# Math Help - Solve y'=y^2 subject to y(0)=yo.

1. ## Solve y'=y^2 subject to y(0)=yo.

So I need to solve y'=y^2 subject to y(0)=yo (ynaught)

So I know I can separate and get y=1/(C-x) or y=1/((1/yo)-x) after applying the condition.

My question is, why can't I just integrate to get y=y^3/3+C? Why do I have to separate?

The next question is to solve A'(t)=6sqrt(A)(9-A) with A(0)=1, and I can't figure out how I would separate this one because I'd get dA/(6sqrt(A)(9-A))=dt, but the LHS is impossible to integrate simply. So can I just take integral 6sqrt(A)(9-A) to do this?

2. Well, it is true that $\int y^2 dy$ = $\frac {\ y^3}{3} +C$, but that is not the answer to the first question you mention because the above integral is not what you are being asked to solve.

In y' = $y^2$, y' is in fact $\frac {\ dy}{dx}$, so you have to move $y^2$ with dy, and get;
$\frac {\ dy}{y^2}$=dx, and then, integrate on both sides : $\int \frac {\ dy}{y^2}$= $\int dx$

Trying to "simply integrate" as you suggest would amount to attempt to the following:
$\int \frac {\ dy}{dx} = \int y^2$
These two expressions are not defined.

3. Ok, I got that now thanks.

The thing is for A'(t)=6sqrt(A)(9-A) with A(0)=1 wouldn't I also have to separate. That gives dA/(6sqrt(A)(9-A))=dt. This is my primary concern...How would the LHS be possible to integrate. Someone told me I can just integrate 6sqrt(A)(9-A), but that wouldn't be separating variables...

My question is, why can't I just integrate to get y=y^3/3+C? Why do I have to separate?

Trying to "simply integrate" as you suggest would amount to attempt to the following:
$\int \frac {\ dy}{dx} = \int y^2$
These two expressions are not defined.
I think there may be a subtle point missed in here, so I'll add one more bit.

twittytwitter you want to integrate y' = y^2, the left side becoming (1/3)y^3 + C, but you aren't integrating over y...

$\displaystyle \frac{dy}{dx} = y^2$

$\displaystyle \int \frac{dy}{dx}~dx = \int y^2~dx$

Notice that we are integrating over x, not y. The LHS is now $\displaystyle \int \frac{dy}{dx}~dx = y + C$, but the RHS is a mess: $\int y^2~dx$

-Dan

Ok, I got that now thanks.

The thing is for A'(t)=6sqrt(A)(9-A) with A(0)=1 wouldn't I also have to separate. That gives dA/(6sqrt(A)(9-A))=dt. This is my primary concern...How would the LHS be possible to integrate. Someone told me I can just integrate 6sqrt(A)(9-A), but that wouldn't be separating variables...
$\dfrac{dA}{\sqrt{A}(9-A)} = 6 \, dt$

left side ... let $u = \sqrt{A}$

$du = \dfrac{dA}{2\sqrt{A}}$

$dA = 2u \, du$

substitute ...

$\dfrac{2u}{u(9-u^2)} \, du = 6 \, dt$

$\dfrac{2}{9-u^2} \, du = 6 \, dt$

$\dfrac{2}{(3-u)(3+u)} \, du = 6 \, dt$

use the method of partial fractions to integrate the left side