# Quick question about Differential Equations.

1. Aug 3, 2011

### ozone

I was working on a simple differential problem which caused me some confusion

original eq =

dy/dx = -x/2y

which can easily be altered into

y dy = (-x/2) dx

This finally transforms into (after antiderivatives are determined)

(y^2/2) = -(x^2/4) + c

While I can see that this logically is equivallent to the problem above I just don't see how the answer was derived. The x side of the equation makes perfect sense to me, but what I can't tell is how the y was derived to be (y^2/2). In my mind it should have just been y...

That aside I can see that the transformation is logical, but I can't figure out how it was derived.

Thanks,
Ozone.

2. Aug 3, 2011

### HallsofIvy

Staff Emeritus
I don't see how you could say that! You say you see how $(x/2) dx$ gives $x^2/4+ c$, I presume by recognizing that
$$\frac{1}{2}\int xdx= \frac{1}{2}\left(\frac{1}{2}x^2\right)+ c= \frac{x^2}{4}+ c$$

Welll, the left side is not different:
$$\int y dy= \frac{1}{2}y^2+ c$$

$$\int y dy= \frac{1}{2}\int x dx$$
$$\frac{1}{2} y^2+ c_1= \frac{1}{4} x^2+ c_2$$

I have written the two constants differently because there is no reason to think they have to be the same. Now subtract $c_1$ from both sides:
$$\frac{1}{2}y^2= \frac{1}{4}x^2+ (c_2- c_1)= \frac{1}{4}x^2+ c$$
where "c" is just $c_2- c_1$.

3. Aug 4, 2011

### ozone

Thanks hallsofivy this is exactly what I was looking for. Also I will be sure to use the forums coding next time :). (I responded to your other post after I made this post).

4. Aug 4, 2011

### matphysik

Set F(x,y)≡-x/2y then 0=dF=(∂F/∂x)dx+(∂F/∂y)dy and, -(∂F/∂x)/(∂F/∂y)=dy/dx. Reversing the steps should answer your question.

5. Aug 6, 2011

### matphysik

I was not able to correct my rough work above, so i shall do it now. You (ozone) say that, "I can't figure out how it was DERIVED".

The derivation follows from simple multivariable calculus. Given dy/dx=-x/2y, rewrite it as 2ydy+xdx=0. Next, introduce F∈C¹(ℝ²) and set F(x,y)=constant (level curves), so that 0=dF=∂F/∂ydy+∂F/∂xdx.

Identify ∂F/∂y=2y (*), and ∂F/∂x=x (**). From (*), F(x,y)=y²+c₁(x) and from (**), F(x,y)=½x²+c₂(y). The obvious choice is, F(x,y)=½x²+y²+constant.