What is the solution for y^{''}+4y=x^2+5cos(2x) ?

The solution for y^{''}+4y=x^2+5cos(2x) is y=c_{1}cos(2x)+c_{2}sin(2x)+(x^2+5xsin(2x))/4-1/8

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y^{''}+4y=x^2+5cos(2x)

y^{^{''}} + 4 y = x^{2} + 5 cos (2 x)

y = c_{1} cos (2 x) + c_{2} sin (2 x) + \frac{x ^{2} + 5 x sin ( 2 x )}{4} - \frac{1}{8}

Solution steps

y^{''} (x) + 4 y = x^{2} + 5 cos (2 x)

Solve linear ODE:

y = c_{1} cos (2 x) + c_{2} sin (2 x) + \frac{x ^{2} + 5 x sin ( 2 x )}{4} - \frac{1}{8}

y = c_{1} cos (2 x) + c_{2} sin (2 x) + \frac{x ^{2} + 5 x sin ( 2 x )}{4} - \frac{1}{8}

\frac{d ^{2} y}{d x ^{2}} - 3 \frac{d y}{d x} + 2 y = 4 x + 6

y^{^{''}} + y^{^{'}} = 3 x^{2}, y (0) = 0, y^{^{'}} (0) = 1

y^{^{''}} - y = sin^{2} (x)

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

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

What is the solution for y^{''}+4y=x^2+5cos(2x) ?
The solution for y^{''}+4y=x^2+5cos(2x) is y=c_{1}cos(2x)+c_{2}sin(2x)+(x^2+5xsin(2x))/4-1/8