Please help me with the following problem. Determine by inspection, two solutions of the given initial-value-problem.

y'=3y^(2/3)
y(0)=0

Question

Please help me with the following problem.  Determine by inspection, two solutions of the given initial-value-problem.

y'=3y^(2/3)
y(0)=0

babyslapmafro · Answer

$$y'=3y ^{2/3}$$
y(0)=0

anonymous · Answer

trivial solution... y = 0
the other solution by inspections would be a bit difficult for me.  been a while since i studied diffy qs.

anonymous · Answer

well consider the power rule... if we had $y=x^n$ we'd get $y'=nx^{n-1}=\dfrac{n}xy$

anonymous · Answer

so here we notice $$3y^{2/3}=3y^{-1/3}y=\frac3{y^{1/3}}y$$ so we can identify $3=n,x=y^{1/3}$ so $y=x^3$ ;-)

babyslapmafro · Answer

I'm having trouble with this for some reason...I'll probably have to go to the tutoring center because I don't really understand what I'm trying to find and I don't follow what you did either...

anonymous · Answer

the whole point is to look at it and guess a nice function -- I gave a huge tip by suggesting the power rule.

anonymous · Answer

he's just using the power rule and substituting.

anonymous · Answer

y'=3y^-2/3,  y(0)=0
$$\frac{ dy }{dx }=3y ^{\frac{ 2 }{ 3 }},separating the variables$$
$$\frac{ dy }{y ^{\frac{ 2 }{3 }} }=3dx$$
integrating both sides
$$\int\limits y ^{\frac{ -2 }{3 }}dy=3x+c ,$$
solve it and find the value of c,given y=0,when x=0