Solving the separable equation, putting it in explicit form

In summary: The interval of validity, or where the solution is defined, would be all values of x except where the denominator is 0, which is at x=-2 and x=3. So the interval of validity would be -2<x<3.
  • #1
shamieh
539
0
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?
 
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  • #2
shamieh said:
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...

$\displaystyle \begin{align*} -\frac{1}{y} &= x - x^2 + 6 \\ -1 &= y \left( x - x^2 + 6 \right) \\ -\frac{1}{x - x^2 + 6} &= y \\ y &= \frac{1}{x^2 - x - 6} \end{align*}$
 
  • #3
Ahh I didn't see that. I've been doing too much math today..

So would the IOV be: $-2< x < 3$
 
  • #4
shamieh said:
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?
 
  • #5
shamieh said:
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:

\(\displaystyle \frac{1}{y}=x^2-x-6\)

Then just invert both sides:

\(\displaystyle y=\frac{1}{x^2-x-6}\)
 

1. What is a separable equation?

A separable equation is an ordinary differential equation that can be written in the form of dy/dx = f(x)g(y), where f(x) and g(y) are functions of x and y respectively. This form allows us to separate the variables and solve for y by integrating both sides.

2. How do you solve a separable equation?

To solve a separable equation, we first rewrite it in the form of dy/dx = f(x)g(y). Then, we separate the variables by dividing both sides by g(y) and integrating both sides with respect to x. This will give us an equation in the form of y = F(x) + C, where F(x) is the antiderivative of f(x) and C is the constant of integration. Finally, we can solve for y by substituting values for x and solving for y.

3. What does it mean to put a separable equation in explicit form?

Putting a separable equation in explicit form means rewriting it in the form of y = f(x), where y is a function of x. This allows us to easily solve for y by plugging in values for x.

4. Can all differential equations be solved using separation of variables?

No, not all differential equations are separable and can be solved using separation of variables. Only equations that can be written in the form of dy/dx = f(x)g(y) are separable and can be solved using this method.

5. What is the significance of the constant of integration in solving separable equations?

The constant of integration, represented by C, is a constant term that is added when solving a separable equation. This constant accounts for all possible solutions that may exist for the given equation. In certain cases, initial conditions may be given to determine the specific value of C, but in other cases, it may remain as an indefinite constant.

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