tut 5103 quiz1

[YUTONG ZHANG]

y^'=(x^2+3y^2)/2xy=x/2y+3y/2x=1/2 (y/x)^(-1)+3/2 (y/x)
v=y/x
v^' x+v=y'
v^' x+v=1/2 v^(-1)+3/2 v
v^' x=1/2 v^(-1)+1/2 v=(1+v^2)/2v
Equation is homogeneous
2v/(1+v^2 ) dv=1/x dx
ln⁡|v^2+1|=ln⁡〖|x|+c〗
v^2+1=cx
y^2/x^2 +1-cx=0
y^2+x^2-〖cx〗^3=0