v(dv)/(dt)-t(v(t)^2+1)=0,v(0)=-1

解答

v \frac{d v}{d t} - t (v (t)^{2} + 1) = 0, v (0) = - 1

- \frac{d v}{d t} + \circ t^{3} v + \circ t

求解步骤

给定

v \frac{d v}{d t} - t (v (t)^{2} + 1) = 0, v (0) = - 1

找到

v \frac{d v}{d t} - t (v (t)^{2} + 1) \circ v

v \frac{d v}{d t} - t (v (t)^{2} + 1) \circ v = v \frac{d v}{d t} - t (v (t)^{2} + 1) \circ v

v : - 1

v : - 1

v : - 1

= (- 1) \frac{d v}{d t} - t (v t^{2} + 1) \circ (- 1)

乘开

(- 1) \frac{d v}{d t} - t (v t^{2} + 1) \circ (- 1) : - \frac{d v}{d t} + \circ t^{3} v + \circ t

= - \frac{d v}{d t} + \circ t^{3} v + \circ t

\frac{d y}{d x} = \frac{x arctan ( x )}{y}

7 x y d x - 4 x d y = 7 x y d y - 8 y d x

\frac{d y}{d x} = y^{2} a - 1

y^{^{'}} = \frac{x}{cos ( y )}, y (1) = 0

x - x^{2} d y - x y d x = 0