• 413
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.

#### 413

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?

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.

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).

• Calculus
Replies
20
Views
2K
• Calculus
Replies
12
Views
1K
• Calculus
Replies
3
Views
110
• Calculus
Replies
1
Views
691
• Calculus
Replies
3
Views
988
• Calculus
Replies
2
Views
2K
• Calculus
Replies
4
Views
78
• Calculus
Replies
3
Views
690
• Calculus
Replies
10
Views
1K
• Calculus
Replies
25
Views
2K