MHB Solving the separable equation, putting it in explicit form

[shamieh]

Find the solution of the given initial value problem in explicit form.
Determine interval which solution is defined. (which i think is the same thing as saying find the interval of validity)

$y' = (1-2x)y^2$ , $y(0) = -1/6$

So here is what I have so far..

$\int y^{-2}dy = x - x^2 + C$

$= \frac{-1}{y} = x-x^2+C$ <-- implicit form

then i plugged $y(0) = -1/6$ and found $C = 6$

But now my question is how do I get my solution in explicit form? If i multiply by a $y$ then I will lose my $y=$ so I'm confused on what to do . I know I need to have $y = x-x^2 + 6$ or something in that form for it to be in explicit form correcT?

 

[Prove It]

Find the solution of the given initial value problem in explicit form.
Determine interval which solution is defined. (which i think is the same thing as saying find the interval of validity)

$y' = (1-2x)y^2$ , $y(0) = -1/6$

So here is what I have so far..

$\int y^{-2}dy = x - x^2 + C$

$= \frac{-1}{y} = x-x^2+C$ <-- implicit form

then i plugged $y(0) = -1/6$ and found $C = 6$

But now my question is how do I get my solution in explicit form? If i multiply by a $y$ then I will lose my $y=$ so I'm confused on what to do . I know I need to have $y = x-x^2 + 6$ or something in that form for it to be in explicit form correcT?

Yes you would lose your y= but you can easily get it back...

$\displaystyle \begin{align*} -\frac{1}{y} &= x - x^2 + 6 \\ -1 &= y \left( x - x^2 + 6 \right) \\ -\frac{1}{x - x^2 + 6} &= y \\ y &= \frac{1}{x^2 - x - 6} \end{align*}$

 

[shamieh]

Ahh I didn't see that. I've been doing too much math today..

So would the IOV be: $-2< x < 3$

 

[Prove It]

Ahh I didn't see that. I've been doing too much math today..

So would the IOV be: $-2< x < 3$

Won't it be valid everywhere except where the denominator is 0?

 

[MarkFL]

Ahh I didn't see that. I've been doing too much math today..

So would the IOV be: $-2< x < 3$

Yes that's correct. :D

Another way to get the explicit solution from what you had is to negate both sides:

$$\frac{1}{y}=x^2-x-6$$

Then just invert both sides:

$$y=\frac{1}{x^2-x-6}$$