2xy^2+2x+(x^2y+4y)y^'=0

解

2 \cdot x \cdot y^{2} + 2 \cdot x + (x^{2} \cdot y + 4 \cdot y) \cdot y^{^{'}} = 0

y = \frac{c _{1} - x ^{4} - 8 x ^{2} - 16}{x ^{4} + 8 x ^{2} + 16}, y = - \frac{c _{1} - x ^{4} - 8 x ^{2} - 16}{x ^{4} + 8 x ^{2} + 16}

解答ステップ

2 \cdot x \cdot y^{2} + 2 \cdot x + (x^{2} \cdot y + 4 \cdot y) \cdot y^{'} (x) = 0

分離可能 ODE を解く:

y = \frac{c _{1} - x ^{4} - 8 x ^{2} - 16}{x ^{4} + 8 x ^{2} + 16}, y = - \frac{c _{1} - x ^{4} - 8 x ^{2} - 16}{x ^{4} + 8 x ^{2} + 16}

y = \frac{c _{1} - x ^{4} - 8 x ^{2} - 16}{x ^{4} + 8 x ^{2} + 16}, y = - \frac{c _{1} - x ^{4} - 8 x ^{2} - 16}{x ^{4} + 8 x ^{2} + 16}