题目

求微分方程(y^2-3x^2)dy+2xydx=0 x=0,y=1时的特解

提问时间：2020-11-01

答案

∵(y²-3x²)dy+2xydx=0
∴((y/x)²-3)dy+2(y/x)dx=0.(1)
设t=y/x,则dy=xdt+tdx
代入(1)得(t²-3)(xdt+tdx)+2tdx=0
==>x(t²-3)dt+(t³-t)dx=0
==>(t²-3)dt/(t-t³)=dx/x
==>[1/(1+t)-1/(1-t)-3/t]dt=dx/x
==>ln│1+t│+ln│1-t│-3ln│t│=ln│x│+ln│C│ (C是积分常数)
==>(1-t²)/t³=Cx
==>(1-(y/x)²)/(y/x)³=Cx
==>(x²-y²)/y³=C
==>x²-y²=Cy³
∵当x=0时,y=1
∴0²-1²=C*1³ ==>C=-1
故原微分方程满足x=0,y=1时的特解是x²-y²=-y³,即x²-y²+y³=0.