(xy-x)y''+xy'^2+yy'-2y'=0,y=ln(xy)验证所给二元方程所确定的函数为(xy-x)y''+xy'^2+yy'-2y'=0,y=ln(xy)验证所给二元方程所确定的函数为所给微分方程的解

Question

(xy-x)y''+xy'^2+yy'-2y'=0,y=ln(xy) 验证所给二元方程所确定的函数为
(xy-x)y''+xy'^2+yy'-2y'=0,y=ln(xy)
验证所给二元方程所确定的函数为所给微分方程的解

Answer 1

y=ln(xy),
y'=1/(xy)*(y+xy') => y'=y/(xy-x)
y''=[y'(xy-x)-y(y+xy'-1)]/(xy-x)^2=-(xy'+y^2-y)/(xy-x)^2
(xy-x)y''=-(xy'+y^2-y)/(xy-x)=(-xy^3+2xy^2-2xy)/(xy-x)^2
xy'^2=xy^2/(xy-x)^2
yy'=y^2/(xy-x)=(xy^3-xy^2)/(xy-x)^2
-2y'=-2y/(xy-x)=(-2xy^2+2xy)/(xy-x)^2
(xy-x)y''+xy'^2+yy'-2y'
=[(-xy^3+2xy^2-2xy)+(xy^2)+(xy^3-xy^2)+(-2xy^2+2xy)]/(xy-x)^2
=0
∴y=ln(xy)符合方程(xy-x)y''+xy'^2+yy'-2y'=0
即y=ln(xy)为上述微分方程的解