What is the solution for x^{''}+x=9cos(4t) ?

The solution for x^{''}+x=9cos(4t) is x=c_{1}cos(t)+c_{2}sin(t)-3/5 cos(4t)

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x^{''}+x=9cos(4t)

x^{^{''}} + x = 9 cos (4 t)

x = c_{1} cos (t) + c_{2} sin (t) - \frac{3}{5} cos (4 t)

Solution steps

x^{''} (t) + x = 9 cos (4 t)

Solve linear ODE:

x = c_{1} cos (t) + c_{2} sin (t) - \frac{3}{5} cos (4 t)

x = c_{1} cos (t) + c_{2} sin (t) - \frac{3}{5} cos (4 t)

y^{^{''}} - 4 y = 3 e^{2 t}

9 y^{^{''}} + 7 y^{^{'}} - y = 24

y^{^{''}} + 16 y = 8 x + 8 sin (x)

y^{^{'''}} + 6 y^{^{''}} + 12 y^{^{'}} + 8 y = 4 e^{- 2 x}

y^{^{''}} - 3 y^{^{'}} = x^{2} - 1

What is the solution for x^{''}+x=9cos(4t) ?
The solution for x^{''}+x=9cos(4t) is x=c_{1}cos(t)+c_{2}sin(t)-3/5 cos(4t)