✔ 最佳答案
(4) xy'=y+2x^3*sin^2(y/x)Let y=ux => y'=u'x+u代入原式裡面:x(u'x+u)=ux+2x^3*sin^2(u)u'x+u=u+2x^2*sin^2(u)u'x=2x^2*sin^2(u)u'=2xsin^2(u) => csc^2(u)du=2xdx∫csc^2(u)du=2∫xdx-cot(u)=-cot(y/x)=x^2+ctan(y/x)=-1/(x^2+c)y/x=-atan[1/(x^2+c)]y(x)=-x*atan[1/(x^2+c)].....ans
(5) xy'=y^2+yLet y=ux => y'=u'x+u代入原式裡面:x(u'x+u)=u^2*x^2+uxu'x+u=u^2*x+uxu'=xu^2 => u'=u^2dx=du/u^2 => ∫dx=∫du/u^2x+c=-1/u=-x/yy(x)=-x/(x+c).....ans