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

Lösung

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

y = (\frac{11}{2} + \frac{13}{10}) e^{- \frac{x}{2}} - \frac{11}{2} e^{- x} - \frac{1}{10} sin (x) - \frac{3}{10} cos (x)

Schritte zur Lösung

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

Solve linear ODE:

y = (\frac{11}{2} + \frac{13}{10}) e^{- \frac{x}{2}} - \frac{11}{2} e^{- x} - \frac{1}{10} sin (x) - \frac{3}{10} cos (x)

y = (\frac{11}{2} + \frac{13}{10}) e^{- \frac{x}{2}} - \frac{11}{2} e^{- x} - \frac{1}{10} sin (x) - \frac{3}{10} cos (x)

y^{^{''}} + 4 y^{^{'}} = 10 sin (2 x)

y^{^{''}} - 4 y^{^{'}} + 3 y = 1, y (0) = 1, y^{^{'}} (0) = 0

(D^{2} + 1) y = sin (x)

y^{^{''}} + y = 16 cos (3 x)

\frac{d ^{2} y}{d x ^{2}} + 2 \frac{d y}{d x} + 5 y = e^{- x}