y^'=(3xsqrt(16+y^2))/y

解

y^{^{'}} = \frac{3 x 16 + y ^{2}}{y}

y = \frac{9 x ^{4} + 12 c _{1} x ^{2} + 4 c _{1}^{2} - 64}{2}, y = - \frac{9 x ^{4} + 12 c _{1} x ^{2} + 4 c _{1}^{2} - 64}{2}

解答ステップ

y^{'} (x) = \frac{3 x 16 + y ^{2}}{y}

分離可能 ODE を解く:

y = \frac{9 x ^{4} + 12 c _{1} x ^{2} + 4 c _{1}^{2} - 64}{2}, y = - \frac{9 x ^{4} + 12 c _{1} x ^{2} + 4 c _{1}^{2} - 64}{2}

y = \frac{9 x ^{4} + 12 c _{1} x ^{2} + 4 c _{1}^{2} - 64}{2}, y = - \frac{9 x ^{4} + 12 c _{1} x ^{2} + 4 c _{1}^{2} - 64}{2}