y^{''}-7y^'+12y=0,y(0)=0,y^'(1)=e^3

Lösung

y^{^{''}} - 7 y^{^{'}} + 12 y = 0, y (0) = 0, y^{^{'}} (1) = e^{3}

y = - \frac{e ^{3}}{- 4 e ^{4} + 3 e ^{3}} e^{4 t} + \frac{e ^{3}}{- 4 e ^{4} + 3 e ^{3}} e^{3 t}

Schritte zur Lösung

y^{''} (t) - 7 y^{'} (t) + 12 y = 0

Solve linear ODE:

y = - \frac{e ^{3}}{- 4 e ^{4} + 3 e ^{3}} e^{4 t} + \frac{e ^{3}}{- 4 e ^{4} + 3 e ^{3}} e^{3 t}

y = - \frac{e ^{3}}{- 4 e ^{4} + 3 e ^{3}} e^{4 t} + \frac{e ^{3}}{- 4 e ^{4} + 3 e ^{3}} e^{3 t}

(2 D^{4} - 3 D^{3} - 2 D^{2}) y = 0

\frac{1}{2} y^{^{''}} + 3 y^{^{'}} + 4 y = 0, y (0) = 0, y^{^{'}} (0) = - \frac{5}{3}

y^{^{''}} - 4 y^{^{'}} - 5 y = 0, y (0) = 4, y^{^{'}} (0) = 14

y^{^{''}} + y = 0, y (0) = 11, y^{^{'}} (0) = 33

\frac{1}{20} y^{^{''}} + 3 y^{^{'}} + 80 y = 0