What is the solution for y^{''}+4y=3x^3 ?

The solution for y^{''}+4y=3x^3 is y=c_{1}cos(2x)+c_{2}sin(2x)+(3x^3)/4-(9x)/8

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y^{''}+4y=3x^3

y^{^{''}} + 4 y = 3 x^{3}

y = c_{1} cos (2 x) + c_{2} sin (2 x) + \frac{3 x ^{3}}{4} - \frac{9 x}{8}

Solution steps

y^{''} (x) + 4 y = 3 x^{3}

Solve linear ODE:

y = c_{1} cos (2 x) + c_{2} sin (2 x) + \frac{3 x ^{3}}{4} - \frac{9 x}{8}

y = c_{1} cos (2 x) + c_{2} sin (2 x) + \frac{3 x ^{3}}{4} - \frac{9 x}{8}

x^{^{''}} + 2 x^{^{'}} + x = e^{- t}

y^{^{''}} - 2 y^{^{'}} - 3 y = cos (2 t)

y^{^{''}} - 5 y^{^{'}} + 6 y = 5 e^{2 x}

y^{^{''}} - 4 y^{^{'}} + 4 y = 2 e^{2 t}

y^{^{'''}} - 5 y^{^{''}} + 6 y^{^{'}} = 0 + 6 sin (x)

What is the solution for y^{''}+4y=3x^3 ?
The solution for y^{''}+4y=3x^3 is y=c_{1}cos(2x)+c_{2}sin(2x)+(3x^3)/4-(9x)/8