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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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    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 is a solution (you need to check the derivative at $t=C$) and they all pass through $(0,0)$ if $C>0$.

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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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    let us [continue this discussion in chat](http://chat.stackexchange.com/rooms/6651/discussion-between-macydanim-and-klara)2012-12-07