anonymous
  • anonymous
How do I solve a differential equation if the equation becomes undefined when I plug in the initial value?
Mathematics
schrodinger
  • schrodinger
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anonymous
  • anonymous
heyllo joseph hru?
anonymous
  • anonymous
Show the equation :D If it's undefined at the initial value, then you probably integrated incorrectly :D
anonymous
  • anonymous
dx/dy = (y-1)^2

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anonymous
  • anonymous
x = 1/3(y-1)^3 + C What are the initial conditions?
anonymous
  • anonymous
y(0) = 1
anonymous
  • anonymous
C = 0
anonymous
  • anonymous
x = 1/3 (y-1)^3
anonymous
  • anonymous
can you show your work in getting the solution to the differential equation?
anonymous
  • anonymous
y(0) = 1 x = 0 y = 1 0 = 1/3(1-1)^3 +C 0 = 0 +C C = 0
anonymous
  • anonymous
excuse me! dy/dx =...
anonymous
  • anonymous
Oh:\[dx/dy = (y-1)^2\] \[\int\limits_{0}^{x}x dx = \int\limits_{0}^{y} (y-1)^2 dy\] Take the antiderivative (in this case, it works) :D \[x = (y-1)^3/3\]
anonymous
  • anonymous
+C
anonymous
  • anonymous
DON'T FORGET THE +C
anonymous
  • anonymous
k
anonymous
  • anonymous
excuse me, i made a mistake, the dy and dx are flipped from what I wrote. Is it OK?
anonymous
  • anonymous
No, then the equation is completely different: \[dy/dx = (y-1)^2\] Limits using initial conditions \[\int\limits_{1}^{y} dy/(y-1)^2 = \int\limits_{0}^{x} xdx\] \[\int\limits_{1}^{y} (y-1)^{-2} dy = x\] Antidifferentiate again:\[-(y-1)^{-1} - - (0)^{-1} = x \] No solution for this diff eq.

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