(x+y)dx+(y-x)dy=0,y(1)=0

Question

(x+y)dx+(y-x)dy=0,y(1)=0

Answer 1

∵令y=xt,则dy=xdt+tdx
代入原方程,化简得 (1-t)dt/(1+t^2)=dx/x
==>arctant-(1/2)ln(1+t^2)=ln│x│+ln│C│ (C是常数)
==>arctant=(1/2)ln(1+t^2)+ln│x│+ln│C│
==>e^(arctant)=Cx√(1+t^2)
==>e^(arctan(y/x))=C√(x^2+y^2)
∴原方程的通解是e^(arctan(y/x))=C√(x^2+y^2)
∵y(1)=0
∴C=1
故所给方程满足初始条件y(1)=0的特解是e^(arctan(y/x))=√(x^2+y^2).