Implicit and explicit solution for a given initial-value problem

In summary: So since they are all factors when you divide by y the y just goes away on the right side. But wouldn't the right side have dx/y? That's why I'm stuck, I'm not sure if it's acceptable!dx can be treated as just another factor on the right side. So since they are all factors when you divide by y the y just goes away on the right side.
  • #1
Cheruby
3
0
The equation is:
x2 dy/dx= y - xy
IC (initial conditions): y(-1) = -1 (This is used to solve for C)
Must first separate the variables x and y and then integrate them and solve for y, but I got stuck...

x2dy = (y - xy)dx
x2dy = y(1-x)dx
dy= y(1-x)dx/x2 <-- not sure what to do with the y now... I figure I could divide everything by y so I could bring it to the left side but the right side would have dx/y

The answer is y = e-(1+1/x)/x
 
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  • #2
Cheruby said:
The equation is:
x2 dy/dx= y - xy
IC (initial conditions): y(-1) = -1 (This is used to solve for C)
Must first separate the variables x and y and then integrate them and solve for y, but I got stuck...

x2dy = (y - xy)dx
x2dy = y(1-x)dx
dy= y(1-x)dx/x2 <-- not sure what to do with the y now... I figure I could divide everything by y so I could bring it to the left side but the right side would have dx/y

The answer is y = e-(1+1/x)/x

You are doing it right, but when you divide by Y you will have dy/y, and that is perfectly acceptable. The right side you will have (1-x)/x^2 dx. Then integrate both sides.
 
  • #3
Jackx said:
You are doing it right, but when you divide by Y you will have dy/y, and that is perfectly acceptable. The right side you will have (1-x)/x^2 dx. Then integrate both sides.

But wouldn't the right side have dx/y? That's why I'm stuck, I'm not sure if it's acceptable!
 
  • #4
dx can be treated as just another factor on the right side. So since they are all factors when you divide by y the y just goes away on the right side.

Now if it was y + (whatever)dx, then yes if you divided by y you would have a dx/y.
 
  • #5
Alright thank you Jackx
 

1. What is the difference between an implicit and explicit solution for a given initial-value problem?

An implicit solution is a general equation that relates the dependent and independent variables in a given initial-value problem. It can be difficult to solve for the dependent variable explicitly. On the other hand, an explicit solution is a specific equation that directly expresses the dependent variable in terms of the independent variable and the initial conditions.

2. When is it necessary to use an implicit solution for a given initial-value problem?

An implicit solution is often used when the explicit solution cannot be found or when it is too complex to compute. It may also be used when the initial conditions are not specified or when the equation is nonlinear and cannot be easily solved for the dependent variable.

3. Is one type of solution better than the other for a given initial-value problem?

Not necessarily. The choice between using an implicit or explicit solution depends on the specific initial-value problem and the goals of the solution. In some cases, an implicit solution may provide more information about the behavior of the system, while in others an explicit solution may be more useful for practical applications.

4. Can an implicit solution be converted to an explicit solution?

Yes, it is possible to convert an implicit solution to an explicit one by solving the equation for the dependent variable. However, this may not always be possible or straightforward, and the resulting explicit solution may be more complex than the original implicit one.

5. Are there any advantages or disadvantages to using implicit or explicit solutions?

The choice between implicit and explicit solutions depends on the specific problem and goals. Implicit solutions may provide more insight into the behavior of the system, but they may be more difficult to compute. Explicit solutions are easier to compute, but may not provide as much information about the system. It is important to carefully consider the trade-offs and choose the most appropriate solution for the given initial-value problem.

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