Question

解微分方程y'-y/x=x2 当y(1)=0时的特解

Answer 1

∵(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。