What is the solution for x^{''}+4x=12*cos(t) ?

The solution for x^{''}+4x=12*cos(t) is x=c_{1}cos(2t)+c_{2}sin(2t)+4cos(t)

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x^{''}+4x=12*cos(t)

x^{^{''}} + 4 x = 12 \cdot cos (t)

x = c_{1} cos (2 t) + c_{2} sin (2 t) + 4 cos (t)

Solution steps

x^{''} (t) + 4 x = 12 \cdot cos (t)

Solve linear ODE:

x = c_{1} cos (2 t) + c_{2} sin (2 t) + 4 cos (t)

x = c_{1} cos (2 t) + c_{2} sin (2 t) + 4 cos (t)

(D^{3} - 2 D^{2} - 5 D + 6) y = e^{4 x}

10 y^{^{''}} + 20 y^{^{'}} + 6260 y = 100

y^{^{''}} + 2 y^{^{'}} - 8 y = 24 e^{2 t}

\frac{d ^{2} y}{d x ^{2}} - 5 \frac{d y}{d x} + 6 y = x^{3} + x

y^{^{''}} + 10 y^{^{'}} + 16 = 0

What is the solution for x^{''}+4x=12*cos(t) ?
The solution for x^{''}+4x=12*cos(t) is x=c_{1}cos(2t)+c_{2}sin(2t)+4cos(t)