I am solving a homogeneous equation $\frac{dy}{dx}= \frac{x^2+xy+y^2}{x^2}$

and have come to this step and I'm stuck now with the integration. I could really use some helpful hints to help me $$ \int \frac{v^2}{1+v+v^2-v^3}dV = \int \frac1x dX = \ln|x| + C $$

  • $\begingroup$ OK, I am going to register my distaste with your acceptance of the answer provided. You asked for hints, which I provided, even though I did work the problem out. Someone else provided a full answer. If that is what you wanted, you should have said so. $\endgroup$ – Ron Gordon Jul 1 '15 at 18:07
  • $\begingroup$ @RonGordon I'm sorry you feel that way. I didn't pick the answer because it had an answer. It just helped me realize what I did wrong in my work. I followed the work and found that when I was subtracting v from both sides I wasn't cancelling the v but finding a common denominator. In the work below they cancelled the v. It helped me find the utterly stupid mistake I was making. $\endgroup$ – Ayoshna Jul 8 '15 at 18:27

$$v=\frac{y}{x}$$. $$y=vx$$. $$\frac{dy}{dx}=v+x\frac{dv}{dx}$$.$$v+x\frac{dv}{dx}=1+v+v^{2}$$ $$x\frac{dv}{dx}=1+v^{2}$$ so $$\frac{dv}{1+v^{2}}=\frac{dx}{x}$$ Hence $\arctan(v)=\ln \vert x \vert +C$


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