# Solution for a first-order differential equation

1. Mar 6, 2017

### toni556

1. The problem statement, all variables and given/known data
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
3. 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: Mar 6, 2017
2. Mar 6, 2017

### haruspex

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.

3. Mar 6, 2017

### toni556

Thanks I corrected them (it was only typing error)

4. Mar 6, 2017

### haruspex

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?

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted