(5x+4)+(5y-5)y^'=0

解

(5 x + 4) + (5 y - 5) y^{^{'}} = 0

y = \frac{10 + 100 - 20 ( - c _{1} + 8 x + 5 x ^{2} )}{10}, y = \frac{10 - 100 - 20 ( - c _{1} + 8 x + 5 x ^{2} )}{10}

解答ステップ

(5 x + 4) + (5 y - 5) y^{'} (x) = 0

分離可能 ODE を解く:

y = \frac{10 + 100 - 20 ( - c _{1} + 8 x + 5 x ^{2} )}{10}, y = \frac{10 - 100 - 20 ( - c _{1} + 8 x + 5 x ^{2} )}{10}

y = \frac{10 + 100 - 20 ( - c _{1} + 8 x + 5 x ^{2} )}{10}, y = \frac{10 - 100 - 20 ( - c _{1} + 8 x + 5 x ^{2} )}{10}