Need help with inital value problem: y'=-y^2 and y(0)=1/2
- Thread starter Pliuo
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What have you been taught about problems like this? What are your thoughts? What have you done so far? Please show us your work even if you feel that it is wrong so we may try to help you. You might also read
http://www.freemathhelp.com/forum/threads/78006-Read-Before-Posting
Hint: if dy/dx = -y2 then y-2 dy = - dx
Jedi Equester
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Think of seperation of variables 
.
\(\displaystyle y' = -y^2 \quad|: (y^2)\)
\(\displaystyle \int \frac{1}{y^2}\;dy = \int-1\;dx\)
.\(\displaystyle y' = -y^2 \quad|: (y^2)\)
\(\displaystyle \int \frac{1}{y^2}\;dy = \int-1\;dx\)
Everything looks fine to me in your workings. A further check of the answer reveals...
\(\displaystyle y\left(x\right)=\frac{1}{x+2}\)
\(\displaystyle y'\left(x\right)=-\frac{1}{\left(x+2\right)^2}=-\left(\frac{1}{x+2}\right)^2\)
\(\displaystyle y'\left(x\right)=-y^2\)
You get back to the original form, so everything checks out.
\(\displaystyle y\left(x\right)=\frac{1}{x+2}\)
\(\displaystyle y'\left(x\right)=-\frac{1}{\left(x+2\right)^2}=-\left(\frac{1}{x+2}\right)^2\)
\(\displaystyle y'\left(x\right)=-y^2\)
You get back to the original form, so everything checks out.
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Generally, I believe one puts the constant only on one side of the equation:
. . . . .\(\displaystyle -\dfrac{1}{y}\, =\, -x\, +\, C\)
Check, by plugging this back into the original exercise.
. . . . .\(\displaystyle -y^2\, =\, \dfrac{-1}{(x\,+\, 2)^2}\)
. . . . .\(\displaystyle \dfrac{dy}{dx}\, =\, \dfrac{d((x\,+\, 2)^{-1})}{dx}\, =\, (-1)(1\, +\, 0)(x\, +\, 2)^{-2}\, =\, \dfrac{-1}{(x\, +\, 2)^2}\)
Do they match?