y+y^4=(y^6+3x)y^'

解

y + y^{4} = (y^{6} + 3 x) y^{^{'}}

y = - 3 \frac{3 c _{1} - 3 x + ( - 3 c _{1} + 3 x ) ^{2} + 12 x}{2}, y = - 3 \frac{3 c _{1} - 3 x - ( - 3 c _{1} + 3 x ) ^{2} + 12 x}{2}

解答ステップ

y + y^{4} = (y^{6} + 3 x) y^{'} (x)

完全 ODE を解く:

y = - 3 \frac{3 c _{1} - 3 x + ( - 3 c _{1} + 3 x ) ^{2} + 12 x}{2}, y = - 3 \frac{3 c _{1} - 3 x - ( - 3 c _{1} + 3 x ) ^{2} + 12 x}{2}

y = - 3 \frac{3 c _{1} - 3 x + ( - 3 c _{1} + 3 x ) ^{2} + 12 x}{2}, y = - 3 \frac{3 c _{1} - 3 x - ( - 3 c _{1} + 3 x ) ^{2} + 12 x}{2}