2y(x^2+4)^2=(5x(y^2+1)^2)/(y^'(x^2+4))

解

2 y (x^{2} + 4)^{2} = \frac{5 x ( y ^{2} + 1 ) ^{2}}{y ^{^{'}} ( x ^{2} + 4 )}

y = \frac{- 4 c _{1} x ^{2} - 16 c _{1} - 2 x ^{2} - 3}{- 5 + 4 c _{1} x ^{2} + 16 c _{1}}, y = - \frac{- 4 c _{1} x ^{2} - 16 c _{1} - 2 x ^{2} - 3}{- 5 + 4 c _{1} x ^{2} + 16 c _{1}}

解答ステップ

2 y (x^{2} + 4)^{2} = \frac{5 x ( y ^{2} + 1 ) ^{2}}{y ^{'} ( x )}

分離可能 ODE を解く:

y = \frac{- 4 c _{1} x ^{2} - 16 c _{1} - 2 x ^{2} - 3}{- 5 + 4 c _{1} x ^{2} + 16 c _{1}}, y = - \frac{- 4 c _{1} x ^{2} - 16 c _{1} - 2 x ^{2} - 3}{- 5 + 4 c _{1} x ^{2} + 16 c _{1}}

y = \frac{- 4 c _{1} x ^{2} - 16 c _{1} - 2 x ^{2} - 3}{- 5 + 4 c _{1} x ^{2} + 16 c _{1}}, y = - \frac{- 4 c _{1} x ^{2} - 16 c _{1} - 2 x ^{2} - 3}{- 5 + 4 c _{1} x ^{2} + 16 c _{1}}