Quick question about Differential Equations.

ozone
Messages
121
Reaction score
0
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.
 
Physics news on Phys.org
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.
 
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).
 
ozone said:
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.

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.
 
matphysik said:
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.


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.
 
Thread 'Direction Fields and Isoclines'
I sketched the isoclines for $$ m=-1,0,1,2 $$. Since both $$ \frac{dy}{dx} $$ and $$ D_{y} \frac{dy}{dx} $$ are continuous on the square region R defined by $$ -4\leq x \leq 4, -4 \leq y \leq 4 $$ the existence and uniqueness theorem guarantees that if we pick a point in the interior that lies on an isocline there will be a unique differentiable function (solution) passing through that point. I understand that a solution exists but I unsure how to actually sketch it. For example, consider a...
Back
Top