y^{''}+14y^'+65y=0

Lösung

y^{^{''}} + 14 y^{^{'}} + 65 y = 0

y = e^{- 7 t} (c_{1} cos (4 t) + c_{2} sin (4 t))

Schritte zur Lösung

y^{''} (t) + 14 y^{'} (t) + 65 y = 0

Solve linear ODE:

y = e^{- 7 t} (c_{1} cos (4 t) + c_{2} sin (4 t))

y = e^{- 7 t} (c_{1} cos (4 t) + c_{2} sin (4 t))

x y^{2} d x + (1 + x^{2} y) d y = 0

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x \to \infty lim (4 cos (x))

\frac{\partial}{\partial x} (2 x y + 2 x)