What is the solution for y^{''}+4y=3x^2 ?

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

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y^{''}+4y=3x^2

y^{^{''}} + 4 y = 3 x^{2}

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

Solution steps

y^{''} (x) + 4 y = 3 x^{2}

Solve linear ODE:

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

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

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

10 y^{^{''}} + 1000 y = 160 cos (8 t), y (0) = 0

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

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

y^{^{''}} + y = 9 e^{- 7 x}, y (0) = 0, y^{^{'}} (0) = 0

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