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

Lösung

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

y = e^{t} (- \frac{( 1 - ( e ^{π} + 1 ) ) cos ( t )}{e ^{π}} - \frac{e ^{π} + 1}{e ^{π}} sin (t))

Schritte zur Lösung

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

Solve linear ODE:

y = e^{t} (- \frac{( 1 - ( e ^{π} + 1 ) ) cos ( t )}{e ^{π}} - \frac{e ^{π} + 1}{e ^{π}} sin (t))

y = e^{t} (- \frac{( 1 - ( e ^{π} + 1 ) ) cos ( t )}{e ^{π}} - \frac{e ^{π} + 1}{e ^{π}} sin (t))

y^{^{''}} - 4 y^{^{'}} + 8 y = 0, y^{^{'}} (0) = 2

y^{^{''}} + 4 y^{^{'}} + 13 y = 0, y (0) = 2, y^{^{'}} (0) = 0

y^{^{''}} + 12 y^{^{'}} + 36 y = 0, y (0) = - 2, y^{^{'}} (0) = 16

4 y^{^{''}} + 4 y^{^{'}} + y = 0, y (0) = 3, y^{^{'}} (0) = - 2

y^{^{''}} + 12 y^{^{'}} + 16 y = 0