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

解答

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

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

求解步骤

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

求解可分离方程:

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

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

y^{^{'}} = e^{t} \cdot y^{2}

y^{^{'}} = 6 y - 5

\frac{d y}{d t} = k y (t)^{1 + c}

y^{^{'}} = (y - 1)^{2} - 0.01

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