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

解

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

解答ステップ

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

分離可能 ODE を解く:

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