6y^{''}+40y^'+54y=0

Lösung

6 y^{^{''}} + 40 y^{^{'}} + 54 y = 0

y = c_{1} e^{\frac{( - 10 + 19 ) t}{3}} + c_{2} e^{- \frac{( 10 + 19 ) t}{3}}

Schritte zur Lösung

6 y^{''} (t) + 40 y^{'} (t) + 54 y = 0

Solve linear ODE:

y = c_{1} e^{\frac{( - 10 + 19 ) t}{3}} + c_{2} e^{- \frac{( 10 + 19 ) t}{3}}

y = c_{1} e^{\frac{( - 10 + 19 ) t}{3}} + c_{2} e^{- \frac{( 10 + 19 ) t}{3}}

\frac{d}{d x} (- k x)

\int cos (x) - x sin (x) d x

d er i v a t i v e e^{- x^{62}}

y^{^{'}} = x - 2 y

s l o p e (1, 4), (- 5, 2)