求方程(y-sinx)dx+tanxdy=o满足条件y(π/6)=1的解万谢.

Question

求方程(y-sinx)dx+tanxdy=o满足条件y(π/6)=1的解
万谢.

Answer 1

dy/dx = cosx - y/tanx;
即y'= cosx - y/tanx
令y1'= cosx ;
y2'= - y2/tanx;
分别解:
解y1,得 y1=sinx +C1;
解y2:即
dy2/dx = - y2/tanx;
分离变量:
dy2/y2 = -dx/tanx;
两边积分得
∫dy2/y2 = -∫dx/tanx
= -∫(cosx/sinx)dx
= -∫(1/sinx)d(sinx);
ln |y2| = -ln |sinx| + ln|C2|;
→y2 = C2/sinx;
则:
y=y1+y2
=sinx +C1 + C2/sinx;
下面求特解:
y(π/6)=1,则:
sin(π/6) +C1 + C2/sin(π/6)=1;
即 1/2+C1 + 2·C2=1;
C1 + 2·C2=1/2 ①;
将x=π/6,y=1代入y'= cosx - y/tanx中得
y'= -√3/2;
对y=sinx +C1 + C2/sinx 求导得:
y'=cosx -C2·cosx/(sin^2 x);
将x=π/6,y=1,及y'= -√3/2代入上式得
-√3/2 = √3/2 -2√3·C2
→C2=1/2;
将C2=1/2代入①中得
C1= -1/2;
将C1= -1/2 ,C2=1/2 代入 y=sinx +C1 + C2/sinx 可得
y=sinx -1/2 + 1/(2·sinx)
y=sinx -1/2 + 1/(2·sinx) 就是原微分方程的解.