anonymous
  • anonymous
Differential equation: Solve the differential equation by regarding y as the independent variable rather than x. (1+2xy)(dy/dx)= 1+y^2
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
So i tried to put it into the format of \[\frac{ dy }{ dx }+P(x)*y=Q(x)\]
anonymous
  • anonymous
(switching basically the x and y components since y is independent) but i keep getting stuck
anonymous
  • anonymous
\[1+2xy=(1+y^2)*\frac{ dx }{dy }\] \[\frac{ dx }{ dy }-\frac{ 2xy }{ 1+y^2}=\frac{ 1 }{ 1+y^2}\]

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anonymous
  • anonymous
so then p(y) = \[\frac{ -2y }{ 1+y^2 }\] if i integrate that i get \[-\ln (y^2+1)\] Then to the e i get \[-(y^2+1)\] so then my equation becomes \[-(y^2+1)x'-2y*x=-1\]
anonymous
  • anonymous
but...i think i did something wrong
anonymous
  • anonymous
the answer should be(solving through the last equation i wrote) \[=1+y^2)x(y)= \frac{ 1 }{ 2 }[y+(1+y^2)(\tan^{-1} (y)+c)]\]
anonymous
  • anonymous
ignore the stuff infront of x(y)...typo
anonymous
  • anonymous
integrating through i get: \[-x(y^2+1)=-y+c\] \[x=\frac{ y+c }{ y^2+1 }\] ..which clearly is wrong
anonymous
  • anonymous
The integrating factor is almost right. You're correct with \[\mu(y)=e^{-\ln(y^2+1)}\] However, this is not equal to \(-(y^2+1)\): \[e^{-\ln(y^2+1)}=e^{\ln(y^2+1)^{-1}}=\frac{1}{y^2+1}\]
anonymous
  • anonymous
dang it i did that on an earlier problem too...alright ill try that real quick
anonymous
  • anonymous
got it, thank you
anonymous
  • anonymous
Yep, you're welcome

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