# [SOLVED] Differentiation

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• January 25th 2009, 01:50 PM
ronaldo_07
[SOLVED] Differentiation
Solve the initial value problem y′= $\frac{x^2}{2y + 1}$, y(0)=−1 by separation of variables.
• January 25th 2009, 01:51 PM
Jhevon
Quote:

Originally Posted by ronaldo_07
Solve the initial value problem y′= $\frac{x^2}{2y + 1}$, y(0)=−1 by separation of variables.

separate the variables

$(2y + 1) y' = x^2$

$\Rightarrow \int (2y + 1)~dy = \int x^2 ~dx$

now finish up
• January 25th 2009, 02:02 PM
ronaldo_07
2yx+x= $\frac{x^3}{3}$ Then rearange to make y=....... then let y{0)=-1 to find the solution right?
• January 25th 2009, 02:12 PM
Jhevon
Quote:

Originally Posted by ronaldo_07
2yx+x= $\frac{x^3}{3}$ Then rearange to make y=....... then let y{0)=-1 to find the solution right?

??

you do realize that on the left side you are integrating with respect to y, right? and what about your +C? you'll need it
• January 25th 2009, 02:31 PM
ronaldo_07
With respect to y, would that be y^2 + y
• January 25th 2009, 02:34 PM
Jhevon
Quote:

Originally Posted by ronaldo_07
With respect to y, would that be y^2 + y

yes
• January 25th 2009, 02:38 PM
ronaldo_07
$y^2+y=\frac{x^3}{3}+c$ Im not sure how to get y on its own at this stage
• January 25th 2009, 02:52 PM
Jhevon
Quote:

Originally Posted by ronaldo_07
$y^2+y=\frac{x^3}{3}+c$ Im not sure how to get y on its own at this stage

you have a quadratic equation in y. it's hard to factor of course, but you can always complete the square, or use the quadratic formula (here, your constant term is $-(x^3/3 + C)$)
• January 25th 2009, 03:10 PM
ronaldo_07
Im still not able to solve for c using quadratic formulae
• January 25th 2009, 03:13 PM
Jhevon
Quote:

Originally Posted by ronaldo_07
Im still not able to solve for c using quadratic formulae

you are not to use the quadratic formula to solve for C, you use y(0) = -1 to do that. the quadratic formula is to solve for y
• January 25th 2009, 03:22 PM
ronaldo_07
c=1? do i need to solve for y to answer this question?
• January 25th 2009, 03:26 PM
Jhevon
Quote:

Originally Posted by ronaldo_07
c=1? do i need to solve for y to answer this question?

actually, no, C is not 1

you do not need to solve for y to find C, however, when a problem says "solve the differential equation", they usually expect you to end up with y = ... or if x is your output variable, x = ...
• January 25th 2009, 03:28 PM
ronaldo_07
can you help im not sure how to finish this
• January 25th 2009, 03:31 PM
Jhevon
Quote:

Originally Posted by ronaldo_07
can you help im not sure how to finish this

y(0) = -1 means when x = 0, y = -1. with this you can find C

the quadratic formula says the solution to a polynomial equation of the form $ay^2 + by + c = 0$ is $y = \frac {-b \pm \sqrt{b^2 - 4ac}}{2a}$
• January 25th 2009, 03:40 PM
ronaldo_07
when I got c=1, I got it because I let -1=Left side $(y^2$+y). So I guess I should have done $(-1)^2$-(-1)= $\frac{(0)}{3}^3$+C. But this would give C to be 0
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