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\begin{document}


{\bf Question}

Find the general solution of the differential equation

$$y'+y^2=1$$

by separation of variables. Examine the same equation by dominant
balance as $x\to +\infty$, comparing the results with the exact
solution.


\vspace{.25in}

{\bf Answer}

$y'+y^2=1 \Rightarrow y'=1-y^2 \Rightarrow \df{y'}{1-y^2}=1$

Therefore $\ds\int\df{dy}{1-y^2}=\ds\int dx \Rightarrow
y=\df{1-Ae^{-2x}}{1+Ae^{-2x}}\ \ A=const$

Dominant balance as $x \to + \infty$: Try

$\un{y'=1} \Rightarrow y=x+c \Rightarrow y^2=O(x^2)$ so
$y'=o(y^2)$. Inconsistent.

$\un{y'=y^2} \Rightarrow \ds\int\df{y'}{y^2}=\ds\int dx
\Rightarrow -\df{1}{y}=x+c \Rightarrow y=-\df{1}{x+c}=o(1)$ as $x
\to + \infty$. Inconsistent.

$\un{y^2=1} \Rightarrow y=\pm 1\ \ y'=0=o(1)$ as $x \to \infty$!

This is the balance.

Therefore $y \sim \pm 1$ as $x \to \infty$.

Second order balance: $y=+1+y_1$ where $y_1=o(1)\ \ast$

(Take $+1$ only)

Then
$\begin{array}{l}(1+y_1)'+(1+y_1)^2=1\\y_1'+1+2y_1+y_1^2=1\\y_1'+2y_1+y_1^2=0
\end{array}$

${}$

\un{Balance}

$\un{y_1'=-2y_1} \Rightarrow \ds\int \df{y_1'}{y'}=-2\ds\int dx
\Rightarrow y_1=Be^{-2x} \Rightarrow y_1^2=O(e^{-4x})=o(e^{-2x})
\rightarrow$ \un{consistent}.

${}$

NB Other choice of -1 leads to inconsistency

$\rightarrow y \sim -1 \Rightarrow y_1=O(e^{2x})=o(e^{4x})$

${}$

Check others:

$\un{y_1=-y_1^2} \Rightarrow \ds\int \df{y_1'}{y_1^2}=-\ds\int dx
\Rightarrow -\df{1}{y_1'}=-x+c \Rightarrow
y_1=O\left(\df{1}{x}\right)\ \ x \to + \infty.$

Hence $y_1^2=o(y_1)$ as $x \to +\infty$.  \un{INCONSISTENT}.

${}$

$\un{2y_1=-y_1^2} \Rightarrow y_1=0$ (gives an exact solution:
$\un{y=\pm 1} \Rightarrow y'=0$ and $y^2=1$) or $y_1=-2$. (This is
not $o(1)$ as assumed$\ \star$)

Therefore we get either $y=\pm 1$ \un{\un{exactly}} \un{or} $y
\sim 1-Be^{-2x}\ \ x \to +\infty$ which is consistent with the
expansion of the exact result if $B=2A$ (as $x \to + \infty$).


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