- #1
edgarpokemon
- 93
- 1
I have been stuck in this problem for two days now. I am starting DE this semester and I want to move ahead. so for this problem I attempted to use x=yv, then v=x/y. so I move dy/dx to the other side of the equation and divide by 2xy both sides, which leads to (3x^2-2y^2)/(2xy)=dx/dy. then if x=vy, then (3x^2-2y^2)/(2xy)=(dv/dy)y+v. then, (3/2)(x/y) -(y/x) -v=(dv/dy)y. then [(3/2)v-(1/v)-v]=(dv/dy)y. There will be a common demonimator of 2v on the right of the equation, leading to (3v^2-2-2v^2)/(2v)=(dv/dy)y. then after algebra, dy/y=(2v/(v^2-2))dv. Integrating leads to y=v^2-2, which is y=(x^2/y^2)-2. but in my book the answer comes with an additional factor of two in the answer. so the right answer should be y=(x^2/2y^2)-2. I found that if divide by only xy on both side of the equations, and later dividing by two when integrating, it will lead me to that right answer, but I don't understand why. there is a similar problem , with the expection that there is no 2 in xy. and I got that one right, but not this one. help please!