9y^{''}+6y^'+y=0,y(-3)=-2,y^'(1)=-1

Lösung

9 y^{^{''}} + 6 y^{^{'}} + y = 0, y (- 3) = - 2, y^{^{'}} (1) = - 1

y = (- 2 + 3 (3 e^{\frac{4}{3}} + 2)) e^{\frac{- t - 3}{3}} + (3 e^{\frac{4}{3}} + 2) e^{\frac{- t - 3}{3}} t

Schritte zur Lösung

9 y^{''} (t) + 6 y^{'} (t) + y = 0

Solve linear ODE:

y = (- 2 + 3 (3 e^{\frac{4}{3}} + 2)) e^{\frac{- t - 3}{3}} + (3 e^{\frac{4}{3}} + 2) e^{\frac{- t - 3}{3}} t

y = (- 2 + 3 (3 e^{\frac{4}{3}} + 2)) e^{\frac{- t - 3}{3}} + (3 e^{\frac{4}{3}} + 2) e^{\frac{- t - 3}{3}} t

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

\int_{0}^{1} \frac{y}{e ^{2 y}} d y

\frac{\partial}{\partial y} (x y^{3} z^{2})

\frac{\partial}{\partial x} (x \cdot Y (x))

a re a 5 - x^{2}, sin (x), - 2.385, 2.025