Quick question about Differential Equations.

In summary: Therefore, setting 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.
  • #1
ozone
122
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.
 
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  • #2
I don't see how you could say that! You say you see how [itex](x/2) dx[/itex] gives [itex]x^2/4+ c[/itex], I presume by recognizing that
[tex]\frac{1}{2}\int xdx= \frac{1}{2}\left(\frac{1}{2}x^2\right)+ c= \frac{x^2}{4}+ c[/tex]

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

[tex]\int y dy= \frac{1}{2}\int x dx[/tex]
[tex]\frac{1}{2} y^2+ c_1= \frac{1}{4} x^2+ c_2[/tex]

I have written the two constants differently because there is no reason to think they have to be the same. Now subtract [itex]c_1[/itex] from both sides:
[tex]\frac{1}{2}y^2= \frac{1}{4}x^2+ (c_2- c_1)= \frac{1}{4}x^2+ c[/tex]
where "c" is just [itex]c_2- c_1[/itex].
 
  • #3
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
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.
 
  • #5
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.
 

1. What are differential equations?

Differential equations are mathematical equations that describe how a quantity changes over time. They involve the use of derivatives, which represent the rate of change of a variable.

2. Why are differential equations important?

Differential equations are used to model and analyze many real-world phenomena, including population growth, chemical reactions, and electrical circuits. They are also essential in physics, engineering, and other scientific fields.

3. What are the types of differential equations?

There are several types of differential equations, including ordinary differential equations, partial differential equations, and stochastic differential equations. Ordinary differential equations involve one independent variable, while partial differential equations involve multiple independent variables. Stochastic differential equations also take into account random factors.

4. How are differential equations solved?

The solution to a differential equation is a function that satisfies the equation. There are various techniques for solving differential equations, including separation of variables, variation of parameters, and using numerical methods.

5. What are the applications of differential equations?

Differential equations have various applications in mathematics, physics, engineering, economics, and other fields. They are used to model and predict the behavior of complex systems, making them a vital tool in scientific research and problem-solving.

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