y^'=(3x^2+2x+1)/(y-2)

解

y^{^{'}} = \frac{3 x ^{2} + 2 x + 1}{y - 2}

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

解答ステップ

y^{'} (x) = \frac{3 x ^{2} + 2 x + 1}{y - 2}

分離可能 ODE を解く:

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

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