Solution for a first-order differential equation

[toni556]

Homework Statement

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

The Attempt at a Solution

change its form to 1/y^2/3 dy/dx =3
integrate both sides with respect to x
then it will be
1/y^2/3 dy = 3dx
now integrate to get y^1/3/(1/3)=3x+c
for y(0)=0 then c=0
then y^1/3/(1/3) =3x then y=x³
Is that right? does I miss any condition or jumped over some steps?

 

[haruspex]

integrate to get y^1/3/3 =3x+c

A couple of errors there. Write 1/y^2/3 as y^a. What is the integral y^a.dy?

then y^1/3/3 =3x then y=x³

That's wrong too, but happens to cancel one of the earlier errors.

 

[toni556]

A couple of errors there. Write 1/y^2/3 as y^a. What is the integral y^a.dy?

That's wrong too, but happens to cancel one of the earlier errors.

Thanks I corrected them (it was only typing error)

 

[haruspex]

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/y^2/3 in the form y^a 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?