Seperable Equations using a substitution for a differential equation

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SUMMARY

The differential equation y' = (y+2x)/(y-2x) is not separable in its original form. However, by using the substitution u = y/x, the equation transforms into a separable form. The new equation allows for the application of separation of variables, leading to a solvable differential equation. The key step involves substituting y' with u' and x, ensuring that y' does not remain in the equation after the substitution.

PREREQUISITES
  • Understanding of differential equations and their classifications
  • Familiarity with substitution methods in solving differential equations
  • Knowledge of the product rule for differentiation
  • Basic algebraic manipulation skills
NEXT STEPS
  • Study the method of substitution in differential equations
  • Learn about separable differential equations and their solutions
  • Explore the product rule in calculus for differentiating products of functions
  • Practice solving various types of differential equations using different substitution techniques
USEFUL FOR

Students studying differential equations, mathematics educators, and anyone looking to deepen their understanding of substitution methods in solving complex differential equations.

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Homework Statement



The d.e

y' = (y+2x)/(y-2x)

is NOT seperable, but if you use a substitution then you obtain a new d.e involving x and u, then the new d.e is seperable... Solve the original d.e by using this change of variable method

Homework Equations


I'm going to use the substitution that u=y/x in the form y=ux


The Attempt at a Solution



y' = (y+2x)/(y-2x)
let y=ux
then

y' = (ux+2x)/(ux-2x)
y'(ux-2x) - (ux+2x) = 0 <--- thus it is seperable
So I can say

d/dx((ux^2)/2 - x^2) - d/dx((ux^2/2) + x^2) = 0

Putting it all together

d/dx[(ux^2)/2 - x^2 - (ux^2)/2 - x^2] = 0
d/dx (-2x^2) = 0

Well that is where I am stuck, how do I solve it from there and what am I trying to get because I've changed the d.e so I'm not sure how the answer from the new d.e will help me find a solution to the original one?
 
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If you are going to substitute y=ux then you also need to substitute y'=(ux)'=u'x+u. That will change the differential equation for y into one for u. You shouldn't have a y' hanging around after the substitution.
 

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