y^'= y/t-1/(y/t)

解答

y^{^{'}} = \frac{y}{t} - \frac{1}{\frac{y}{t}}

y = t - 2 ln (t) + c_{1}, y = - t - 2 ln (t) + c_{1}

求解步骤

y^{'} (t) = \frac{y}{t} - \frac{1}{\frac{y}{t}}

求解贝努力方程:

y = t - 2 ln (t) + c_{1}, y = - t - 2 ln (t) + c_{1}

y = t - 2 ln (t) + c_{1}, y = - t - 2 ln (t) + c_{1}

t an g e n t f (x) = x + \frac{2}{x}, at x = 2

\int \frac{50}{x ^{0.25}} d x

\frac{d}{d y} (x e^{x - y})

\int_{0}^{6} π (3 x)^{2} d x

\int \frac{( ln ( u ) ) ^{8}}{u} d u