4 x^{3} - 6 x

\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

y^{^{''}} + y = 2 cos (x), y (0) = 1

What is the solution for y^{''}-4y=16x^3,y(0)=2,y^'(0)=-14 ?
The solution for y^{''}-4y=16x^3,y(0)=2,y^'(0)=-14 is y=-e^{2x}+3e^{-2x}-4x^3-6x