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

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

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

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

y = c_{1} cos (2 x) + c_{2} sin (2 x) + 2 x^{2} - 1

Solution steps

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

Solve linear ODE:

y = c_{1} cos (2 x) + c_{2} sin (2 x) + 2 x^{2} - 1

y = c_{1} cos (2 x) + c_{2} sin (2 x) + 2 x^{2} - 1

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

\int (7 x^{2}) e x^{3} + 1 d x

f (x) = - 4 sin (4 x)

(- 4, - 2), (5, - 7)

y = x + \frac{6}{x}, (3, 5)

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