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

Lösung

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

y = e^{\frac{t}{2}} (5 cos (\frac{7 t}{2}) + \frac{5}{7} sin (\frac{7 t}{2}))

Schritte zur Lösung

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

Solve linear ODE:

y = e^{\frac{t}{2}} (5 cos (\frac{7 t}{2}) + \frac{5}{7} sin (\frac{7 t}{2}))

y = e^{\frac{t}{2}} (5 cos (\frac{7 t}{2}) + \frac{5}{7} sin (\frac{7 t}{2}))

\frac{d ^{2} y}{d x ^{2}} + 9 \frac{d y}{d x} + 20 y = 0

y^{^{''}} + 12 y^{^{'}} + 27 y = 0, y (0) = 11, y^{^{'}} (0) = - 81

4 y^{^{''}} - 4 y - 3 y = 0

\frac{d ^{2} x}{d t ^{2}} + 10 \frac{d x}{d t} + 25 x = 0

y^{^{''}} + 10 y^{^{'}} + 16 y = 0, y (0) = 9, y^{^{'}} (0) = - 60