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

Lösung

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

y = e^{- \frac{t}{2}} (cos (\frac{3 t}{2}) - \frac{1}{3} sin (\frac{3 t}{2}))

Schritte zur Lösung

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

Solve linear ODE:

y = e^{- \frac{t}{2}} (cos (\frac{3 t}{2}) - \frac{1}{3} sin (\frac{3 t}{2}))

y = e^{- \frac{t}{2}} (cos (\frac{3 t}{2}) - \frac{1}{3} sin (\frac{3 t}{2}))

\frac{\partial}{\partial x} (ln (x^{4} + y^{3}))

\frac{d}{d x} (- \frac{2 x}{( 2 + x ^{2} ) ^{2}})

x \to 1 lim (2 x - x^{2} - x^{3})

\frac{\partial}{\partial x} (2 x^{2} - 4)

\frac{\partial}{\partial x} (1 + x y)