What is the solution for x^{''}+8x=16cos(4t) ?

The solution for x^{''}+8x=16cos(4t) is x=c_{1}cos(2^{1+1/2}t)+c_{2}sin(2^{1+1/2}t)-2cos(4t)

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x^{''}+8x=16cos(4t)

x^{^{''}} + 8 x = 16 cos (4 t)

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

Solution steps

x^{''} (t) + 8 x = 16 cos (4 t)

Solve linear ODE:

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

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

What is the solution for x^{''}+8x=16cos(4t) ?
The solution for x^{''}+8x=16cos(4t) is x=c_{1}cos(2^{1+1/2}t)+c_{2}sin(2^{1+1/2}t)-2cos(4t)