(dy)/(dt)= 1/((2y+1))

Lösung

\frac{d y}{d t} = \frac{1}{( 2 y + 1 )}

y = \frac{- 1 + 1 - 4 ( - c _{1} - t )}{2}, y = \frac{- 1 - 1 - 4 ( - c _{1} - t )}{2}

Schritte zur Lösung

\frac{d y}{d t} = \frac{1}{( 2 y + 1 )}

Solve separable ODE:

y = \frac{- 1 + 1 - 4 ( - c _{1} - t )}{2}, y = \frac{- 1 - 1 - 4 ( - c _{1} - t )}{2}

y = \frac{- 1 + 1 - 4 ( - c _{1} - t )}{2}, y = \frac{- 1 - 1 - 4 ( - c _{1} - t )}{2}

l a pl a ce t r an s f or m 3 \cdot e^{- 2 t} + 2 \cdot sin (3 t)

\int_{- 16}^{0} \frac{1}{9 - x} d x

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x \to 0 lim (\frac{x - 15}{x - 5})

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