Question

可降阶的二阶微分方程

xy‘‘=y‘+xsin(y‘/x)，要过程特别是化成齐次方程之后对u的积分

Answer 1

y'=p, p'=p/x+sinp/x
u=p/x,p=ux,p'=u+u'x u+u'x=u+sinu
du/sinu=dx/x
ln(tanu/2)=lnx+lnC1
tanu/2=C1x
u=2arctan(C1x)
y'=p=ux=2xarctan(C1x)
y=∫2xarctan(C1x)dx
=∫arctan(C1x)dx^2
=x^2arctan(C1x)-∫C1x^2/(1+(C1x)^2)
=x^2arctan(C1x)-(1/C1)∫(C1^2x^2+1-1)/(1+(C1x)^2)
=x^2arctan(C1x)-(x/C1)+(1/C1^2)arctanC1x+C2

追问

du/sinu=dx/x到ln(tanu/2)=lnx lnc1是怎么弄出来的

追答

两边积分

追问

左边是怎么积出tan的，还是u/2求解释

追答

∫du/sinu
=∫du/(2sinu/2cosu/2)
=∫du/(2tanu/2(cos(u/2))^2
=∫(1/tanu/2)dtanu/2
=ln|tanu/2|+c