x^{''}+4x=0

Lösung

x^{^{''}} + 4 x = 0

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

Schritte zur Lösung

x^{''} (t) + 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)

n = 1 \sum \infty \frac{1}{n + 1 ^{- n}}

5 y^{^{''}} + 15 y^{^{'}} - 20 y = 0

\int (x^{2} - x + 1) d x

\int \frac{6}{x ^{2} x ^{2} + 16} d x

x \to 2 lim (x + 1)