What is the solution for (d^2x)/(dt^2)+4x=0 ?

The solution for (d^2x)/(dt^2)+4x=0 is x=c_{1}cos(2t)+c_{2}sin(2t)

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(d^2x)/(dt^2)+4x=0

\frac{d ^{2} x}{d t ^{2}} + 4 x = 0

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

Solution steps

\frac{d ^{2} x}{d t ^{2}} + 4 x = 0

Solve linear ODE:

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

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

x^{^{''}} + 16 x = 0, x (0) = 0, x^{^{'}} (0) = 5

y^{^{''}} + 12 y^{^{'}} + 36 y = 0, y (0) = 3, y^{^{'}} (0) = 0

y^{^{''''}} - 42 y^{^{'''}} + 441 y^{^{''}} = 0

{\frac{d ^{2} y}{d x ^{2}} + 4 \frac{d y}{d x} + 20 y}^{2} = 0

3 y^{^{'''}} + 17 y^{^{''}} + 31 y^{^{'}} + 14 y = 0

What is the solution for (d^2x)/(dt^2)+4x=0 ?
The solution for (d^2x)/(dt^2)+4x=0 is x=c_{1}cos(2t)+c_{2}sin(2t)