Solution for a first-order differential equation

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Homework Help Overview

The discussion revolves around a first-order initial value problem (IVP) involving the differential equation dy/dx = 3y^(2/3) with the condition y(0) = 0. Participants are exploring potential solutions and the integration process involved in solving the equation.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to change the form of the equation and integrate both sides, leading to a proposed solution. Some participants question the correctness of the integration steps and suggest rewriting terms for clarity. Others inquire about the implications of dividing by expressions and the need to check conditions when doing so.

Discussion Status

The discussion is ongoing, with participants providing feedback on the integration process and pointing out potential errors. There is an emphasis on ensuring all conditions are considered, particularly regarding the initial value and the implications of division in the steps taken.

Contextual Notes

Participants are tasked with finding at least two solutions to the IVP, which adds complexity to the discussion. The original poster's approach raises questions about the handling of the initial condition and the integration process.

toni556
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Homework Statement


determine by inspection at least two
solutions of the given first-order IVP
dy/dx = 3y2/3
y(0)=0
2. Equations:
integral xa dx= xa+1/(a+1)+constant

The Attempt at a Solution


change its form to 1/y2/3 dy/dx =3
integrate both sides with respect to x
then it will be
1/y2/3 dy = 3dx
now integrate to get y1/3/(1/3)=3x+c
for y(0)=0 then c=0
then y1/3/(1/3) =3x then y=x3
Is that right? does I miss any condition or jumped over some steps?
 
Last edited:
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toni556 said:
integrate to get y1/3/3 =3x+c
A couple of errors there. Write 1/y2/3 as ya. What is the integral ya.dy?
toni556 said:
then y1/3/3 =3x then y=x3
That's wrong too, but happens to cancel one of the earlier errors.
 
haruspex said:
A couple of errors there. Write 1/y2/3 as ya. What is the integral ya.dy?

That's wrong too, but happens to cancel one of the earlier errors.
Thanks I corrected them (it was only typing error)
 
toni556 said:
Thanks I corrected them (it was only typing error)
That corrected one of the errors in the integration step. One remains. As I suggested, rewrite 1/y2/3 in the form ya before trying to integrate it.

You are asked for two solutions. Whenever you divide by an expression, what should you check for? At what step did you do a division?
 

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