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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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# Singular solutions Separable Eq. and IVP

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