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I'd love your help with finding the function $y$ which satisfies: y'=y^a, $y(a)=a-2$, for $a \in \mathbb{N}$

This is what I did:

\begin{align*} \int \frac{y'}{y^a}dx&=\int 1dx\\ \frac{y^{-a+1}}{-a+1}+c_1&=x+c_2\\ y^{-a+1}&=(x+C)(-a+1), \end{align*} where $C=c_2-c_1$, and for $a=1$ there's no solution.

So I get $y=\left(\frac{1}{(x+c)(1-a)}\right)^{a-1},$ finding $c$ is not pleasant.

I assume that something is wrong, Am I suppose leave $y$ in the way that I find it after finding $c$?

Thanks a lot!

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    Shouldn't your general solution for $a\ne1$ be $y=((x+c)(1-a))^{1/(1-a)}$?2012-03-16

2 Answers 2

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There is nothing wrong, except abandoning poor $a=1$, which though a little special gives no problems. When we do the details we will see there is a problem at $a=2$.

Quite quickly (for $a\ne 1$) we reach $\frac{y^{-a+1}}{-a+1}=x+C.$ It is best to find $C$ now. Put $x=a$. We get $\frac{(a-2)^{-a+1}}{-a+1}=a+C.$ Now we know $C$, except when $a=2$ (one cannot divide by $0$). So for $a=2$ there is no solution that satisfies the initial condition.

There is no trouble if $a=1$. True, the above general formula does not quite work. But if we integrate, we get $\ln(|y|)=x+C.$ Put $x=1$. We can now solve for $C$, and end up with $y=-e^{x-1}$. Alternately, we end up with y=C'e^x, and conclude that C'=-e^{-1}.

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    Right, I missed it. Thanks a lot!2012-03-16
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I presume for each $a$ your need to find one function $y_a$.

Your working seems essentially correct, but you can re-write as

$\frac{1}{(1-a)y^{a-1}} -x = C$

Plug in the required values of $x$ and $y$ and find $C$.