What is the solution for y^{''}+4y=5sin(3x) ?

The solution for y^{''}+4y=5sin(3x) is y=c_{1}cos(2x)+c_{2}sin(2x)-sin(3x)

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y^{''}+4y=5sin(3x)

y^{^{''}} + 4 y = 5 sin (3 x)

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

Solution steps

y^{''} (x) + 4 y = 5 sin (3 x)

Solve linear ODE:

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

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

y^{^{''}} + y - \frac{π}{2} = 0

y^{^{''''}} - 2 y^{^{''}} + y = 100 cos (3 x)

y^{^{''}} - 3 y^{^{'}} - 40 y = 168 e^{2 t}

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

y^{^{''}} + b y = a

What is the solution for y^{''}+4y=5sin(3x) ?
The solution for y^{''}+4y=5sin(3x) is y=c_{1}cos(2x)+c_{2}sin(2x)-sin(3x)