What is the solution for x^{''}+x=cos(3t) ?

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

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x^{''}+x=cos(3t)

x^{^{''}} + x = cos (3 t)

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

Solution steps

x^{''} (t) + x = cos (3 t)

Solve linear ODE:

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

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

y^{^{'''}} - 5 y^{^{''}} + 6 y^{^{'}} = 5 + 4 sin (x)

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

8 y^{^{''}} - 32 y = 16 cos (x)

y^{^{''}} - 6 y^{^{'}} + 5 y = x, y (0) = e^{x}

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

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