Differential Equation (Homogeneous / scale-invariant

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SUMMARY

The discussion focuses on testing the scale invariance of the differential equation (x+y²)dy + ydx = 0. The user explores substitutions x = αx and y = αⁿx, ultimately determining that n = 1/2. The user then substitutes y = v√x but encounters confusion regarding the differentiation of dy. The correct expression for dy is established as dy = (v/(2√x))dx + √xdv, clarifying the differentiation process.

PREREQUISITES
  • Understanding of first-order differential equations
  • Familiarity with scale invariance in mathematical contexts
  • Basic differentiation techniques in calculus
  • Knowledge of substitution methods in differential equations
NEXT STEPS
  • Study the concept of scale invariance in differential equations
  • Learn about substitution methods in solving differential equations
  • Explore advanced differentiation techniques, particularly in the context of variable substitutions
  • Review the theory behind homogeneous functions and their applications in differential equations
USEFUL FOR

Students in mathematics or physics courses, particularly those studying differential equations and mathematical methods, as well as educators seeking to clarify concepts of scale invariance and differentiation.

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Homework Statement


Test the following equation to show that they are scale invariant. Find their general solutions (It is not necessary to do the anti-derivative.)

(x+y^2)dy+ydx=0

I believe what my tutorial wants me to do is to check for homogeneity. I'm not sure though. This is not a Differential equations class, it is a math methods in physics class. The tutorial we use titles this section "Scale-invariant first-order differential equations".

The Attempt at a Solution



First I make the following substitutions:

x=\alpha{x}

y=\alpha^{n}x

I then use the substitutions in the DiffEq and solve for n so that the weight of each term is equal. I find that:

n=\frac{1}{2}

Based off that, I now make the following substitution:

y=v\sqrt{x}

From there things begin to get messy. I am unsure about 1 step so far: I am not sure what dy equates to. In the case that n=1, dy = vdx+xdv. I am not sure how to get dy using n=1/2.

Is it:

dy= vdx+{\sqrt{x}}dv

or:

dy= \frac{v}{2\sqrt{x}}+{\sqrt{x}dv

Or something else completely?

I guess I am asking what is dx in the equations above? Is it the derivative of x^1/2 or is it just dx?
 
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So it was something else completely. It is:

dy = \frac{v}{2\sqrt{x}}dx+{\sqrt{x}dv

I think I misunderstood basic differentiation notation.
 

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