What is the solution for y^{''}+4y=xcos(2x) ?

The solution for y^{''}+4y=xcos(2x) is y=c_{1}cos(2x)+c_{2}sin(2x)+(x^2sin(2x))/8+(xcos(2x))/(16)

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

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

y = c_{1} cos (2 x) + c_{2} sin (2 x) + \frac{x ^{2} sin ( 2 x )}{8} + \frac{x cos ( 2 x )}{16}

Solution steps

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

Solve linear ODE:

y = c_{1} cos (2 x) + c_{2} sin (2 x) + \frac{x ^{2} sin ( 2 x )}{8} + \frac{x cos ( 2 x )}{16}

y = c_{1} cos (2 x) + c_{2} sin (2 x) + \frac{x ^{2} sin ( 2 x )}{8} + \frac{x cos ( 2 x )}{16}

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

y^{^{''}} + 2 y^{^{'}} - 5 y = - 2

4 y^{^{''}} + 16 y = 1, y (0) = - 2, y^{^{'}} (0) = 0

y^{^{''}} + y = t^{3}

y^{^{''}} - 8 y^{^{'}} + 16 y = x

What is the solution for y^{''}+4y=xcos(2x) ?
The solution for y^{''}+4y=xcos(2x) is y=c_{1}cos(2x)+c_{2}sin(2x)+(x^2sin(2x))/8+(xcos(2x))/(16)