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\begin{document}


{\bf Question}

Obtain the first two terms of an asymptotic expansion for
solutions of the following differential equations as $x\to
+\infty$ using the method of dominant balance.

\begin{description}
\item[(a)]
$y'+y^4=x$

\item[(b)]
$y'+y^{\frac{1}{2}}=x^2$

\end{description}




\vspace{.25in}

{\bf Answer}

\begin{description}
\item[(a)]
$\ \ y'+y^4=x$

$\stackrel{\uparrow}{(1)}\ \ \stackrel{\uparrow}{(2)}\ \
\stackrel{\uparrow}{(3)}$

Look for dominant balance solution as $x\to+\infty$.

\un{(1) and (3)}

$y'=x \Rightarrow y=\df{x^2}{2}+c$

But then $y^4=O(x^8)$ as $x\to +\infty$ which is \un{inconsistent}
as we've assumed $y^4=o(y')$ and $o(x)$.

\un{(1) and (2)}

$y'=-y^4 \Rightarrow \ds\int\df{dy}{y^4}=-\ds\int dx \Rightarrow
-\df{1}{3y^3}=-x+c$

$\Rightarrow y^3=\df{1}{3(x+c)} \Rightarrow
y=\df{1}{[3(x+c)]^{\frac{1}{3}}}=o(x)$ as $x\to +\infty$ Therefore
\un{inconsistent}.

\un{(2) and (3)}

$y^4=x \Rightarrow y=\pm x^{\frac{1}{4}} \Rightarrow
y'=O(x^{-\frac{3}{4}})$ as $x\to +\infty$ so $y'=o(y^4)$ and
$o(x)\ \surd\surd$

This is a good balance. Hence $y \sim \pm x^{\frac{1}{4}},\ \pm
ix^{\frac{1}{4}}$ as $x\to +\infty$ (boundary conditions would
determine)

\item[(b)]
$\ \ y'+y^{\frac{1}{2}}=x^2$

$\stackrel{\uparrow}{(1)}\ \ \stackrel{\uparrow}{(2)}\ \
\stackrel{\uparrow}{(3)}$

Look for dominant balance solutions as $x \to +\infty$.

\un{(1) and (3)}

$y'=x^2 \Rightarrow y=\df{x^3}{3}+c \Rightarrow
y^{\frac{1}{2}}=O(x^{\frac{3}{2}})\ \ x\to +\infty$

and $x^{\frac{3}{2}}=o(x^2)$ so this \un{looks} consistent.

We must still check the other possible balances to make sure
there's no ambiguity.

\un{(1) and (2)}

$y'+y^\frac{1}{2}=0 \Rightarrow \ds\int\df{dy}{\sqrt{y}}=-\ds\int
dx \Rightarrow 2y^{\frac{1}{2}}=-x+c \Rightarrow
y=\df{(x-c)^2}{4}$

Therefore $y^\frac{1}{2}=O(x)\ \ x\to +\infty$ but therefore

$y^{\frac{1}{2}}=o(x^2)$ an inconsistency.

\un{(2) and (3)}

$y^{\frac{1}{2}}=x^2 \Rightarrow y=x^4 \Rightarrow y'=4x^3$ so
$y^\frac{1}{2}=o(y')$ an inconsistency.

Hence $y \sim \df{x^3}{3}+c\ \ \rm{as}\ x\to +\infty$.
\end{description}



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