工程數學問題

gaexp254432

xy'=y^2+y  令(y/x=u)
xy'=y+2x^3sin^2(y/x) 令(y/x=u)
該如何解
謝謝各路大神

u06m4rmp4

xy'=y^2+y

x*dy/dx=y^2+y
dy/(y^2+y)=dx/x
兩邊積分
ln(y)-ln(y+1)=ln(x)
y/(y+1)=x
y=x/(1-x)

u06m4rmp4

xy'=y+2x^3sin²(y/x)

dy/dx=u+x*du/dx chain rule

x*u+x²*du/dx=x*u+2*x³*(sin(u))²
du/(sin(u))²=2*x dx
-1/tan(u)=x²
tan(u)=-1/x²)
u=arctan(-1/x²)
y=x*u=x*arctan(-1/x²)