How do I solve a differential equation with the given initial condition?

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Discussion Overview

The discussion revolves around solving a differential equation of the form dy/dx = (3x^2)(e^-y) with a given initial condition y(0)=1. Participants explore methods for separation of variables, integration, and the implications of the initial condition on the solution.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Homework-related
  • Debate/contested

Main Points Raised

  • One participant presents the equation dy/dx = (3x^2)(e^-y) and seeks assistance on how to proceed after separating variables.
  • Another participant suggests integrating both sides after rewriting the equation as e^y dy = 3x^2 dx.
  • A subsequent reply proposes that the solution could be y = ln(x^3 + C), but this is met with uncertainty.
  • Further discussion clarifies the application of the initial condition y(0)=1, leading to the conclusion that C should equal e, resulting in y = ln(x^3 + e).
  • Another participant raises a different differential equation dy/dx = 4(2y-1) and suggests a solution form, prompting a correction about the integration process.
  • There is a playful exchange regarding the username "stupidmath," with some participants correcting the spelling and expressing appreciation for the help provided.
  • One participant cautions against a misunderstanding of logarithmic properties in the context of the solution.

Areas of Agreement / Disagreement

Participants generally agree on the method of separation of variables and the integration process, but there are disagreements regarding the interpretation of the initial condition and the correct form of the solution. The discussion remains unresolved on some aspects, particularly concerning the handling of logarithmic identities.

Contextual Notes

Some participants express uncertainty about the placement of parentheses in the solution and the implications of logarithmic properties, indicating potential limitations in their understanding of these mathematical concepts.

Who May Find This Useful

Students and individuals interested in differential equations, particularly those seeking clarification on separation of variables and initial conditions in solving such equations.

Mitchtwitchita
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dy/dx = (3x^2)(e^-y)

1/(e^-y) dy = 3x^2 dx
1/(e^-y) = x^3 + C

Can anybody please give me any help or advice on to where to go from here? I'm having a really tough time with this one.
 
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[tex]\frac{dy}{dx}=3x^{2}e^{-y}[/tex]

this diff. eq. can be solved with separation of variables.So:

[tex]e^{y}dy=3x^{2}dx[/tex] now integrate both sides and solve for y.
 
So, y = ln(x^3+C)?
 
Mitchtwitchita said:
So, y = ln(x^3+C)?

Looks like it is right.
 
Thanks stupidmath! Now, if I wanted to solve, for example y(0)=1

ln (C) = y - ln(x^3)
ln (C) = 1 - 0
C = e^1 ?

y = ln(x^3) + e^1 ?
 
Well, to impose the initial condition y(0)=1, you actually here have that when x=0, y=1.

So it means:

[tex]1=ln(c)=>C=e[/tex]

so our particular solution would be

[tex]y=ln(x^{3}+e)[/tex]

You just misplaced the parentheses on your final answer.
 
Ah, thanks a bunch stupidmath!
 
One last question stupidmath. wouldn't dy/dx = 4(2y-1) end up being y = Ce^-32x?
 
It is not "stupidmath". Beware of his ire!

You should be able to do this one yourself now. If dy/dx= 4(2y-1) then dy/(2y-1)= 4dx.
Integrating on both sides (1/2)ln|2y-1|= 4x+ C so ln|2y-1|= 8x+ C' (C'= 2C). Then 2y-1= C"e^(8c).
 
  • #10
It is suttypud-math, not stupidmath!
 
  • #11
Oh, I apologize for the faux pas sutupidmath. I also appreciate the help, thanks.
 
  • #12
arildno said:
It is suttypud-math, not stupidmath!

Is there any meaning to this?:smile:
 
  • #13
Mitchtwitchita said:
Thanks stupidmath! Now, if I wanted to solve, for example y(0)=1

ln (C) = y - ln(x^3)
ln (C) = 1 - 0
C = e^1 ?

y = ln(x^3) + e^1 ?


Remember that ln(a + b) =/= ln(a) + ln(b), which is what you did here.
 

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