Question

积分问题求解

曲线上点P的切线与x轴的夹角为a，P与原点的连线与x轴夹角为b，a=2b，求曲线方程。

Answer 1

解：设曲线上P点的坐标是(x,y)。
∵曲线上点P的切线与x轴的夹角为a
∴dy/dx=tana
∵P与原点的连线与x轴夹角为b
∴tanb=y/x
∵a=2b
∴tana=tan(2b)=2tanb/(1-tan²b)=2(y/x)/(1-(y/x)²)
即dy/dx=2(y/x)/(1-(y/x)²)........(1)
现在来求解微分方程(1):
设t=y/x,则dy/dx=xdt/dx+t
代入(1)得，xdt/dx+t=2t/(1-t²)
==>xdt/dx=t(1+t²)/(1-t²)
==>(1-t²)dt/[t(1+t²)]=dx/x
==>[1/t-2t/(1+t²)]dt=dx/x
==>ln│t│-ln(1+t²)=ln│x│+ln│C│ (C是积分常数)
==>ln│t/(1+t²)│=ln│Cx│
==>t/(1+t²)=Cx
==>(y/x)/[1+(y/x)²]=Cx
==>xy/(x²+y²)=Cx
==>y/(x²+y²)=C
即微分方程(1)的通解是y/(x²+y²)=C (C是积分常数)
故所求曲线方程是y/(x²+y²)=C (C是积分常数)。

Answer 2

P(x,y)为曲线y上任意一点；由已知条件知道
y'=tana;y/x=tanb;又a=2b;
y'=tana=tan2b=2tanb/(1-(tanb)^2);
令u=y/x;则y=ux;u为关于x的函数；
则y'=xu'+u;
于是
xu'+u=2u/(1-u^2);
整理后得到
(1-u^2)/u(1+u^2)*du=dx/x;
两边积分
∫(1-u^2)/u(1+u^2)*du=∫dx/x
即
u/(1+u^2)=Cx;
然后用y/x换掉u即可

Answer 3

设：y=f(x)
tanb = y/x
y' = tana = tan2b = 2tanb/(1 -（tanb）^2) = 2(y/x)/[1-(y/x)^2]
令：y/x = u , y'=u+xu'
u+xu' = 2u/(1-u^2)
(1-u^2)/(u+u^3) du = 1/x dx

-(1-1/u^2)/(1/u+u) du = dlnx
-1/(1/u+u) d(u+1/u) = dlnx

-ln(u+1/u) = lnx + C1
u+1/u = 1/c2x
x^2+y^2 = cy

Answer 4

“曲线上点P”这一条件应该是“曲线上任意点P”吧？？