What is the solution for y^{''}+4y=6sin(2x) ?

The solution for y^{''}+4y=6sin(2x) is y=c_{1}cos(2x)+c_{2}sin(2x)-(3xcos(2x))/2

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y^{''}+4y=6sin(2x)

y^{^{''}} + 4 y = 6 sin (2 x)

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

Solution steps

y^{''} (x) + 4 y = 6 sin (2 x)

Solve linear ODE:

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

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

(5 x + 13 y) d x + (13 x + 6 y) d y = 0, y (0) = 2

\frac{d}{d x} (- e^{x} (x + 5))

\int sec (2) (x) d x

\frac{d ^{2}}{d x ^{2}} (5 - x^{2})

\frac{5 x - 1}{x + 5}

What is the solution for y^{''}+4y=6sin(2x) ?
The solution for y^{''}+4y=6sin(2x) is y=c_{1}cos(2x)+c_{2}sin(2x)-(3xcos(2x))/2