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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{'''}} - 5 y^{^{''}} + 3 y^{^{'}} + 9 = 0

y^{^{''}} - 2 y^{^{'}} + y = - \frac{1}{x ^{2}} e^{x}

y^{^{''}} + \frac{1}{8} y^{^{'}} + 4 y = 3 cos (6 t)

y^{^{''}} + 7 y^{^{'}} + 12 y = 21 e^{3 t}

y^{^{''}} + 3^{2} y = 6 \cdot sin (3 x)

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