What is the solution for x^{''}+23.04x=2cos(5t) ?

The solution for x^{''}+23.04x=2cos(5t) is x=c_{1}cos((24t)/5)+c_{2}sin((24t)/5)-50/49 cos(5t)

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x^{''}+23.04x=2cos(5t)

x^{^{''}} + 23.04 x = 2 cos (5 t)

x = c_{1} cos (\frac{24 t}{5}) + c_{2} sin (\frac{24 t}{5}) - \frac{50}{49} cos (5 t)

Solution steps

x^{''} (t) + 23.04 x = 2 cos (5 t)

Solve linear ODE:

x = c_{1} cos (\frac{24 t}{5}) + c_{2} sin (\frac{24 t}{5}) - \frac{50}{49} cos (5 t)

x = c_{1} cos (\frac{24 t}{5}) + c_{2} sin (\frac{24 t}{5}) - \frac{50}{49} cos (5 t)

y^{^{''}} + 4 y = 9, t y (0) = 0, y^{^{'}} (0) = 7

y^{^{''}} + 8 y^{^{'}} + 12 y = 1 - H (t)

y^{^{'''}} - 11 y^{^{''}} + 28 y^{^{'}} = 36 e^{x}, y (0) = 25

y^{^{''}} - 2 y^{^{'}} + 5 y = e x sin (x)

(D^{3} + D^{2} - 4 D - 4) y = e^{2 x} sin (3 x)

What is the solution for x^{''}+23.04x=2cos(5t) ?
The solution for x^{''}+23.04x=2cos(5t) is x=c_{1}cos((24t)/5)+c_{2}sin((24t)/5)-50/49 cos(5t)