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

Lösung

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

y = e^{t} (- cos (t) + sin (t))

Schritte zur Lösung

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

Solve linear ODE:

y = e^{t} (- cos (t) + sin (t))

y = e^{t} (- cos (t) + sin (t))

y^{^{'''}} - 9 y^{^{''}} + 23 y^{^{'}} - 15 y = 0

y^{^{''}} + y^{^{'}} - 2 y = 0, y (0) = 1, y^{^{'}} (0) = 4

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

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

\frac{d ^{2} y}{d x ^{2}} - 6 \frac{d y}{d x} + 13 y = 0