Question

求解高数，求下列微分方程在给定的初始条件下的特解

Answer 1

let
u= y/x
du/dx = -y/x^2 + (1/x). dy/dx
= -(1/x)u + (1/x). dy/dx
x.du/dx =-u + dy/dx
dy/dx = u +x.du/dx
y' = y/x + tan(y/x)

u +x.du/dx = u + tanu
x.du/dx = tanu
∫du/tanu = ∫dx/x
∫ (cosu/sinu )du = ∫dx/x
ln|sinu| =ln|x| + C'
sinu = Cx
y(1) =π/4
C =√2/2

ie
sinu = (√2/2) x

sin(y/x) =(√2/2) x
y/x= arcsin[(√2/2) x]
y= x. arcsin[(√2/2) x]