8x^{''}+8x^'+1/4 x=0,x^'(0)=1,x(0)=2

Lösung

8 x^{^{''}} + 8 x^{^{'}} + \frac{1}{4} x = 0, x^{^{'}} (0) = 1, x (0) = 2

x = - \frac{( 56 - ( 4 14 - 7 ) ( 4 + 14 ) ) ( 14 + 4 ) e ^{\frac{( - 4 + 14 ) t}{8}}}{14} - \frac{4 14 - 7}{7} e^{- \frac{( 4 + 14 ) t}{8}}

Schritte zur Lösung

8 x^{''} (t) + 8 x^{'} (t) + \frac{1}{4} x = 0

Solve linear ODE:

x = - \frac{( 56 - ( 4 14 - 7 ) ( 4 + 14 ) ) ( 14 + 4 ) e ^{\frac{( - 4 + 14 ) t}{8}}}{14} - \frac{4 14 - 7}{7} e^{- \frac{( 4 + 14 ) t}{8}}

x = - \frac{( 56 - ( 4 14 - 7 ) ( 4 + 14 ) ) ( 14 + 4 ) e ^{\frac{( - 4 + 14 ) t}{8}}}{14} - \frac{4 14 - 7}{7} e^{- \frac{( 4 + 14 ) t}{8}}

\int u^{\frac{3}{2}} d u

\int x cos^{2} (x) d x

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

t \to 0 + lim (t (ln (t))^{2})

d er i v a t i v e 3 cos^{2} (x)