Michael Maggard

2022-01-20

Find y which satisfies:
${y}^{\prime }={y}^{a},y\left(a\right)=a-2$, for $a\in N$

Serita Dewitt

This is what I did:
$\int \frac{{y}^{\prime }}{{y}^{a}}dx=\int 1dx$
$\frac{{y}^{-a+1}}{-a+1}+{c}_{1}=x+{c}_{2}$
${y}^{-a+1}=\left(x+C\right)\left(-a+1\right)$,
where $C={c}_{2}-{c}_{1}$, and for $a=1$ there's no solution.
So I get $y={\left(\frac{1}{\left(x+c\right)\left(1-a\right)}\right)}^{a-1}$,
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?

Alex Sheppard

I presume for each a your need to find one function ${y}_{a}$.
Your working seems essentially correct, but you can re-write as
$\frac{1}{\left(1-a\right){y}^{a-1}}-x=C$
Plug in the required values of x and y and find C.

RizerMix

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 $\frac{\left(a-2{\right)}^{-a+1}}{-a+1}=a+C$ 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 $\mathrm{ln}\left(|y|\right)=x+C$. Put $x=1$. We can now solve for C, and end up with $y=-{e}^{x-1}$. Alternately, we end up with \(y=Ce^{x}),