Question

if dy/dx = 2y^2 and if y=-1 when x=1, then when x=2, y=

Answer 1

I know the answer if that helps.. -1/3 but I need to learn how to get there

Answer 2

@Jhannybean help please

Answer 3

getting anywhere?

Answer 4

with wolfram I get
http://www.wolframalpha.com/input/?i=solve+dy%2Fdx+%3D+2y^2%2C++y%281%29+%3D+-1

The particular solution is

y = 1 / (1 - 2x)

y(2) = 1 / ( 1 - 2*2 ) = 1 / (1 - 4) = 1 / -3 = -1/3

Answer 5

@Jhannybean
on the fourth line you introduced an x

going over steps

dy/(2y^2) = dx
is correct

the next step should be

1/2 * integral dy / y^2 = integral dx

Answer 6

Oh I see, I see.

Answer 7

Let me fix that.

Answer 8

do you have to write all that over again? would be nice if we could just edit past posts

Answer 9

or do you use a notepad file before you submit

Answer 10

the quote feature :)

Answer 11

if its in Latex you can quote it ? i was wondering about that, i saw people quoting in red

Answer 12

Nvm, already covered.

Answer 13

yep :)

Answer 14

Yeah, I arbitrarily added it in by mistake.

Answer 15

no biggie. Errors, typos, errata, etc. are sometimes useful to learn from

Answer 16

\[\begin{align}
\&:~\frac{dy}{2y^2} = dx
\\&:~\int \frac{dy}{2y^2} = \int dx
\\&:~\frac{1}{2}\int \frac{1}{y^2}dy = x + C
\\&:~\frac{1}{2} \int y^{-2}dy =x+C
\\&: ~ -\frac{1}{2y} =x+C \ \ \ \sf \text{when x = 1, y = -1}
\\&: ~ \frac{1}{2} = 1 + C
\\&: ~ \frac{1}{2}-1 = C
\\&: ~ -\frac{1}{2} = C \\
\\&: ~ -\frac{1}{2y} = 2-\frac{1}{2}
\\&: ~ -\frac{1}{2y} = \frac{3}{2}
\\&:~ -2y =\frac{2}{3}
\\&:~ \boxed{y=-\dfrac{1}{3}}
\end{align}\]

Answer 17

There we go.

Answer 18

^^