Babyslapmafro
  • Babyslapmafro
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
Mathematics
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katieb
  • katieb
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Babyslapmafro
  • Babyslapmafro
\[y'=3y ^{2/3}\] y(0)=0
anonymous
  • anonymous
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
  • anonymous
well consider the power rule... if we had \(y=x^n\) we'd get \(y'=nx^{n-1}=\dfrac{n}xy\)

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anonymous
  • anonymous
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
  • Babyslapmafro
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
  • anonymous
the whole point is to look at it and guess a nice function -- I gave a huge tip by suggesting the power rule.
anonymous
  • anonymous
he's just using the power rule and substituting.
anonymous
  • anonymous
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

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