What is the solution for y^{''}+4y=8x^2+7x ?

The solution for y^{''}+4y=8x^2+7x is y=c_{1}cos(2x)+c_{2}sin(2x)+2x^2+(7x)/4-1

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y^{''}+4y=8x^2+7x

y^{^{''}} + 4 y = 8 x^{2} + 7 x

y = c_{1} cos (2 x) + c_{2} sin (2 x) + 2 x^{2} + \frac{7 x}{4} - 1

Solution steps

y^{''} (x) + 4 y = 8 x^{2} + 7 x

Solve linear ODE:

y = c_{1} cos (2 x) + c_{2} sin (2 x) + 2 x^{2} + \frac{7 x}{4} - 1

y = c_{1} cos (2 x) + c_{2} sin (2 x) + 2 x^{2} + \frac{7 x}{4} - 1

y^{^{''}} - 4 y^{^{'}} = x

10 y^{^{''}} + 90 y^{^{'}} + \frac{1}{0.005} y = 500 sin (x)

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

y^{^{''}} + 3 y^{^{'}} = 9

y^{^{''}} - 4 y^{^{'}} = e^{4 x}

What is the solution for y^{''}+4y=8x^2+7x ?
The solution for y^{''}+4y=8x^2+7x is y=c_{1}cos(2x)+c_{2}sin(2x)+2x^2+(7x)/4-1