Solving an Initial Value Problem Analytically

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Discussion Overview

The discussion revolves around solving an initial value problem given by the differential equation $y'=\dfrac{1+3x^2}{3y^2-6y}$ with the initial condition $y(0)=1$. Participants explore the analytical solution, integration steps, and the implications of the solution's domain.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant presents the initial value problem and attempts to separate variables for integration.
  • Another participant agrees on the separation of variables and prompts for the integration step.
  • A participant claims to have integrated and found $y^3 - 3 y^2 = x^3 + x + c$, suggesting a specific value for the constant $c$ based on the initial condition.
  • Another participant discusses the implications of the solution, noting that the denominator $3y^2 - 6y$ leads to restrictions on the values of $y$, specifically that $y$ cannot equal 0 or 2.
  • One participant mentions observing a vertical slope at $x=1$ when graphing the solution.
  • A later reply clarifies that the condition $|x|< 1$ is tied to the initial condition at $x=0$ and discusses how different initial conditions would affect the domain of the solution.

Areas of Agreement / Disagreement

Participants generally agree on the steps taken to solve the problem and the restrictions on the solution's domain due to the nature of the differential equation. However, there is no consensus on the implications of the solution's domain for different initial conditions, as different scenarios are proposed without resolution.

Contextual Notes

The discussion highlights limitations related to the assumptions about the values of $y$ and the corresponding restrictions on $x$. The dependence on the initial condition and how it affects the domain of the solution is also noted but remains unresolved.

karush
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Solve the initial value problem
$y'=\dfrac{1+3x^2}{3y^2-6y},
\quad y(0)=1$

Solving analytically
$3y^2-6y\ dy = 1+3x^2 \ dx$

so far hopefully...
 
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Well, I would write it as $(3y^2- 6y)dy= (1+ 3x^2)dx$, but, yes, the differential equation can be "separated" so that y is on one side and x on the other.

Now, integrate both sides. Have you done that?
 
$y^3 - 3 y^2 =x^3 + x + c$

ok i think we can plug in $x=0$ and $y=1$ so
$1-3=0+0+c$
$c=-2$

the book ans was
$y^3−3y^2−x−x^3+2=0,\quad |x|<1$
but i don't kmow how they got the interval
 
Last edited:
That requires some careful thought! The original problem was $y'= \frac{1+ 3x^2}{3y^2- 6y}$. The denominator is $3y^2- 6y= 3y(y- 2)$ so y cannot be 0 or 2. You have, correctly, that $y^3- 3y^2- x^3- x+ 2= 0$. If y= 0, that is $-x^3- x+ 2$ which is 0 when x= 1. If y= 2, that is $8- 12- x^4- x+ 2= -x^3- x- 2$ which is 0 when x= -1. Since y cannot be 0 or 2, x cannot be -1 or 1.
 
Last edited:
mahalo
very helpful

when I graphed it i noticed vertical slope at 1 but,,,
 
Notice that the specific condition, |x|< 1, depends also on the fact that the initial condition is at x=0 which is between -1 and 1.

If we had exactly the same differential equation with condition that y(2) be a given value, since 2> 1, we would have to restrict the solution to "x> 1".

If we had exactly the same differential equation with condition that y(-3) be a given value, since -3< -1, we would have to restrict the solution is "x< -1".
 

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