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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

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

y^{^{''}} + 6 y^{^{'}} + 5 y = 36 e^{x}, y (0) = - 2

y^{^{''}} + 3 y^{^{'}} + 2 y = e^{t}, y (0) = 1, y (0) = - 1

y^{^{''}} + y = e^{t}

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

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