anonymous
  • anonymous
Solve the equation for y in terms of x: 3y + 2y^2 = -e^(-x) - e^(x) + 7
Mathematics
katieb
  • katieb
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dinamix
  • dinamix
its not differential equation ?
anonymous
  • anonymous
Ill post the entire question and maybe that'll help
anonymous
  • anonymous
Find the solution to the initial value problem in explicit form \[y' = \frac{e^{-x} - e^{x}}{3 + 4y}\]

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anonymous
  • anonymous
y(0) = 1
anonymous
  • anonymous
freckles
  • freckles
y'*y doesn't equal y^2
freckles
  • freckles
\[(3+4y) dy=(e^{-x}-e^{x} )dx\] equation is seperable integrate both sides
freckles
  • freckles
oh wait are you saying you think you found the solution to that?
anonymous
  • anonymous
yes i believe so and now i need it in explicit form
anonymous
  • anonymous
I believe i have found the correct value for the constant c given the initial value as well
freckles
  • freckles
your solution is correct
freckles
  • freckles
don't believe you can find explicit form
anonymous
  • anonymous
the answer in the back of the book has done it some how.
freckles
  • freckles
hmmm well you have a quadratic in terms of y
freckles
  • freckles
\[2y^2-3y+e^{-x}+e^{x}-7=0 \\ y=\frac{-(-3) \pm \sqrt{(-3)^2-4(2)(e^{-x}+e^{x}-7)}}{2(2)}\]
freckles
  • freckles
though the implicit form looks much better to me
anonymous
  • anonymous
that looks like the answer, how'd you do that?!
freckles
  • freckles
it is a quadratic in terms of y
freckles
  • freckles
all i did was use the quadratic formula
freckles
  • freckles
a=2 b=-3 c=e^(-x)+e^x-7
anonymous
  • anonymous
ohhhhh
anonymous
  • anonymous
i gotcha now
anonymous
  • anonymous
Thank you!!!!!!
freckles
  • freckles
np

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