高等数学：微分方程（x^2+2xy-y^2）dx+(y^2+2xy-x^2)dy=0.x=1时,y=六分之派.求特解
2019-04-14
高等数学：微分方程
（x^2+2xy-y^2）dx+(y^2+2xy-x^2)dy=0.x=1时,y=六分之派.求特解
优质解答
x^2+2xy-y^2+(y^2+2xy-x^2) y'(x)=0
令y=u*x,则y'(x)=u(x)+x*u'(x);带入上式化简,有
x^2+2xy-y^2+(y^2+2xy-x^2)*(u+x*u')=0
(u^2+1)*(u+1)+(u^2+2u-1)*x*u'=0
dx/x=-(u^2+2u-1)/((u^2+1)*(u+1))*du
dx/x=(1/(u+1)-2u/(u^2+1))*du
ln x=ln(u+1)-ln(u^2+1)+_C1
x=_C1*(u+1)/(u^2+1)
(y^2+x^2)=_C1(x+y)
令x=0,y=PI/6,有_C1=PI/6.
即有特解(y^2+x^2)=PI/6*(x+y)
x^2+2xy-y^2+(y^2+2xy-x^2) y'(x)=0
令y=u*x,则y'(x)=u(x)+x*u'(x);带入上式化简,有
x^2+2xy-y^2+(y^2+2xy-x^2)*(u+x*u')=0
(u^2+1)*(u+1)+(u^2+2u-1)*x*u'=0
dx/x=-(u^2+2u-1)/((u^2+1)*(u+1))*du
dx/x=(1/(u+1)-2u/(u^2+1))*du
ln x=ln(u+1)-ln(u^2+1)+_C1
x=_C1*(u+1)/(u^2+1)
(y^2+x^2)=_C1(x+y)
令x=0,y=PI/6,有_C1=PI/6.
即有特解(y^2+x^2)=PI/6*(x+y)