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

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

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y^{''}+y=3xsin(2x)

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

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

Solution steps

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

Solve linear ODE:

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

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

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

y^{^{''}} + 5 y = 4, y (0) = 0, y^{^{'}} (0) = 2

y^{^{''}} - y^{^{'}} - 20 y = - 56 t + 40 t^{2}

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

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

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