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

Lösung

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

y = e^{t} t

Schritte zur Lösung

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

Solve linear ODE:

y = e^{t} t

y = e^{t} t

2 y^{^{''}} + 10 y^{^{'}} + 8 y = 0, y^{^{'}} (0) = 0, y (0) = 5

7 y^{^{''}} - 8 y^{^{'}} + y = 0, y (0) = 7, y^{^{'}} (0) = \frac{1}{7}

y^{^{''}} - 4 y^{^{'}} + 3 y = 0, y (0) = \frac{3}{5}, y^{^{'}} (0) = \frac{1}{5}

y^{^{''}} - 12 y^{^{'}} + 36 y = 0, y (0) = 5, y^{^{'}} (0) = 0

y^{^{''}} + 3 y = 4 y^{^{'}}