y^{''}+5y^'+ay=0,y(0)=1,y^'(0)=7

Lösung

y^{^{''}} + 5 \cdot y^{^{'}} + a \cdot y = 0, y (0) = 1, y^{^{'}} (0) = 7

y = e^{\frac{t ( - 5 + 25 - 4 a )}{2}} (1 + \frac{- - 4 a + 25 + 19}{2 25 - 4 a}) - \frac{e ^{\frac{t ( - 5 - 25 - 4 a )}{2}} ( - - 4 a + 25 + 19 )}{2 25 - 4 a}

Schritte zur Lösung

y^{''} (t) + 5 \cdot y^{'} (t) + a \cdot y = 0

Solve linear ODE:

y = e^{\frac{t ( - 5 + 25 - 4 a )}{2}} (1 + \frac{- - 4 a + 25 + 19}{2 25 - 4 a}) - \frac{e ^{\frac{t ( - 5 - 25 - 4 a )}{2}} ( - - 4 a + 25 + 19 )}{2 25 - 4 a}

y = e^{\frac{t ( - 5 + 25 - 4 a )}{2}} (1 + \frac{- - 4 a + 25 + 19}{2 25 - 4 a}) - \frac{e ^{\frac{t ( - 5 - 25 - 4 a )}{2}} ( - - 4 a + 25 + 19 )}{2 25 - 4 a}

5 y^{^{''}} + c y^{^{'}} + 45 y = 0, y (0) = - 8, y^{^{'}} (0) = 0

(d^{5} - d^{3}) \cdot y = 0

y^{^{''}} = k y (x)

4 \frac{d ^{2} x}{d t ^{2}} + c \frac{d x}{d t} + k (x) = 0

y^{^{''}} + 10 y^{^{'}} + 25 y = 0, y (0) = - 4, y^{^{'}} (0) = 18