y^{''}+3y^'-5y=0

Lösung

y^{^{''}} + 3 y^{^{'}} - 5 y = 0

y = c_{1} e^{\frac{( - 3 + 29 ) t}{2}} + c_{2} e^{\frac{( - 3 - 29 ) t}{2}}

Schritte zur Lösung

y^{''} (t) + 3 y^{'} (t) - 5 y = 0

Solve linear ODE:

y = c_{1} e^{\frac{( - 3 + 29 ) t}{2}} + c_{2} e^{\frac{( - 3 - 29 ) t}{2}}

y = c_{1} e^{\frac{( - 3 + 29 ) t}{2}} + c_{2} e^{\frac{( - 3 - 29 ) t}{2}}

(d^{2} + 2 d + 1) y = 0

16 y^{^{''}} + 16 y^{^{'}} + 3 y = 0

(d^{3} - 2 d^{2} - 3 d) y = 0

y^{^{''}} - y^{^{'}} - 6 y = 0, y (ln^{2} (x)) = 5, y (0) = 1

d (x) = 0