What is the solution for (d^2h)/(dt^2)=-kh ?

The solution for (d^2h)/(dt^2)=-kh is h=c_{1}cos(sqrt(k)t)+c_{2}sin(sqrt(k)t)

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(d^2h)/(dt^2)=-kh

\frac{d ^{2} h}{d t ^{2}} = - kh

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

Solution steps

\frac{d ^{2} h}{d t ^{2}} = - kh

Solve linear ODE:

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

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

x \to 2 lim (2 x^{2} - 3 x + 1)

\int \frac{x ^{2}}{4} - 1 d x

y^{^{''}} + 36 y = - 36 sec (6 t)

\frac{d}{d x} (e^{x} + 1)

(arctan (x))^{2}

What is the solution for (d^2h)/(dt^2)=-kh ?
The solution for (d^2h)/(dt^2)=-kh is h=c_{1}cos(sqrt(k)t)+c_{2}sin(sqrt(k)t)