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I need to show that the DE $y'=y^{\alpha}$, where $\alpha$ is a constant with $0<\alpha<1$, has infinitely many solutions passing through the point $(0,0)$. Also I need four of such solutions. Thank you for your help! Klara

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    This smells like homework. But *you*, do you have any idea?2012-12-06
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    I think that considering "the solution at a point 'c' " is equivalent to considering a boundary value problem which always have a unique solution but the trivial or singular solution(if exist). Also, your problem (considering the condition (0,0)) is a BV. I could not see what is the matter of being puzzled.2013-07-09

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Here how you can proceed. First note that $y=0$ is a solution. Next, you use integration to get $$ \int_0^y\frac{dy}{y^{\alpha}}=t+C_1. $$ Note that on the left hand side you have an improper integral which converges if $0<\alpha<1$ -- this is the reason the solutions are not unique at the point $(0,0)$. Assuming that $0<\alpha<1$ you get $$ y=K(t-C)^{\frac{1}{1-\alpha}} $$ also a solution to your equation. Here $K$ is a constant which depends on $\alpha$.

Here is a first example of a solution $$ y_1(t)=\begin{cases} 0&t<0,\\ Kt^{\frac{1}{1-\alpha}}&t\geq 0. \end{cases} $$ The only thing you need to check that it has a continuous derivative at $t=0$, which can be checked by direct calculation.

Generalizing, any function of the form $$ y_1(t)=\begin{cases} 0&t$t=C$) and they all pass through $(0,0)$ if $C>0$.

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    @ artem thank you, this is exactly the solution my professor put on the board, I wander why is it important that y_1 should have an continous derivative at C and not only a lets say continuous solution at c? Thank you again.2012-12-10
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    @Klara There should be a continuous derivative at $C$ because otherwise this function would not be a solution (by definition, a solution is a function which you can plug in into your equation and get an identity).2012-12-10
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You have $$ y' = \frac{dy}{dx} = y^{\alpha}, \ y(0)=0, \ 0<\alpha<1$$ From this we can start \begin{align} \frac{dy}{dx} &= y^{\alpha} \\ \Leftrightarrow \frac{1}{ y^{\alpha}} dy &= dx \\ \Leftrightarrow \int \frac{1}{ y^{\alpha}} dy &= \int 1 dx \end{align} Since this really does smell like homework like Siminore said, you should try from this point by your own. Remember the integration constant and fit that with the initial condition $y(0)=0$.

Edit: Okay if this isn't HW I shall continue.

Taking the above we get: \begin{align} \int \frac{1}{ y^{\alpha}} dy &= \int 1 dx \\ \Leftrightarrow \frac{y^{1-\alpha}}{1-\alpha}+c &=x \, \text{ as } \, 0<\alpha<1 \\ \Leftrightarrow y^{1-\alpha}&=(x-c)\cdot(1-\alpha) \\ \Leftrightarrow y &= ((x-c)\cdot(1-\alpha))^{\frac{1}{1-\alpha}} \end{align}

Now we also have to check whether our initial condition is fulfilled. $$ y(0) =((0-c)\cdot(1-\alpha))^{\frac{1}{1-\alpha}}=0$$ only for $c=0$, so $$ y(x) = (x\cdot(1-\alpha))^{\frac{1}{1-\alpha}} $$ is the unique solution of the above ODE.

Edit: It is not the unique solution, since $y(t)=0 \ \forall t$ also solves the problem.

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    Thank you for your comment, This is not HW.I know how to do what you wrote upstairs. I had this on an prior exam and now I need to study the exams for possible similar problem in final and since I did not do it right to begin with, I posted it here for help.2012-12-06
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    This DE does have infinite number of solutions.My professor said that the combination of these two solutions at point c is a solutions as well. And we need to show that this cmobination solution is differentiable at c and since c can have any value in the real line then there are infinite number of solutions possible. Thank you!2012-12-07
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    "_and that the combination of these two solutions at point c is a solutions as well_" Which two solutions to you mean? I am not sure whether I am missing something here. Maybe somebody can check my answer?2012-12-07
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    hm and obvious $y(t)=0 \forall t$ is a solution as well.2012-12-07
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    yes the y=0 is the other one, but why showing that the combined solution is just continuous at c is not enough?2012-12-07
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    sorry i don't quite understand your question. And having two solutions isn't the same as having infinite solution. And finally I am no longer sure whether my answer is correct2012-12-07
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    let us [continue this discussion in chat](http://chat.stackexchange.com/rooms/6651/discussion-between-macydanim-and-klara)2012-12-07