xy'-y=√2x-x^2, y(1)=0
1.微分方程xy'-y-√y^2-x^2=0的通解为?2.设y=arctan(e^x)-ln√(e^2x)/(e^2x+1),求y'|x=1时的值.
(xy'-y)/x^2=√(y^2-x^2)/x^2
d(y/x)=√[(y/x)^2-1]dx/x
secu=y/x
dsecu/tanu=dln|x|
secudu=dln|x|
dln|secu+tanu|=dln|x|
secu+tanu=Cx
通解y/x+√((y/x)^2-1)=Cx
2
y'=e^x/(1+e^2x) -2e^2x/(1+e^2x)^2 *(1/2√(e^2x/(e^2x+1)))*(1/√(e^2x/(1+e^2x)))
=e^x/(1+e^2x)-e^2x/(1+e^2x) *1/(e^2x/(1+e^2x))
=e^x/(1+e^2x)-1
xy'-y=√2x-x^2, y(1)=0
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请问当x,y都趋向0时,xy\/√(x^2+y^2)的极限是多少?
如果x=y，极限=1\/sqrt(2)
求解常微分方程 xy(y-xy')=x+yy',y(0)=(1\/2)*根号2
方程两边同时除以xy后，用分离变量法
