What is the solution for y^{''}+4y=4cos(2x)+6cos(x) ?

The solution for y^{''}+4y=4cos(2x)+6cos(x) is y=c_{1}cos(2x)+c_{2}sin(2x)+xsin(2x)+2cos(x)

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y^{''}+4y=4cos(2x)+6cos(x)

y^{^{''}} + 4 y = 4 cos (2 x) + 6 cos (x)

y = c_{1} cos (2 x) + c_{2} sin (2 x) + x sin (2 x) + 2 cos (x)

Solution steps

y^{''} (x) + 4 y = 4 cos (2 x) + 6 cos (x)

Solve linear ODE:

y = c_{1} cos (2 x) + c_{2} sin (2 x) + x sin (2 x) + 2 cos (x)

y = c_{1} cos (2 x) + c_{2} sin (2 x) + x sin (2 x) + 2 cos (x)

y^{^{''}} + 5 y^{^{'}} + 4 y = sec (x)

y^{^{''}} + 4 y = cos (t), y^{^{'}} (0) = 1, y (0) = 1

y^{^{''}} - 4 y = (x + 1) \cdot e^{x}

y^{^{''}} - 4 y = 16 x^{3}, y (0) = 2, y^{^{'}} (0) = - 14

\frac{d ^{2} y}{d x ^{2}} + \frac{d y}{d x} = tan (2 x)

What is the solution for y^{''}+4y=4cos(2x)+6cos(x) ?
The solution for y^{''}+4y=4cos(2x)+6cos(x) is y=c_{1}cos(2x)+c_{2}sin(2x)+xsin(2x)+2cos(x)