Separable equations-check my answers please

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In summary: You are right in saying that there are no solutions that intersect y= 1. That is a consequence of the fact that y= 1 and y= -1 are solutions. Those are the only solutions that intersect y= 1.
  • #1

413

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dy/dx= (y^2 -1)/x

1. Give the general equation of the curves that satisfy this equation.

2. Show that the straight lines y=1 and y=-1 are also solutions

3. Do any of the curves you found in 1) intersect y=1?

My Ans:
1. The general solution i found out to be x^2 + C = (y-1)/(y+1), correct?

2. Does that mean y=1 is a solution and y=-1 is not because it will have a denominator of zero?

3. The only intersection is at x=0, y=1 , since x can't be 0, there are no curves intersecting y=1

Am i doing anything wrong here?
 
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  • #2
y=-1 is a valid solution as you can see from the ODE. y=1 as well. For the solution curves they are horizontal asymptotes.

Daniel.
 
  • #3
413 said:
dy/dx= (y^2 -1)/x

1. Give the general equation of the curves that satisfy this equation.

2. Show that the straight lines y=1 and y=-1 are also solutions

3. Do any of the curves you found in 1) intersect y=1?

My Ans:
1. The general solution i found out to be x^2 + C = (y-1)/(y+1), correct?
2x= {1(y+1)-1(y-1)}/(y+1)^2 dy/dx= 2/(y+1)^2 dy/dx so
dy/dx= x(y+1)^2. Doesn't look anything like dy/dx= (y^2-1)/x to me!
I assume you got dy/(y^2-1)= dx/x. How did you integrate that? Be careful with your constant of integration. If ln a= ln b+ C then a= eCb= C'b where C'= eC. The point is that the constant is now multiplying not added to b.

2. Does that mean y=1 is a solution and y=-1 is not because it will have a denominator of zero?
No, "Show that the straight lines y=1 and y=-1 are also solutions" implies that both y= 1 and y= -1 are solutions! There is no "denominator of zero" because the equation as given, dy/dx= (y^2 -1)/x, has only x as denominator. y= 1 or y= -1 can't make the denominator 0. If you are referring to the y+1 in the denominator of your general solutions, the whole point of this exercise is to show you that there may be solutions to a d.e. that your "general" solution from integrating does not give.
Suppose y= 1 for all x. What is dy/dx? What is (y^2- 1)/x when y= 1? Same questions for y= -1.

3. The only intersection is at x=0, y=1 , since x can't be 0, there are no curves intersecting y=1
Recalculate your "general solution" and try this again. In fact, none of the solutions intersect y= 1 but not because of any problem at x= 0.
 

1. What are separable equations?

Separable equations are differential equations that can be separated into two parts, one involving only the dependent variable and the other only involving the independent variable.

2. How do I know if an equation is separable?

An equation is separable if it can be written in the form f(y)dy = g(x)dx, where f and g are functions of y and x respectively.

3. What is the process for solving a separable equation?

The general process for solving a separable equation is to first separate the variables, then integrate both sides, and finally solve for the dependent variable.

4. Are there any special cases for separable equations?

Yes, there are a few special cases for separable equations such as when the dependent variable is raised to a power, or when there is a constant term on one side of the equation.

5. Can you give an example of solving a separable equation?

Sure, for example, let's solve the equation dy/dx = x/y. First, we can separate the variables to get ydy = xdx. Then we can integrate both sides to get (y^2)/2 = (x^2)/2 + C. Finally, we can solve for y to get y = ± √(x^2 + 2C).

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