# It looks simple, but I can’t do it

1. Dec 2, 2007

### tomcenjerrym

Can anyone help me how to make the following equation each variable x and y together with dx and dy respectively?

$$y'=\frac {(4x^2 + y^2)} {xy}$$

2. Dec 2, 2007

### Xevarion

Try letting $$z=y/x$$ and creating a differential equation for $$z$$ using the one for $$y$$.

3. Dec 3, 2007

### HallsofIvy

Staff Emeritus
y'= dy/dx so this is
$$\frac{dy}{dx}= \frac{4x^2+ y^2}{xy}$$
$$xydy= (4x^2+ y^2)dx$$
$$-(4x^2+y^2)dx+ xydy= 0$$
Is that what you meant?

You could also let v= y/x and make the substitution y= xv. In that case, dy= xdv+ vdx and the equation becomes
$$(-4x^2+ x^2v^2)dx+ x(vx)(xdv+ vdx)= 0[/itex] [tex]x^2(v^2- 4)dx+ x^2vdv+ x^2v^2dx= 0[/itex] [tex]x^2(2v^2-4)dx+ x^2vdv= 0[/itex] As long as x is not 0 we can divide by $x^2$ [tex](2v^2-4)dx+ vdv= 0$$
$$dx+ \frac{v}{2v^2- 4}dv= 0$$
Is that what you mean?