y^'=((y+3)/(4x+5))^3

解答

y^{^{'}} = (\frac{y + 3}{4 x + 5})^{3}

y = \frac{- 64 x ^{2} - 160 x - 100}{- 1 + 128 c _{1} x ^{2} + 320 c _{1} x + 200 c _{1}} - 3, y = - \frac{- 64 x ^{2} - 160 x - 100}{- 1 + 128 c _{1} x ^{2} + 320 c _{1} x + 200 c _{1}} - 3

求解步骤

y^{'} (x) = (\frac{y + 3}{4 x + 5})^{3}

求解可分离方程:

y = \frac{- 64 x ^{2} - 160 x - 100}{- 1 + 128 c _{1} x ^{2} + 320 c _{1} x + 200 c _{1}} - 3, y = - \frac{- 64 x ^{2} - 160 x - 100}{- 1 + 128 c _{1} x ^{2} + 320 c _{1} x + 200 c _{1}} - 3

y = \frac{- 64 x ^{2} - 160 x - 100}{- 1 + 128 c _{1} x ^{2} + 320 c _{1} x + 200 c _{1}} - 3, y = - \frac{- 64 x ^{2} - 160 x - 100}{- 1 + 128 c _{1} x ^{2} + 320 c _{1} x + 200 c _{1}} - 3

\int y (y^{2} + 1)^{5} d y

x \to - 6 lim (\frac{( x + 7 )}{x + 6})

\int \frac{6 e ^{2 x}}{e ^{2 x} + 12 e ^{x} + 27} d x

\int x (x^{2} + x) d x

\int x (a - b x^{2}) d x