5x^{''}+5x^'+6x=0,x^'(0)=0.45,x(0)=-0.3

Lösung

5 x^{^{''}} + 5 x^{^{'}} + 6 x = 0, x^{^{'}} (0) = 0.45, x (0) = - 0.3

x = e^{- \frac{t}{2}} (- 0.3 cos (\frac{19 t}{2 5}) + 0.30779 \dots sin (\frac{19 t}{2 5}))

Schritte zur Lösung

5 x^{''} (t) + 5 x^{'} (t) + 6 x = 0

Solve linear ODE:

x = e^{- \frac{t}{2}} (- 0.3 cos (\frac{19 t}{2 5}) + 0.30779 \dots sin (\frac{19 t}{2 5}))

x = e^{- \frac{t}{2}} (- 0.3 cos (\frac{19 t}{2 5}) + 0.30779 \dots sin (\frac{19 t}{2 5}))

\int \frac{cos ( x )}{x} d x

\frac{\partial}{\partial x} (x z^{2})

\int \frac{5}{6 + 5 x} d x

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

(x, y) \to (0, 0) lim (x y^{2})