# Thread: [SOLVED] Differentiation

1. ## [SOLVED] Differentiation

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

2. 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

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

4. 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

5. With respect to y, would that be y^2 + y

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

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

8. 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 $-(x^3/3 + C)$)

9. Im still not able to solve for c using quadratic formulae

10. 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

11. c=1? do i need to solve for y to answer this question?

12. 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 = ...

13. can you help im not sure how to finish this

14. 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 $y = \frac {-b \pm \sqrt{b^2 - 4ac}}{2a}$

15. 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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