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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} + 9 y = cos (8 x)

(d^{2} - 2 d + 1) y = x^{2} e^{2} x

y^{^{''}} - 6 y^{^{'}} + 9 y = 18 x + 5

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

y^{^{''}} - y^{^{'}} = cos (2 x)

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