Singular solutions Separable Eq. and IVP

In summary, the conversation discusses the relationship between singular solutions and initial value problems. The general solution for a separable equation may assume a particular condition, leading to a potential asymptotic behavior. A singular solution cannot be obtained from a general solution. However, it is possible for a particular solution to be derived from a general solution that is not defined for the given initial condition. This is because the general solution can still be extended to include the initial condition, as long as the solution is continuous.
  • #1
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Hi,

I am a little bit confused about singular solutions and their relationship with IVP's and decided to ask you.


As far as I understood, the IVP's could be in a form:
y' = f(x,y)
y(x0) = y0

To obtain a general solution, we could use separable eq. method. I have learned that sometimes when we separate the variables, we assume a particular condition to be valid. (that might have an asymptotic behaviour. )

Also, as far as I know, a singular solution can not be obtained from a general solution.

My impression was that the general solution shouldn't have worked for y = y0 if we derived it assuming y is not equal to y0. However, that example confuses me;

y' = 2 (y^1/2)
y(0) = 0


By the sep. eq. method and integrating both sides;

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

I have found that y = (x-c)^2. However, while separating, we actually assume that y is not equal to 0.

What's confusing is, at this general solution, when I plug y = 0 and x = 0; I get a particular solution; which is y = x^2 and that perfectly works well! It does not make sense to get a particular solution which suggests y= 0 when x = 0 using a general solution that is not defined for y = 0. (Because I had assumed y is not equal to 0 while I was separating the variables. )
 
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  • #2
It is sufficient to solve for y with y not 0, then extend the solution to y= 0 since, in order that the derivative exist, y must be continuous.

But did you notice that y(x)= 0 for all x is also a solution? In fact, if you set y(x)= (x- a)2 for x< a< 0, y(x)= 0 for a< x< b (where b> 0), y(x)= (x-b)2 for 0< b< x is also a solution to the differential equation satisfying y(0)= 0.
 

1. What is a singular solution?

A singular solution is a special type of solution to a differential equation that cannot be obtained by solving the equation using standard techniques. It arises when the general solution to the equation contains a term with a logarithmic or inverse trigonometric function that results in a singularity, or a value that is undefined, at a particular point.

2. What is a separable equation?

A separable equation is a type of first-order differential equation that can be written in the form dy/dx = f(x)g(y). This means that the equation can be separated into two parts, one containing only x and the other containing only y. This type of equation can be solved by integrating both sides with respect to the corresponding variable.

3. How do you know if a differential equation can be solved using separable equations?

A differential equation can be solved using separable equations if it can be written in the form dy/dx = f(x)g(y). In other words, the equation must be able to be separated into two parts, one containing only x and the other containing only y. If this is not the case, then the equation cannot be solved using separable equations.

4. What is an initial value problem (IVP)?

An initial value problem (IVP) is a type of differential equation that involves finding the specific solution that satisfies both the equation and a given set of initial conditions. These initial conditions typically involve specifying the value of the dependent variable at a particular point.

5. How do you solve a singular solution separable equation with an initial value problem?

To solve a singular solution separable equation with an initial value problem, you first need to find the general solution to the equation using standard techniques. Then, you can use the given initial conditions to determine the specific solution that satisfies both the equation and the initial conditions. This may require some manipulation of the general solution to account for the singularity at a particular point.

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