What is the solution for x^{''}+9x=5cos(t) ?

The solution for x^{''}+9x=5cos(t) is x=c_{1}cos(3t)+c_{2}sin(3t)+5/8 cos(t)

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x^{''}+9x=5cos(t)

x^{^{''}} + 9 x = 5 cos (t)

x = c_{1} cos (3 t) + c_{2} sin (3 t) + \frac{5}{8} cos (t)

Solution steps

x^{''} (t) + 9 x = 5 cos (t)

Solve linear ODE:

x = c_{1} cos (3 t) + c_{2} sin (3 t) + \frac{5}{8} cos (t)

x = c_{1} cos (3 t) + c_{2} sin (3 t) + \frac{5}{8} cos (t)

y^{^{''}} - 3 y^{^{'}} + 7 y = 3 e^{3 t}

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

y^{^{''}} + 4 y^{^{'}} + 8 y = 24, y^{^{'}} (0) = 0, y (0) = 0

y^{^{''}} + 18 y^{^{'}} - y = t^{- 1} e^{t}

D B = 10 lo g_{10} (\frac{0.1}{1 \cdot 1 0 ^{- 12}})

What is the solution for x^{''}+9x=5cos(t) ?
The solution for x^{''}+9x=5cos(t) is x=c_{1}cos(3t)+c_{2}sin(3t)+5/8 cos(t)