What is the solution for (d^2y)/(dx^2)+y=xcos(2x) ?

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

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(d^2y)/(dx^2)+y=xcos(2x)

\frac{d ^{2} y}{d x ^{2}} + 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

\frac{d ^{2} y}{d x ^{2}} + 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^{^{''}} + 2 y^{^{'}} = x^{2} + 3 x^{2} e^{- 2 x} + e^{2 x} + 1

y^{^{''}} - 8 y^{^{'}} + 12 y = e^{2 x} (x^{2} - 1)

y^{^{''}} + y^{^{'}} - 12 y = 2 t e - 4 t, y (0) = - 5, y^{^{'}} (0) = - 4

y^{^{''}} + 25 y = 5 tan^{2} (5 x) cos (5 x)

y^{^{''}} + y = 2 sin (2 t)

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