# Thread: Rewrite DE as linear equation

1. ## Rewrite DE as linear equation

Hi,

Is it possible to rewrite the equations

(a) $\displaystyle x'=\begin{cases} \frac{x^{2}-1}{x-1} & x\neq1\\ 2 & x=1\end{cases}$

(b) $\displaystyle x'=\begin{cases} \frac{x^{4}-1}{x^2-1} & x\neq1\\ 2 & x=1\end{cases}$
as linear equations?

Can someone give me a hint, please?

Honey$\displaystyle \pi$

2. Originally Posted by HoneyPi
Hi,

Is it possible to rewrite the equations

(a) $\displaystyle x'=\begin{cases} \frac{x^{2}-1}{x-1} & x\neq1\\ 2 & x=1\end{cases}$

(b) $\displaystyle x'=\begin{cases} \frac{x^{4}-1}{x^2-1} & x\neq1\\ 2 & x=1\end{cases}$
as linear equations?

Can someone give me a hint, please?

Honey$\displaystyle \pi$
I'm hoping you can see that

$\displaystyle \frac{x^2 - 1}{x - 1} = \frac{(x + 1)(x - 1)}{x - 1}$

$\displaystyle = x + 1$.

As $\displaystyle x \to 1, f(x) \to 2$.

So you can rewrite this function as

$\displaystyle f(x) = x + 1$ for all $\displaystyle x$.

You should also be able to see that

$\displaystyle \frac{x^4 - 1}{x^2 - 1} = \frac{(x^2 + 1)(x^2 - 1)}{x^2 - 1}$

$\displaystyle = x^2 + 1$.

As $\displaystyle x \to 1, f(x) \to 2$.

So you can rewrite the function as

$\displaystyle f(x) = x^2 + 1$ for all $\displaystyle x$.

So, you can rewrite the first as a linear function, and the second as a quadratic function.