Question

常微分方程问题？

第14题怎么做呀？

Answer 1

(14)
(1+y^2)dx-x(1+x)ydy =0
整理方程
(1+y^2)dx =x(1+x)ydy
y/(1+y^2) dy = dx/[x(1+x)]
两边求积分
∫y/(1+y^2) dy = ∫dx/[x(1+x)]
(1/2)∫d(1+y^2)/(1+y^2) dy = ∫[1/x-1/(1+x)] dx
(1/2)ln|1+y^2| = ln|x/(1+x)| +C'
1+y^2 = C.[x/(1+x)]^2
y= √{ -1+[x/(1+x)]^2}
得出结果
(1+y^2)dx-x(1+x)ydy =0 的通解
y= √{ -1+[x/(1+x)]^2}

Answer 2

14. 微分方程即 (1+y^2)dx = x(1+x)ydy
可分离变量为 ydy/(1+y^2) = dx/[x(1+x)]
(1/2)d(1+y^2)/(1+y^2) = [1/x - 1/(1+x)]dx
d(1+y^2)/(1+y^2) = 2[1/x - 1/(1+x)]dx
ln(1+y^2) = 2[lnx-ln(1+x)]+ lnC
1+y^2 = Cx^2/(1+x)^2