y^'=((2y+3)^2)/((4x+5)^2)

解

y^{^{'}} = \frac{( 2 y + 3 ) ^{2}}{( 4 x + 5 ) ^{2}}

y = - \frac{4 x}{- 1 + 16 c _{1} x + 20 c _{1}} - \frac{24 c _{1} x}{- 1 + 16 c _{1} x + 20 c _{1}} - \frac{30 c _{1}}{- 1 + 16 c _{1} x + 20 c _{1}} - \frac{7}{2 ( - 1 + 16 c _{1} x + 20 c _{1} )}

解答ステップ

y^{'} (x) = \frac{( 2 y + 3 ) ^{2}}{( 4 x + 5 ) ^{2}}

分離可能 ODE を解く:

y = - \frac{4 x}{- 1 + 16 c _{1} x + 20 c _{1}} - \frac{24 c _{1} x}{- 1 + 16 c _{1} x + 20 c _{1}} - \frac{30 c _{1}}{- 1 + 16 c _{1} x + 20 c _{1}} - \frac{7}{2 ( - 1 + 16 c _{1} x + 20 c _{1} )}

y = - \frac{4 x}{- 1 + 16 c _{1} x + 20 c _{1}} - \frac{24 c _{1} x}{- 1 + 16 c _{1} x + 20 c _{1}} - \frac{30 c _{1}}{- 1 + 16 c _{1} x + 20 c _{1}} - \frac{7}{2 ( - 1 + 16 c _{1} x + 20 c _{1} )}