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

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

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y^{''}+4y=xcos(x)

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

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

Solution steps

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

Solve linear ODE:

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

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

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

y^{^{''}} + y^{^{'}} - 6 y = 6 x^{3} - 3 x^{2} + 12 x

y^{^{''}} + 6 y^{^{'}} + 12 y = 763 e^{2 t} cos (7 t)

y^{^{''}} + y = \frac{2}{cos ( x )}, y (0) = 2, y^{^{'}} (0) = 2

y^{^{''}} + y = 4 x + 10 sin (x)

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