xy’+x+sin(x+y)=0 求解该积分方程 .急.

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xy’+x+sin(x+y)=0 求解该积分方程 .急.

xy’+x+sin(x+y)=0 求解该积分方程 .急.
xy’+x+sin(x+y)=0 求解该积分方程 .
急.

xy’+x+sin(x+y)=0 求解该积分方程 .急.
∵xy’+x+sin(x+y)=0
==>x(y'+1)=-sin(x+y)
==>x(dy/dx+dx/dx)=-sin(x+y)
==>xd(x+y)/dx=-sin(x+y)
==>d(x+y)/sin(x+y)=-dx/x
==>sin(x+y)d(x+y)/sin²(x+y)=-dx/x
==>d[cos(x+y)]/[1-cos²(x+y)]=dx/x
==>{1/[1+cos(x+y)]+1/[1-cos(x+y)]}d[cos(x+y)]=2dx/x
==>ln[1+cos(x+y)]-ln[1-cos(x+y)]=2ln│x│+ln│C│ (C是积分常数)
==>[1+cos(x+y)]/[1-cos(x+y)]=Ce^(2x)
==>cos(x+y)=[Ce^(2x)-1]/[Ce^(2x)+1]
∴原方程的通解是cos(x+y)=[Ce^(2x)-1]/[Ce^(2x)+1] (C是积分常数).

xy’+x+sin(x+y)=0
令x+y=p
两边求导得
1+y'=p'
代入得
x(p'-1)+x+sinp=0
xp'+sinp=0
dp/sinp=-dx/x
两边积分得
-1/2ln(1+p)+1/2ln(1-p)=-lnx+C