y^'=(3xsqrt(1+y^2))/y

解答

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

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

求解步骤

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

求解可分离方程:

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

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

\frac{d y}{y} = d x (x - 1)

\frac{d y}{d x} y = ln (7) x^{2}

\frac{d y}{d x} = (x - 5) e^{- 2 y}, y (2) = ln (5)

\frac{d y}{d x} = x^{2} + x^{2} y^{2}

\frac{d y}{d x} + 4 y = 3, y (0) = 0