y^{''}-3y^'+2y=sin(3x),y(0)=1,y^'(0)=0

Lösung

y^{^{''}} - 3 y^{^{'}} + 2 y = sin (3 x), y (0) = 1, y^{^{'}} (0) = 0

y = (- \frac{17}{10} + \frac{121}{130}) e^{2 x} + \frac{17}{10} e^{x} - \frac{7}{130} sin (3 x) + \frac{9}{130} cos (3 x)

Schritte zur Lösung

y^{''} (x) - 3 y^{'} (x) + 2 y = sin (3 x)

Solve linear ODE:

y = (- \frac{17}{10} + \frac{121}{130}) e^{2 x} + \frac{17}{10} e^{x} - \frac{7}{130} sin (3 x) + \frac{9}{130} cos (3 x)

y = (- \frac{17}{10} + \frac{121}{130}) e^{2 x} + \frac{17}{10} e^{x} - \frac{7}{130} sin (3 x) + \frac{9}{130} cos (3 x)

y^{^{''}} + 2 y^{^{'}} + 2 y = (cos (x)) e^{- x}

y^{^{''}} - 4 y = 96 x^{2} e^{2 x} + 4 e^{- 2 x}

x^{^{''}} + x^{^{'}} = t

(d^{2} - 1) y = x^{2} e^{x}

x^{^{''}} + 4 x^{^{'}} - 5 x = 4 t^{2}