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

The solution for y^{''}+y=xsin(2x) is y=c_{1}cos(x)+c_{2}sin(x)-(xsin(2x))/3-4/9 cos(2x)

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

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

y = c_{1} cos (x) + c_{2} sin (x) - \frac{x sin ( 2 x )}{3} - \frac{4}{9} cos (2 x)

Solution steps

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

Solve linear ODE:

y = c_{1} cos (x) + c_{2} sin (x) - \frac{x sin ( 2 x )}{3} - \frac{4}{9} cos (2 x)

y = c_{1} cos (x) + c_{2} sin (x) - \frac{x sin ( 2 x )}{3} - \frac{4}{9} cos (2 x)

y^{^{''}} + 4 y = 3 cos (2.1 x)

y^{^{''}} - 2 y^{^{'}} + 2 y = e^{x}, y (\frac{π}{2}) = 0, y (0) = 0

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

y^{^{''}} + 16 y = 6 sin (4 t), y (0) = 2, y^{^{'}} (0) = 0

y^{^{''}} - 5 y^{^{'}} + 6 y = 6 x + 2 - 3 e^{2}

What is the solution for y^{''}+y=xsin(2x) ?
The solution for y^{''}+y=xsin(2x) is y=c_{1}cos(x)+c_{2}sin(x)-(xsin(2x))/3-4/9 cos(2x)