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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

x \to 0 lim (\frac{e ^{x} - 1}{x})

y = 3 x^{2} + 5 x - 1

\frac{\partial}{\partial z} (sin (yz))

\int 12 sin^{2} (x) cos^{3} (x) d x

\frac{d}{d x} (tan (2 x + 4))

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