(dr)/(dq)=(50-4q)e^{-r/5}

解答

\frac{d r}{d q} = (50 - 4 q) e^{- \frac{r}{5}}

r = 5 ln (10 q + \frac{- 2 q ^{2} + c _{1}}{5})

求解步骤

\frac{d r}{d q} = (50 - 4 q) e^{- \frac{r}{5}}

求解可分离方程:

r = 5 ln (10 q + \frac{- 2 q ^{2} + c _{1}}{5})

r = 5 ln (10 q + \frac{- 2 q ^{2} + c _{1}}{5})

\frac{d r ( θ )}{r ( θ ) ^{2}} = \frac{r ( θ ) d θ}{θ ^{2}}

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

\frac{d y}{d x} = \frac{sin ^{2} ( y )}{1 - x ^{2}}

v^{2} x^{2} d x + x^{2} d x - 2 x^{2} v^{2} d x - 2 x^{3} v d v = 0

\frac{d y}{d x} y = 2 3^{x}