y^{''}-2y^'+2y=sin(t),y(0)=0,y^'(0)=1

Lösung

y^{^{''}} - 2 y^{^{'}} + 2 y = sin (t), y (0) = 0, y^{^{'}} (0) = 1

y = e^{t} (- \frac{2}{5} cos (t) + \frac{6}{5} sin (t)) + \frac{1}{5} sin (t) + \frac{2}{5} cos (t)

Schritte zur Lösung

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

Solve linear ODE:

y = e^{t} (- \frac{2}{5} cos (t) + \frac{6}{5} sin (t)) + \frac{1}{5} sin (t) + \frac{2}{5} cos (t)

y = e^{t} (- \frac{2}{5} cos (t) + \frac{6}{5} sin (t)) + \frac{1}{5} sin (t) + \frac{2}{5} cos (t)

y^{^{''}} - 5 y^{^{'}} + 6 y = 3 x^{2} + 5 x

y^{^{''}} + 2 y^{^{'}} - 3 y = 5 e^{2 x}, y (0) = 0, y^{^{'}} (0) = a

y^{^{''}} + 4 y = sin^{22} (x)

y^{^{''}} - 3 y^{^{'}} = 2 - 6 x

y^{^{'''}} - 2 y^{^{''}} + y^{^{'}} = 1 + x e^{x}