0
$\begingroup$

$$y'=\frac{x^3+y^2\sqrt{x^2+y^2}}{xy\sqrt{x^2+y^2}}$$

How should I approach this?

I thought about using $z=\sqrt{x^2+y^2}$ but it does not seems to help.

I have tried to simplify the equation to:

$$\frac{x^3+y^2\sqrt{x^2+y^2}}{xy\sqrt{x^2+y^2}}=\frac{x^3}{xy\sqrt{x^2+y^2}}+\frac{y^2\sqrt{x^2+y^2}}{xy\sqrt{x^2+y^2}}=\frac{x^2}{y\sqrt{x^2+y^2}}+\frac{y}{x}$$

And then use $z=\frac{y}{x}$

  • 0
    Are you trying to solve this differential equation? If so, change the title from "Evaluate..." to "Solve...". – 2017-01-23
  • 0
    @SeanRoberson Done – 2017-01-23
  • 1
    @gbox: I would try $y = v x$ at first glance or perhaps polar coordinates as a second approach. – 2017-01-23
  • 0
    set $x=\cos(\phi(t)),y=\sin(\phi(t))$ – 2017-01-23

2 Answers 2

4

If you take $y=vx$ $$\frac{x^3+y^2\sqrt{x^2+y^2}}{xy\sqrt{x^2+y^2}}=\frac{x^3}{xy\sqrt{x^2+y^2}}+\frac{y^2\sqrt{x^2+y^2}}{xy\sqrt{x^2+y^2}}=\frac{x^2}{y\sqrt{x^2+y^2}}+\frac{y}{x}\\\frac{x^2}{y.x.\sqrt{1+\dfrac{y^2}{x^2}}}+\frac{y}{x}=\\\frac{x}{y\sqrt{1+\dfrac{y^2}{x^2}}}+\frac{y}{x}=\\\dfrac{\frac1v}{\sqrt{1+v^2}}+v$$ so now $$y'=v+xv'=\dfrac{\frac1v}{\sqrt{1+v^2}}+v$$ $v$ cancel from both sides and there is no hard operation ...after

1

The equation has the form $P(x,y)dx +Q(x,y)dy=0$ with $P,Q$ homgeneous funcions of the same degree ($3$ in this case). So, $y=vx$ transforms the equation to one of separate variables.

Edit. See also Khosrotash's answer.