Question

(x²＋2xy－y²)dx＋(y²＋2xy－x²)dy＝0,y | x＝1,y＝1求微分方程特解

求全过程

Answer 1

dy/dx=(x²+2xy-y²)/(x²-2xy-y²)
令y=ux，
则u+du/dx=(1+u-u²)/(1-u-u²)
算出u然后算出y然后特解