austinL
  • austinL
Find the solution of the given initial value problem in explicit form. \(y\prime=(1-2x)y^2~,~y(0)=\frac{-1}{6}\)
Mathematics
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SOLVED
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jamiebookeater
  • jamiebookeater
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austinL
  • austinL
I am not even sure where to begin, it isn't in any form that I am familiar with.
Loser66
  • Loser66
\[\frac{dy}{y^2}= (1-2x)dx\]integral both sides
austinL
  • austinL
\(\large\int \dfrac{dy}{y^2}=?\) What? How the...... I feel silly now, but...

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More answers

Loser66
  • Loser66
yup, \[\int y^{-2}dy=?\]
austinL
  • austinL
Aha, Now I see. Yep, I feel silly.
Loser66
  • Loser66
ok,
Loser66
  • Loser66
@CarlosGP watch him out, please. I have to go somewhere to eat something. Cannot stay
austinL
  • austinL
\(\large{-\frac{1}{y}+C=x-x^2+C}\) This time I actually need to solve for y, yes?
anonymous
  • anonymous
If you put C on both sides they will cancel each other:. Put just one C and calculate it using y(0)=-1/6
austinL
  • austinL
\(\large{y(x)=\frac{1}{x(x-1)}+C}\)
anonymous
  • anonymous
either way is ok. leaving y(x) alone is not always possible
austinL
  • austinL
That can't be right.
anonymous
  • anonymous
now do x=0 and get the "C" that makes y(0)=-1/6
austinL
  • austinL
But... the denominator can't be zero. I am lost now.
anonymous
  • anonymous
.Because the solution is: \[y(x)=\frac{ 1 }{ x(x-1)+C }\]
anonymous
  • anonymous
It is what you get from: \[-\frac{ 1 }{ y }=x-x^2+C\]
austinL
  • austinL
\(\large{y(x)=\dfrac{1}{x(x-1)-6}}\) Is this correct?
anonymous
  • anonymous
that is ok. Now check that\[y'=(1-2x)y^2\]
austinL
  • austinL
Actually, \(\large{y(x)=\dfrac{1}{x^2-x-6}}\) And that is the solution that my text book gives! :D Thanks very much guys!
anonymous
  • anonymous
you are welcome

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