anonymous
  • anonymous
Differential Equation Problem: How Can i solve the Initial value problem. I am not sure if to use integrating factor? Equation below.
Mathematics
schrodinger
  • schrodinger
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anonymous
  • anonymous
\[2y \prime -y^2 + 4 =0, y(0) = 0\]
amistre64
  • amistre64
what methods are available to you?
anonymous
  • anonymous
integrating factory and separable....

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amistre64
  • amistre64
2 dy/dx = y^2-4 looks separable to me ... not sure if it makes life easier tho
anonymous
  • anonymous
Yep, i can work with that.
anonymous
  • anonymous
I'll post the result soon.
anonymous
  • anonymous
It does not. It is complex.
anonymous
  • anonymous
|dw:1443368763254:dw|
Loser66
  • Loser66
Use partial fraction.
Loser66
  • Loser66
|dw:1443368964936:dw|
anonymous
  • anonymous
I'm using that, A = -1/2 and b = 1/2
anonymous
  • anonymous
is it safe to say....|dw:1443369358723:dw|
anonymous
  • anonymous
but as y (0) = 0. I noticed that the equation initial condition will bring take the solution to.... e^(2c) = -1, but c does not exist since the ln(-1) is undefined.
amistre64
  • amistre64
as per the wolf ... might help you see where to go.
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amistre64
  • amistre64
A and B are +- 1/4 right?
amistre64
  • amistre64
i ignored the 2 lol
amistre64
  • amistre64
dy/(y^2-4) = dx/2 ln((y-2)/(y+2))/4 = x/2 +C ln((y-2)/(y+2)) = 2x +4C (y-2)/(y+2) = e^(2x+4C) = C e^(2x) x=0,y=0 -1 = C
amistre64
  • amistre64
y-2 = -e^(2x) (y+2) y+ye^(2x) = -2e^(2x)+2 y(1+e^(2x)) = -2(1-e^(2x)) ...
amistre64
  • amistre64
mighta factored the rhs off by a negative in my head
anonymous
  • anonymous
Thanks all.

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