x^{''}+2x=0

Lösung

x^{^{''}} + 2 x = 0

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

Schritte zur Lösung

x^{''} (t) + 2 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)

d er i v a t i v e y = x^{3} - 3 x + 1

\frac{\partial}{\partial x} (sin (x) + x^{2} e^{y} - 1)

d er i v a t i v e f (x) = \frac{ln ( x )}{x ^{3}}

x \to 0.01 lim (a x^{2} - x + b)

\int cos (x) (9 + 5 sin^{2} (x)) d x