微分方程dy/dx=x/y+k怎么求解?
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微分方程dy/dx=x/y+k怎么求解?
![微分方程dy/dx=x/y+k怎么求解?](/uploads/image/z/7770029-5-9.jpg?t=%E5%BE%AE%E5%88%86%E6%96%B9%E7%A8%8Bdy%2Fdx%3Dx%2Fy%2Bk%E6%80%8E%E4%B9%88%E6%B1%82%E8%A7%A3%3F)
令y/x=u y=ux y'=u+xu'
代入原式得:u+u'x=1/u+k
u'x=1/u+k-u
du/(1/u+k-u)=dx
udu/(1+ku-u^2)=dx
(2u-k+k)du/(1+ku-u^2)=2dx
(2u-k)du/(1+ku-u^2)+kdu/(1+ku-u^2)=2dx
两边积分得:
-ln|1+ku-u^2|+∫kdu/(1+k^2/4-(u-k/2)^2)=x^2
通解为:-ln|1+ku-u^2|+(2k/√(4+k^2)ln|(u-k/2+√(4+k^2)/2)/(u-k/2-√(4+k^2)/2)|=x^2+C
其中u=y/x
代入原式得:u+u'x=1/u+k
u'x=1/u+k-u
du/(1/u+k-u)=dx
udu/(1+ku-u^2)=dx
(2u-k+k)du/(1+ku-u^2)=2dx
(2u-k)du/(1+ku-u^2)+kdu/(1+ku-u^2)=2dx
两边积分得:
-ln|1+ku-u^2|+∫kdu/(1+k^2/4-(u-k/2)^2)=x^2
通解为:-ln|1+ku-u^2|+(2k/√(4+k^2)ln|(u-k/2+√(4+k^2)/2)/(u-k/2-√(4+k^2)/2)|=x^2+C
其中u=y/x