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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} - 5 y^{^{'}} = 25 (H (t - 2) - H (t - 4))

x^{^{''}} - 3 x^{^{'}} - 15.25 x = e^{t}

(D^{2} + 3 D + 2) y = sin (e^{x})

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

y^{^{''}} + 1000 y^{^{'}} + 2500 y = 30

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