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

Beginner2022-01-20Added 41 answers

This is what I did:

$\int \frac{{y}^{\prime}}{{y}^{a}}dx=\int 1dx$

$\frac{{y}^{-a+1}}{-a+1}+{c}_{1}=x+{c}_{2}$

${y}^{-a+1}=(x+C)(-a+1)$ ,

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?

where

So I get

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

Beginner2022-01-21Added 36 answers

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.

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

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

RizerMix

Expert2022-01-27Added 583 answers

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

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I Completed the square on the bottom but what do you do now?

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inverse laplace transform - with symbolic variables:

$F(s)=\frac{2{s}^{2}+(a-6b)s+{a}^{2}-4ab}{({s}^{2}-{a}^{2})(s-2b)}$

My steps:

$F(s)=\frac{2{s}^{2}+(a-6b)s+{a}^{2}-4ab}{(s+a)(s-a)(s-2b)}$

$=\frac{A}{s+a}+\frac{B}{s-a}+\frac{C}{s-2b}+K$

$K=0$

$A=F(s)\ast (s+a)$