验证给定函数是其对应微分方程的解:xyy"+x(y')^2-yy'=0,x^2/C1+y^2/C2=1xyy"+x(y')^2-yy'=0 ,x^2/C1+y^2/C2=1
问题描述:
验证给定函数是其对应微分方程的解:xyy"+x(y')^2-yy'=0,x^2/C1+y^2/C2=1
xyy"+x(y')^2-yy'=0 ,x^2/C1+y^2/C2=1
答
x^2/C1+y^2/C2=1
两边对x求导:
2x/c1+2yy'/c2=0
x/c1=-yy'/c2
(yy')/x=-c2/c1
两边对x求导:
[(y'^2+yy'')x-yy']/x^2=0
xyy''+x(y')^2-yy'=0