Differential Equation Exact Solution

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SUMMARY

The discussion focuses on solving the differential equation (2x - y) + (2y - x) dy/dx = 0 with the initial condition y(1) = 3. The solution involves finding the implicit function x^2 - yx + y^2 = c, where c is determined to be 7. The participants explore the transformation to the explicit form y = x + sqrt(28 - 3x^2)/2, clarifying the steps to derive this from the quadratic equation y^2 + (-x)y + (x^2 - 7) = 0. The discussion emphasizes the importance of correctly applying the quadratic formula to find the roots for y.

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  • Understanding of first-order differential equations
  • Familiarity with implicit and explicit functions
  • Knowledge of quadratic equations and the quadratic formula
  • Basic calculus concepts, including derivatives and initial value problems
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  • Study the method of characteristics for solving first-order PDEs
  • Learn about implicit function theorem applications in differential equations
  • Explore the quadratic formula and its applications in solving equations
  • Investigate the stability and validity of solutions to differential equations
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Students and educators in mathematics, particularly those focusing on differential equations, as well as anyone seeking to deepen their understanding of implicit and explicit solutions in calculus.

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Homework Statement


(2x - y) + (2y-x) dy/dx = 0
y(1) = 3
Solve and determine where the solution is approximately valid.


Homework Equations





The Attempt at a Solution



Ux = 2x - y
Uy = 2y - x

U = x^2 - yx + h(y)
Uy = 2y -x = -x + h'(y)
h'(y) = 2y
h(y) = y^2

x^2 - yx + y^2 = c
1 - 3 + 3^2 = c = 7

I'm not really sure where to go from here,
x^2 - yx + y^2 = 7
However, the answer is quite a bit different:
y = x + sqrt( 28 - 3x^2)/2 and I can't quite see the step toward that. (x-y)^2 = 7-xy but I still can't seem to solve for y
 
Last edited:
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I have only read the last line of your post, so I don't know if your steps are right, but regarding solving for y in (x-y)^2 = 7-xy, notice that solving this equation for y(x) is the same as finding the roots of the quadratic equation y^2 + (-x)y + (x^2-7) = 0.
 
And, in this case "b2- 4ac"= (-x)2- 4(1)(x[sup[2]- 7)= x2- 4x2+ 28= 28- 3x2. That's where that came from.
 

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