Question

(x^2+2xy-y^2)dx+(y^2+2xy-x^2)dy=0 当x =1时 y =5 求特接 要详细过成

Answer 1

解：∵(x^2+2xy-y^2)dx+(y^2+2xy-x^2)dy=0
==>(1+2(y/x)-(y/x)²)dx+((y/x)²+2(y/x)-1)dy=0
∴设u=y/x，则dy=xdu+udx
∵代入上式得(1+2u-u²)dx+(u²+2u-1)(xdu+udx)=0
==>(u³+u²+u+1)dx+x(u²+2u-1)du=0
==>(1-2u-u²)du/(u³+u²+u+1)=dx/x
==>[1/(u+1)-2u/(u²+1)]du=dx/x
==>ln│u+1│-ln(u²+1)=ln│x│+ln│C│ (C是积分常数)
==>(u+1)/(u²+1)=Cx
==>((y/x)+1)/((y/x)²+1)=Cx
==>(x+y)/(x²+y²)=C
∴原方程的通解是 (x+y)/(x²+y²)=C （C是积分常数）
∵ 当x =1时，y =5
∴C=3/13 ==>(x+y)/(x²+y²)=3/13
==>3(x²+y²)-13(x+y)=0
故所求特解是3(x²+y²)-13(x+y)=0。

Answer 2

原式=d{[(x+y)(x-y)+2xy]x+[(x+y)(y-x)+2xy]
=d{[(x+y)(x-y)+2xy]x-[(x+y)(x-y)-2xy]}
=d*4xy
当x =1,y =5时
20d=0
d=0