What is the solution for y^{''}+16y=6sin(4x) ?

The solution for y^{''}+16y=6sin(4x) is y=c_{1}cos(4x)+c_{2}sin(4x)-(3xcos(4x))/4

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y^{''}+16y=6sin(4x)

y^{^{''}} + 16 y = 6 sin (4 x)

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

Solution steps

y^{''} (x) + 16 y = 6 sin (4 x)

Solve linear ODE:

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

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

y^{^{''}} + y = sin (x), y (0) = 0

y^{^{'''}} - y^{^{''}} + y^{^{'}} - y = x e^{x} - e^{x} + 7

(D^{2} - D - 2) y = 36 x e^{2 x}

y^{^{''}} - 2 y^{^{'}} - 8 y = 2 sin (x) + (12 x - 10) e^{4 x}

\frac{d ^{2} y}{d x ^{2}} - 3 \frac{d y}{d x} - 4 y = 2 sin (x)

What is the solution for y^{''}+16y=6sin(4x) ?
The solution for y^{''}+16y=6sin(4x) is y=c_{1}cos(4x)+c_{2}sin(4x)-(3xcos(4x))/4