(40x-24)dx+(87y+33)dy=0

解答

(40 x - 24) d x + (87 y + 33) d y = 0

y = \frac{- 66 + 4356 - 348 ( - c _{1} - 48 x + 40 x ^{2} )}{174}, y = \frac{- 66 - 4356 - 348 ( - c _{1} - 48 x + 40 x ^{2} )}{174}

求解步骤

(40 x - 24) d x + (87 y + 33) d y = 0

求解可分离方程:

y = \frac{- 66 + 4356 - 348 ( - c _{1} - 48 x + 40 x ^{2} )}{174}, y = \frac{- 66 - 4356 - 348 ( - c _{1} - 48 x + 40 x ^{2} )}{174}

y = \frac{- 66 + 4356 - 348 ( - c _{1} - 48 x + 40 x ^{2} )}{174}, y = \frac{- 66 - 4356 - 348 ( - c _{1} - 48 x + 40 x ^{2} )}{174}

u^{^{'}} = 5 - 0.5 H (t)

f^{^{'}} (x) f (x) = x^{2} + 10 x - 5

y (4 - x^{2}) d y = (4 + y^{2}) d x

\frac{d y}{d x} + 2 x y = 25 x

x y^{^{'}} + 4 y + 8 = 0