What is the solution for 1/2 y^{''}+10y=8cos(6t) ?

The solution for 1/2 y^{''}+10y=8cos(6t) is y=c_{1}cos(2sqrt(5)t)+c_{2}sin(2sqrt(5)t)-cos(6t)

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1/2 y^{''}+10y=8cos(6t)

\frac{1}{2} y^{^{''}} + 10 y = 8 cos (6 t)

y = c_{1} cos (25 t) + c_{2} sin (25 t) - cos (6 t)

Solution steps

\frac{1}{2} y^{''} (t) + 10 y = 8 cos (6 t)

Solve linear ODE:

y = c_{1} cos (25 t) + c_{2} sin (25 t) - cos (6 t)

y = c_{1} cos (25 t) + c_{2} sin (25 t) - cos (6 t)

2 y^{^{''}} + 24 y^{^{'}} + \frac{1}{5} y = 12

0.1 q^{^{''}} + \frac{1}{1 0 ^{- 3}} q = 120 cos (101 t)

y^{^{''}} - 6 y^{^{'}} - 6 y = 9 x

(D^{3} + D) y = sec (x)

y^{^{''}} - 9 y^{^{'}} + 121 y = 11 sin (11 t)

What is the solution for 1/2 y^{''}+10y=8cos(6t) ?
The solution for 1/2 y^{''}+10y=8cos(6t) is y=c_{1}cos(2sqrt(5)t)+c_{2}sin(2sqrt(5)t)-cos(6t)