austinL
  • austinL
Solve the given differential equation. \(y\prime = \dfrac{(3x^2-1)}{3+2y)}\) Where would I begin?
Mathematics
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
As simple as this:\[y'=\frac{ dy }{ dx }=\frac{ 3x^2-1 }{ 3+2y } \rightarrow (3+2y)dy=(3x^2-1)dx\]You can integrate
austinL
  • austinL
Oh my goodness. That makes gobs more sense.
austinL
  • austinL
Ok, so now I have: \(y^2+3y+C=x^3-x+C\) Now, I haven't a clue what to do next, combine like terms and solve for y?

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austinL
  • austinL
\(\large{y=\pm \dfrac{-(\sqrt{4x^3-4x+9}+3)}{2}}\) That makes absolutely zero sense.
austinL
  • austinL
+C on the end of course....
Loser66
  • Loser66
lol!!! Actually, you don't need to find y out, just let it there as y ^2 +3y -x^3 +x = C that's it
austinL
  • austinL
Many typos there. \(y^2+3y-x^3+x=C~,~y\ne\frac{-3}{2}\)
Loser66
  • Loser66
it's ok, looks good to me
austinL
  • austinL
Woohoo, very nice!
Loser66
  • Loser66
hahahaha... glad to see "Woohoo..." wudwud... You are gud
anonymous
  • anonymous
If you leave the solution like: \[3y+y^2-x^3+x=C\]is perfectly valid

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