Determining a solution by inspection of a DE

[aaronfue]

Determine, by inspection, at least two solutions of the given IVP.

y'= 3y^2/3


Would someone kindly provide an explanation that makes more sense than my textbook?

I'm sure that one answer is y=0. How can I quickly identify another?

Thanks for the help.

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[MarkFL]

It's fairly easy to see that \(\displaystyle y=x^3\) is another solution.

 

[DrPhil]

Determine, by inspection, at least two solutions of the given IVP.

y'= 3y^2/3

If this is an IVP, what was the given initial value? If it was y(0)=0, then everything is happy.

I hope that "by inspection" allows you to scribble some notes on the back of an envelope - I had to imagine the separation of variables and get that pesky "3" in the right place before I could agree with MarkFL!

 

[HallsofIvy]

Determine, by inspection, at least two solutions of the given IVP.

y'= 3y^2/3


Would someone kindly provide an explanation that makes more sense than my textbook?

I'm sure that one answer is y=0. How can I quickly identify another?

Thanks for the help.

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It may not be "by inspection" or as quick as you would like but we can separate that as
\(\displaystyle \frac{dy}{y^{2/3}}= 3dx\) and integrating both sides
\(\displaystyle 3y^{1/3}= 3x+ C\) so that \(\displaystyle y= (x+ c)^3\) (c= C/3)
in particular, taking c= 0, \(\displaystyle y= x^3\).