What is the solution for y^{''}+4y=3*sin(2x) ?

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

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y^{''}+4y=3*sin(2x)

y^{^{''}} + 4 y = 3 \cdot sin (2 x)

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

Solution steps

y^{''} (x) + 4 y = 3 \cdot sin (2 x)

Solve linear ODE:

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

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

y^{^{''}} - y = 5 x + 2

y^{^{''}} + 2 y^{^{'}} - 3 y = e^{2 x} - 2 x

\frac{d ^{3} y}{d x ^{3}} + y = cos (x)

y^{^{'''}} - y = e^{x} - x + 16

y^{^{'''}} - 5 y^{^{''}} + 9 y = t cos (2 t)

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