4
$\begingroup$

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!

  • 2
    Shouldn't your general solution for $a\ne1$ be $y=((x+c)(1-a))^{1/(1-a)}$?2012-03-16

2 Answers 2

2

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 $C'=-e^{-1}$.

  • 0
    Thank you for the answer. How do you get the equation with $\ln$ for $a=1$? Don't we get $y^0=0$ which leads us to no solution for this case?2012-03-16
  • 1
    Remember from early integration that $\int \frac{dt}{t}=\ln(|t|)+C$. The case $k=-1$ is the only exception to the general formula for $\int t^k\,dt$.2012-03-16
  • 0
    Right, I missed it. Thanks a lot!2012-03-16
3

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 $C$.