Question

高等数学求微分方程的通解

Answer 1

y'+x=√(x^2+y)
设y=x^2u
dy=2xudx+x^2du
2xudx+x^2du+xdx=x√(1+u)dx
2udx+xdu+dx=√(1+u)dx
xdu=[√(1+u)-2u-1]dx
du/[√(1+u)-2u-1] =dx/x
ln|x|=∫du/[√(1+u)-2u-1]
=∫2√(u+1)d√(u+1)/[√(1+u)-2√(1+u)^2+1]
=∫-2√(u+1)d√(u+1)[/(2√(1+u)+1)(√(1+u)-1)]
=(-2/3)∫d√(u+1)/(2√(1+u)+1) -(2/3)d√(1+u)/(∫√(1+u)-1
=(-1/3)ln|2√(1+u)+1| -(2/3)ln|√(1+u)-1| +C
通解
ln|x|=(-1/3)ln|2√(1+y/x^2) +1| -(2/3)ln|√(1+y/x^2) -1| +C

Answer 2

设y=kx^2, 比较系数，
2k+1=sqrt(k+1),
4k^2+3k=0,
k=0, -3/4

Answer 3

不会哈哈哈