What is the solution for y^{''}+16y=2cos(4x) ?

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

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y^{''}+16y=2cos(4x)

y^{^{''}} + 16 y = 2 cos (4 x)

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

Solution steps

y^{''} (x) + 16 y = 2 cos (4 x)

Solve linear ODE:

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

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

y^{^{''}} - 10 y^{^{'}} + 25 y = 48 e^{x}

y^{^{''}} - y^{^{'}} - 2 y = 1 - x, y (0) = 0, y^{^{'}} (0) = 0

y^{^{''}} - 2 y^{^{''}} + y = 4 e^{x} + 4 e^{- x}

2 y^{^{''}} + 4 y = 6 t, y (0) = 0, y^{^{'}} (0) = 0

y^{^{''}} + 4 y = t^{2} + 6 e^{t}, y (0) = 0, y^{^{'}} (0) = 0

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