What is the solution for (d^2+4)y=cos(2x)+cos(4x) ?

The solution for (d^2+4)y=cos(2x)+cos(4x) is y=c_{1}cos(2x)+c_{2}sin(2x)+(xsin(2x))/4-1/12 cos(4x)

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(d^2+4)y=cos(2x)+cos(4x)

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

y = c_{1} cos (2 x) + c_{2} sin (2 x) + \frac{x sin ( 2 x )}{4} - \frac{1}{12} cos (4 x)

Solution steps

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

Solve linear ODE:

y = c_{1} cos (2 x) + c_{2} sin (2 x) + \frac{x sin ( 2 x )}{4} - \frac{1}{12} cos (4 x)

y = c_{1} cos (2 x) + c_{2} sin (2 x) + \frac{x sin ( 2 x )}{4} - \frac{1}{12} cos (4 x)

y^{^{'''}} - y^{^{''}} - 25 y^{^{'}} + 25 y = 9 - e^{x} + e^{5 x}

y^{^{''}} + 4 y^{^{'}} + 13 y = t e^{- t}, y (0) = 0, y^{^{'}} (0) = 4

D (T) = - 2

y^{^{''}} + 9 y = e^{t}, y (0) = 0

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

What is the solution for (d^2+4)y=cos(2x)+cos(4x) ?
The solution for (d^2+4)y=cos(2x)+cos(4x) is y=c_{1}cos(2x)+c_{2}sin(2x)+(xsin(2x))/4-1/12 cos(4x)