2y^{''}+y^'-10y=0,y(1)=5,y^'(1)=2

Lösung

2 y^{^{''}} + y^{^{'}} - 10 y = 0, y (1) = 5, y^{^{'}} (1) = 2

y = e^{\frac{4 t - 9}{2}} (5 e^{\frac{5}{2}} - \frac{16 e ^{\frac{5}{2}}}{9}) + \frac{16 e ^{\frac{5}{2}}}{9} e^{- \frac{5 t}{2}}

Schritte zur Lösung

2 y^{''} (t) + y^{'} (t) - 10 y = 0

Solve linear ODE:

y = e^{\frac{4 t - 9}{2}} (5 e^{\frac{5}{2}} - \frac{16 e ^{\frac{5}{2}}}{9}) + \frac{16 e ^{\frac{5}{2}}}{9} e^{- \frac{5 t}{2}}

y = e^{\frac{4 t - 9}{2}} (5 e^{\frac{5}{2}} - \frac{16 e ^{\frac{5}{2}}}{9}) + \frac{16 e ^{\frac{5}{2}}}{9} e^{- \frac{5 t}{2}}

\frac{d}{d x} ((\frac{1}{x ^{2}} - \frac{3}{x ^{4}}) (x + 5 x^{3}))

3 y^{^{'}} + 15 y = 180

\int_{0}^{2} \frac{1}{8 + 2 t ^{2}} d t

\int (arccos (x)) d x

\frac{d}{d x} (x + x^{2})