3.2 HW2

3.2.4 problem 3.35
3.2.5 problem 3.39(h)
3.2.6 problem 3.42(a)
3.2.7 problem 3.49(c)

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3.2.1 problem 3.27 (page 138)

Problem derive 3.4.28. Below is a screen shot from the book giving 3.4.28 at page 88, and the context it is used in before solving the problem

Solution

For $$n^{th}$$ order ODE, $$S_{0}\left ( x\right )$$ is given by$S_{0}\left ( x\right ) \thicksim \omega \int ^{x}Q\left ( t\right ) ^{\frac{1}{n}}dt$ And (page 497, textbook) $$S_{1}\left ( x\right ) \thicksim \frac{1-n}{2n}\ln \left ( Q\left ( x\right ) \right ) +c\tag{10.2.11}$$ Therefore\begin{align*} y\left ( x\right ) & \thicksim \exp \left ( S_{0}+S_{1}\right ) \\ & \thicksim \exp \left ( \omega \int ^{x}Q\left ( t\right ) ^{\frac{1}{n}}dt+\frac{1-n}{2n}\ln \left ( Q\left ( x\right ) \right ) +c\right ) \\ & \thicksim c\left [ Q\left ( x\right ) \right ] ^{\frac{1-n}{2n}}\exp \left ( \omega \int ^{x}Q\left ( t\right ) ^{\frac{1}{n}}dt\right ) \end{align*}

Note: I have tried other methods to proof this, such as a proof by induction. But was not able to after many hours trying. The above method uses a given formula which the book did not indicate how it was obtained. (see key solution)

3.2.2 Problem 3.33(b) (page 140)

Problem Find leading behavior as $$x\rightarrow 0^{+}$$ for $$x^{4}y^{\prime \prime \prime }-3x^{2}y^{\prime }+2y=0$$

Solution Let \begin{align*} y\left ( x\right ) & =e^{S\left ( x\right ) }\\ y^{\prime } & =S^{\prime }e^{S}\\ y^{\prime \prime } & =S^{\prime \prime }e^{S}+\left ( S^{\prime }\right ) ^{2}e^{S}\\ y^{\prime \prime \prime } & =S^{\prime \prime \prime }e^{S}+S^{\prime \prime }S^{\prime }e^{S}+2S^{\prime }S^{\prime \prime }e^{S}+\left ( S^{\prime }\right ) ^{3}e^{S} \end{align*}

Hence the ODE becomes$$x^{4}\left [ S^{\prime \prime \prime }+3S^{\prime }S^{\prime \prime }+\left ( S^{\prime }\right ) ^{3}\right ] -3x^{2}S^{\prime }=-2\tag{1}$$ Now, we deﬁne $$S\left ( x\right )$$ as sum of a number of leading terms, which we try to ﬁnd$S\left ( x\right ) =S_{0}\left ( x\right ) +S_{1}\left ( x\right ) +S_{2}\left ( x\right ) +\cdots$ Therefore (1) becomes (using only two terms for now $$S=S_{0}+S_{1}$$)\begin{align} \left \{ S_{0}+S_{1}\right \} ^{\prime \prime \prime }+3\left \{ \left ( S_{0}+S_{1}\right ) \left ( S_{0}+S_{1}\right ) ^{\prime \prime }\right \} +\left \{ \left ( S_{0}+S_{1}\right ) ^{\prime }\right \} ^{3}-\frac{3}{x^{2}}\left \{ S_{0}+S_{1}\right \} ^{\prime } & =-\frac{2}{x^{4}}\nonumber \\ \left \{ S_{0}^{\prime \prime \prime }+S_{1}^{\prime \prime \prime }\right \} +3\left \{ \left ( S_{0}+S_{1}\right ) \left ( S_{0}^{\prime \prime }+S_{1}^{\prime \prime }\right ) \right \} +\left \{ S_{0}^{\prime }+S_{1}^{\prime }\right \} ^{3}-\frac{3}{x^{2}}\left ( S_{0}^{\prime }+S_{1}^{\prime }\right ) & =-\frac{2}{x^{4}}\nonumber \\ \left \{ S_{0}^{\prime \prime \prime }+S_{1}^{\prime \prime \prime }\right \} +3\left \{ S_{0}^{\prime \prime }S_{0}^{\prime }+S_{0}^{\prime \prime }S_{1}^{\prime }+S_{1}^{\prime \prime }S_{0}^{\prime }\right \} +\left \{ \left ( S_{0}^{\prime }\right ) ^{3}+3\left ( S_{0}^{\prime }\right ) ^{2}S_{1}^{\prime }+3S_{0}^{\prime }\left ( S_{1}^{\prime }\right ) ^{2}\right \} -\frac{3}{x^{2}}\left \{ S_{0}^{\prime }+S_{1}^{\prime }\right \} & =-\frac{2}{x^{4}}\tag{2} \end{align}

Assuming that $$S_{0}^{\prime }\ggg S_{1}^{\prime },S_{0}^{\prime \prime \prime }\ggg S_{1}^{\prime \prime \prime },\left ( S_{0}^{\prime }\right ) ^{3}\ggg 3\left ( S_{0}^{\prime }\right ) ^{2}S_{1}^{\prime }$$ then equation (2) simpliﬁes to$S_{0}^{\prime \prime \prime }+3S_{0}^{\prime \prime }S_{0}^{\prime }+\left ( S_{0}^{\prime }\right ) ^{3}-\frac{3}{x^{2}}S_{0}^{\prime }\thicksim -\frac{2}{x^{4}}$ Assuming $$\left ( S_{0}^{\prime }\right ) ^{3}\ggg S_{0}^{\prime \prime \prime },\left ( S_{0}^{\prime }\right ) ^{3}\ggg 3S_{0}^{\prime \prime }S_{0}^{\prime },\left ( S_{0}^{\prime }\right ) ^{3}\ggg \frac{3}{x^{2}}S_{0}^{\prime }$$ (which we need to verify later), then the above becomes$\left ( S_{0}^{\prime }\right ) ^{3}\thicksim -\frac{2}{x^{4}}$ Veriﬁcation2

Since $$S_{0}^{\prime }\thicksim \left ( \frac{-2}{x^{4}}\right ) ^{\frac{1}{3}}=\frac{1}{x^{\frac{4}{3}}}$$ then $$S_{0}^{\prime \prime }\thicksim \frac{1}{x^{\frac{7}{3}}}$$ and $$S_{0}^{\prime \prime \prime }\thicksim \frac{1}{x^{\frac{10}{3}}}$$. Now we need to verify the three assumptions made above, which we used to obtain $$S_{0}^{\prime }$$.\begin{align*} \left ( S_{0}^{\prime }\right ) ^{3} & \ggg S_{0}^{\prime \prime \prime }\\ \frac{1}{x^{4}} & \ggg \frac{1}{x^{\frac{10}{3}}} \end{align*}

Yes.\begin{align*} \left ( S_{0}^{\prime }\right ) ^{3} & \ggg 3S_{0}^{\prime \prime }S_{0}^{\prime }\\ \frac{1}{x^{4}} & \ggg \left ( \frac{1}{x^{\frac{7}{3}}}\right ) \left ( \frac{1}{x^{\frac{4}{3}}}\right ) \\ \frac{1}{x^{4}} & \ggg \left ( \frac{1}{x^{\frac{11}{3}}}\right ) \end{align*}

Yes.\begin{align*} \left ( S_{0}^{\prime }\right ) ^{3} & \ggg \frac{3}{x^{2}}S_{0}^{\prime }\\ \frac{1}{x^{4}} & \ggg \left ( \frac{1}{x^{2}}\right ) \frac{1}{x^{\frac{4}{3}}}\\ \frac{1}{x^{4}} & \ggg \frac{1}{x^{\frac{10}{3}}} \end{align*}

Yes. Assumed balance ise veriﬁed. Therefore\begin{align*} \left ( S_{0}^{\prime }\right ) ^{3} & \thicksim -\frac{2}{x^{4}}\\ S_{0}^{\prime } & \thicksim \omega x^{\frac{-4}{3}} \end{align*}

Where $$\omega ^{3}=-2$$. Integrating\begin{align*} S_{0} & \thicksim \omega \int x^{\frac{-4}{3}}dx\\ & \thicksim \omega \int x^{-\frac{4}{3}}dx\\ & \thicksim -3\omega x^{\frac{-1}{3}} \end{align*}

Where we ignored the constant of integration since subdominant. To ﬁnd leading behavior, we go back to equation (2) and now solve for $$S_{1}$$.$\left \{ S_{0}^{\prime \prime \prime }+S_{1}^{\prime \prime \prime }\right \} +3\left \{ S_{0}^{\prime \prime }S_{0}^{\prime }+S_{0}^{\prime \prime }S_{1}^{\prime }+S_{1}^{\prime \prime }S_{0}^{\prime }\right \} +\left \{ \left ( S_{0}^{\prime }\right ) ^{3}+3\left ( S_{0}^{\prime }\right ) ^{2}S_{1}^{\prime }+3S_{0}^{\prime }\left ( S_{1}^{\prime }\right ) ^{2}\right \} -\frac{3}{x^{2}}\left \{ S_{0}^{\prime }+S_{1}^{\prime }\right \} =-\frac{2}{x^{4}}$ Moving all known quantities (those which are made of $$S_{0}$$ and its derivatives) to the RHS and simplifying, gives$\left \{ S_{1}^{\prime \prime \prime }\right \} +3\left \{ S_{0}^{\prime \prime }S_{1}^{\prime }+S_{1}^{\prime \prime }S_{0}^{\prime }\right \} +\left \{ 3\left ( S_{0}^{\prime }\right ) ^{2}S_{1}^{\prime }+3S_{0}^{\prime }\left ( S_{1}^{\prime }\right ) ^{2}\right \} -\frac{3S_{1}^{\prime }}{x^{2}}\thicksim -S_{0}^{\prime \prime \prime }-3S_{0}^{\prime \prime }S_{0}^{\prime }+\frac{3}{x^{2}}S_{0}^{\prime }$ Now we assume the following (then will verify later)\begin{align*} 3\left ( S_{0}^{\prime }\right ) ^{2}S_{1}^{\prime } & \ggg 3S_{0}^{\prime }\left ( S_{1}^{\prime }\right ) ^{2}\\ 3\left ( S_{0}^{\prime }\right ) ^{2}S_{1}^{\prime } & \ggg S_{1}^{\prime \prime \prime }\\ 3\left ( S_{0}^{\prime }\right ) ^{2}S_{1}^{\prime } & \ggg S_{1}^{\prime \prime }S_{0}^{\prime }\\ 3\left ( S_{0}^{\prime }\right ) ^{2}S_{1}^{\prime } & \ggg S_{0}^{\prime \prime }S_{1}^{\prime }\\ 3\left ( S_{0}^{\prime }\right ) ^{2}S_{1}^{\prime } & \ggg S_{1}^{\prime \prime }S_{0}^{\prime } \end{align*}

Hence

$$3\left ( S_{0}^{\prime }\right ) ^{2}S_{1}^{\prime }-\frac{3S_{1}^{\prime }}{x^{2}}\thicksim -S_{0}^{\prime \prime \prime }-3S_{0}^{\prime \prime }S_{0}^{\prime }+\frac{3S_{0}^{\prime }}{x^{2}}\tag{3}$$

But  \begin{align*} S_{0}^{\prime } & \thicksim \omega x^{\frac{-4}{3}}\\ \left ( S_{0}^{\prime }\right ) ^{2} & \thicksim \omega ^{2}x^{\frac{-8}{3}}\\ S_{0}^{\prime \prime } & \thicksim -\frac{4}{3}\omega x^{\frac{-7}{3}}\\ S_{0}^{\prime \prime \prime } & \thicksim \frac{28}{9}\omega x^{\frac{-10}{3}} \end{align*}

Hence (3) becomes\begin{align*} 3\left ( \omega ^{2}x^{\frac{-8}{3}}\right ) S_{1}^{\prime }-\frac{3S_{1}^{\prime }}{x^{2}} & \thicksim \frac{28}{9}\omega x^{\frac{-10}{3}}+3\left ( \frac{4}{3}\omega ^{2}x^{\frac{-7}{3}}x^{\frac{-4}{3}}\right ) +\frac{3\omega x^{\frac{-4}{3}}}{x^{2}}\\ 3\omega ^{2}x^{\frac{-8}{3}}S_{1}^{\prime }-3x^{-2}S_{1}^{\prime } & \thicksim \frac{28}{9}\omega x^{\frac{-10}{3}}+4\omega ^{2}x^{\frac{-11}{3}}+3\omega x^{\frac{-10}{3}} \end{align*}

For small $$x$$, $$x^{\frac{-8}{3}}S_{1}^{\prime }\ggg x^{-2}S_{1}^{\prime }$$ and $$x^{\frac{-11}{3}}\ggg x^{\frac{-10}{3}}$$, then the above simpliﬁes to

\begin{align*} 3\omega ^{2}x^{\frac{-8}{3}}S_{1}^{\prime } & \thicksim 4\omega ^{2}x^{\frac{-11}{3}}\\ S_{1}^{\prime } & \thicksim \frac{4}{3}x^{-1} \end{align*}

$S_{1}\thicksim \frac{4}{3}\ln x$ Where constant of integration was dropped, since subdominant.

Veriﬁcation Using $$S_{0}^{\prime }\backsim x^{\frac{-4}{3}},\left ( S_{0}^{\prime }\right ) ^{2}\backsim x^{\frac{-8}{3}},S_{0}^{\prime \prime }\backsim x^{\frac{-7}{3}},S_{1}^{\prime }\thicksim \frac{1}{x},S_{1}^{\prime \prime }\thicksim \frac{1}{x^{2}},S_{1}^{\prime \prime \prime }\thicksim \frac{1}{x^{3}}$$\begin{align*} 3\left ( S_{0}^{\prime }\right ) ^{2}S_{1}^{\prime } & \ggg 3S_{0}^{\prime }\left ( S_{1}^{\prime }\right ) ^{2}\\ x^{\frac{-8}{3}}\frac{1}{x} & \ggg x^{\frac{-4}{3}}\frac{1}{x^{2}}\\ x^{\frac{-8}{3}} & \ggg x^{\frac{-7}{3}} \end{align*}

Yes.\begin{align*} 3\left ( S_{0}^{\prime }\right ) ^{2}S_{1}^{\prime } & \ggg S_{1}^{\prime \prime \prime }\\ x^{\frac{-8}{3}}\frac{1}{x} & \ggg \frac{1}{x^{3}}\\ \frac{1}{x^{\frac{8}{3}}} & \ggg \frac{1}{x^{2}} \end{align*}

Yes.\begin{align*} 3\left ( S_{0}^{\prime }\right ) ^{2}S_{1}^{\prime } & \ggg S_{1}^{\prime \prime }S_{0}^{\prime }\\ x^{\frac{-8}{3}}\frac{1}{x} & \ggg \frac{1}{x^{2}}x^{\frac{-4}{3}}\\ x^{\frac{-8}{3}} & \ggg x^{\frac{-7}{3}} \end{align*}

Yes.\begin{align*} 3\left ( S_{0}^{\prime }\right ) ^{2}S_{1}^{\prime } & \ggg S_{0}^{\prime \prime }S_{1}^{\prime }\\ x^{\frac{-8}{3}}\frac{1}{x} & \ggg x^{\frac{-7}{3}}\frac{1}{x}\\ x^{\frac{-8}{3}} & \ggg x^{\frac{-7}{3}} \end{align*}

Yes\begin{align*} 3\left ( S_{0}^{\prime }\right ) ^{2}S_{1}^{\prime } & \ggg S_{1}^{\prime \prime }S_{0}^{\prime }\\ x^{\frac{-8}{3}}\frac{1}{x} & \ggg \frac{1}{x^{2}}x^{\frac{-4}{3}}\\ x^{\frac{-8}{3}} & \ggg x^{\frac{-7}{3}} \end{align*}

Yes. All veriﬁed. Leading behavior is\begin{align*} y\left ( x\right ) & \backsim e^{S_{0}\left ( x\right ) +S_{1}\left ( x\right ) }\\ & =\exp \left ( c\omega x^{\frac{-1}{3}}+\frac{4}{3}\ln x\right ) \\ & =x^{\frac{4}{3}}e^{c\omega x^{\frac{-1}{3}}} \end{align*}

I now wanted to see how Maple solution to this problem compare with the leading behavior near $$x=0$$. To obtain a solution from Maple, one have to give initial conditions a little bit removed from $$x=0$$ else no solution could be generated. So using arbitrary initial conditions at $$x=\frac{1}{100}$$ a solution was obtained and compared to the above leading behavior. Another problem is how to select $$c$$ in the above leading solution. By trial and error a constant was selected. Here is screen shot of the result. The exact solution generated by Maple is very complicated, in terms of hypergeom special functions.

ode:=x^4*diff(y(x),x\$3)-3*x^2*diff(y(x),x)+2*y(x);
pt:=1/100:
ic:=y(pt)=500,D(y)(pt)=0,(D@@2)(y)(pt)=0:
sol:=dsolve({ode,ic},y(x)):

3.2.3 problem 3.33(c) (page 140)

Problem Find leading behavior as $$x\rightarrow 0^{+}$$ for $$y^{\prime \prime }=\sqrt{x}y$$

Solution

Let $$y\left ( x\right ) =e^{S\left ( x\right ) }$$. Hence\begin{align*} y\left ( x\right ) & =e^{S_{0}\left ( x\right ) }\\ y^{\prime }\left ( x\right ) & =S_{0}^{\prime }e^{S_{0}}\\ y^{\prime \prime } & =S_{0}^{\prime \prime }e^{S_{0}}+\left ( S_{0}^{\prime }\right ) ^{2}e^{S_{0}}\\ & =\left ( S_{0}^{\prime \prime }+\left ( S_{0}^{\prime }\right ) ^{2}\right ) e^{S_{0}} \end{align*}

Substituting in the ODE gives$$S_{0}^{\prime \prime }+\left ( S_{0}^{\prime }\right ) ^{2}=\sqrt{x}\tag{1}$$ Assuming $$S_{0}^{\prime \prime }\thicksim \left ( S_{0}^{\prime }\right ) ^{2}$$ then (1) becomes$S_{0}^{\prime \prime }\thicksim -\left ( S_{0}^{\prime }\right ) ^{2}$

Let $$S_{0}^{\prime }=z$$ then the above becomes $$z^{\prime }=-z^{2}$$. Hence $$\frac{dz}{dx}\frac{1}{z^{2}}=-1$$ or $$\frac{dz}{z^{2}}=-dx$$. Integrating $$-\frac{1}{z}=-x+c$$ or $$z=\frac{1}{x+c_{1}}$$. Hence $$S_{0}^{\prime }=\frac{1}{x+c_{1}}$$. Integrating again gives

$S_{0}\left ( x\right ) \thicksim \ln \left \vert x+c_{1}\right \vert +c_{2}$

Veriﬁcation

$$S_{0}^{\prime }=\frac{1}{x+c_{1}},\left ( S_{0}^{\prime }\right ) ^{2}=\frac{1}{\left ( x+c_{1}\right ) ^{2}},S_{0}^{\prime \prime }=\frac{-1}{\left ( x+c_{1}\right ) ^{2}}$$

\begin{align*} & S_{0}^{\prime \prime }\ggg x^{\frac{1}{2}}\\ & \frac{1}{\left ( x+c_{1}\right ) ^{2}}\ggg x^{\frac{1}{2}} \end{align*}

Yes for $$x\rightarrow 0^{+}.$$

\begin{align*} & \left ( S_{0}^{\prime }\right ) ^{2}\ggg x^{\frac{1}{2}}\\ & \frac{1}{\left ( x+c_{1}\right ) ^{2}}\ggg x^{\frac{1}{2}} \end{align*}

Yes for $$x\rightarrow 0^{+}.$$ Veriﬁed. Controlling factor is \begin{align*} y\left ( x\right ) & \thicksim e^{S_{0}\left ( x\right ) }\\ & \thicksim e^{\ln \left \vert x+c_{1}\right \vert +c_{2}}\\ & \thicksim Ax+B \end{align*}

3.2.4 problem 3.35

Problem: Obtain the full asymptotic behavior for small $$x$$ of solutions to the equation$x^{2}y^{\prime \prime }+\left ( 2x+1\right ) y^{\prime }-x^{2}\left ( e^{\frac{2}{x}}+1\right ) y=0$ Solution

Let $$y\left ( x\right ) =e^{S\left ( x\right ) }$$. Hence\begin{align*} y\left ( x\right ) & =e^{S_{0}\left ( x\right ) }\\ y^{\prime }\left ( x\right ) & =S_{0}^{\prime }e^{S_{0}}\\ y^{\prime \prime } & =S_{0}^{\prime \prime }e^{S_{0}}+\left ( S_{0}^{\prime }\right ) ^{2}e^{S_{0}}\\ & =\left ( S_{0}^{\prime \prime }+\left ( S_{0}^{\prime }\right ) ^{2}\right ) e^{S_{0}} \end{align*}

Substituting in the ODE gives\begin{align*} x^{2}\left ( S_{0}^{\prime \prime }+\left ( S_{0}^{\prime }\right ) ^{2}\right ) e^{S_{0}}+\left ( 2x+1\right ) S_{0}^{\prime }e^{S_{0}}-x^{2}\left ( e^{\frac{2}{x}}+1\right ) e^{S_{0}\left ( x\right ) } & =0\\ x^{2}\left ( S_{0}^{\prime \prime }+\left ( S_{0}^{\prime }\right ) ^{2}\right ) +\left ( 2x+1\right ) S_{0}^{\prime }-x^{2}\left ( e^{\frac{2}{x}}+1\right ) & =0\\ x^{2}\left ( S_{0}^{\prime \prime }+\left ( S_{0}^{\prime }\right ) ^{2}\right ) +\left ( 2x+1\right ) S_{0}^{\prime } & =x^{2}\left ( e^{\frac{2}{x}}+1\right ) \\ S_{0}^{\prime \prime }+\left ( S_{0}^{\prime }\right ) ^{2}+\frac{\left ( 2x+1\right ) }{x^{2}}S_{0}^{\prime } & =e^{\frac{2}{x}}+1 \end{align*}

Assuming balance \begin{align*} \left ( S_{0}^{\prime }\right ) ^{2} & \thicksim \left ( e^{\frac{2}{x}}+1\right ) \\ \left ( S_{0}^{\prime }\right ) ^{2} & \thicksim e^{\frac{2}{x}}\\ S_{0}^{\prime } & \thicksim \pm e^{\frac{1}{x}} \end{align*}

Where $$1$$ was dropped since subdominant to $$e^{\frac{1}{x}}$$ for small $$x$$.

Veriﬁcation Since$$\ \left ( S_{0}^{\prime }\right ) ^{2}\thicksim e^{\frac{2}{x}}$$ then $$S_{0}^{\prime }\thicksim e^{\frac{1}{x}}$$ and $$S_{0}^{\prime \prime }\thicksim -\frac{1}{x^{2}}e^{\frac{1}{x}}$$, hence \begin{align*} & \left ( S_{0}^{\prime }\right ) ^{2}\ggg S_{0}^{\prime \prime }\\ e^{\frac{2}{x}} & \ggg \frac{1}{x^{2}}e^{\frac{1}{x}}\\ e^{\frac{1}{x}} & \ggg \frac{1}{x^{2}} \end{align*}

Yes, As $$x\rightarrow 0^{+}$$

\begin{align*} \left ( S_{0}^{\prime }\right ) ^{2} & \ggg \frac{\left ( 2x+1\right ) S_{0}^{\prime }}{x^{2}}\\ e^{\frac{2}{x}} & \ggg \frac{\left ( 2x+1\right ) }{x^{2}}e^{\frac{1}{x}}\\ e^{\frac{1}{x}} & \ggg \frac{\left ( 2x+1\right ) }{x^{2}}\\ e^{\frac{1}{x}} & \ggg \frac{2}{x}+\frac{1}{x^{2}} \end{align*}

Yes as $$x\rightarrow 0^{+}$$. Veriﬁed. Hence both assumptions used were veriﬁed OK. Hence\begin{align*} S_{0}^{\prime } & \thicksim \pm e^{\frac{2}{x}}\\ S_{0} & \thicksim \pm \int e^{\frac{1}{x}}dx \end{align*}

Since the integral do not have closed form, we will do asymptotic expansion on the integral. Rewriting $$\int e^{\frac{1}{x}}dx$$ as $$\int \frac{e^{\frac{1}{x}}}{\left ( -x^{2}\right ) }\left ( -x^{2}\right ) dx$$. Using $$\int udv=uv-\int vdu$$, where $$u=-x^{2},dv=\frac{-e^{\frac{1}{x}}}{x^{2}}$$, gives $$du=-2x$$ and $$v=e^{\frac{1}{x}}$$, hence\begin{align} \int e^{\frac{1}{x}}dx & =uv-\int vdu\nonumber \\ & =-x^{2}e^{\frac{1}{x}}-\int -2xe^{\frac{1}{x}}dx\nonumber \\ & =-x^{2}e^{\frac{1}{x}}+2\int xe^{\frac{1}{x}}dx \tag{1} \end{align}

Now we apply integration by parts on $$\int xe^{\frac{1}{x}}dx=\int x\frac{e^{\frac{1}{x}}}{-x^{3}}\left ( -x^{3}\right ) du=\int \frac{e^{\frac{1}{x}}}{-x^{2}}\left ( -x^{3}\right ) du$$, where $$u=-x^{3},dv=\frac{e^{\frac{1}{x}}}{-x^{2}}$$, hence $$du=-3x^{2},v=e^{\frac{1}{x}}$$, hence we have\begin{align*} \int xe^{\frac{1}{x}}dx & =uv-\int vdu\\ & =-x^{3}e^{\frac{1}{x}}+\int 3x^{2}e^{\frac{1}{x}}dx \end{align*}

Substituting this into (1) gives\begin{align*} \int e^{\frac{1}{x}}dx & =-x^{2}e^{\frac{1}{x}}+2\left ( -x^{3}e^{\frac{1}{x}}+\int 3x^{2}e^{\frac{1}{x}}dx\right ) \\ & =-x^{2}e^{\frac{1}{x}}-2x^{3}e^{\frac{1}{x}}+6\int 3x^{2}e^{\frac{1}{x}}dx \end{align*}

And so on. The series will become\begin{align*} \int e^{\frac{1}{x}}dx & =-x^{2}e^{\frac{1}{x}}-2x^{3}e^{\frac{1}{x}}+6x^{4}e^{\frac{1}{x}}+24x^{5}e^{\frac{1}{x}}+\cdots +n!x^{n+1}e^{\frac{1}{x}}+\cdots \\ & =-e^{\frac{1}{x}}\left ( x^{2}+2x^{3}+6x^{4}+\cdots +n!x^{n+1}+\cdots \right ) \end{align*}

Now as $$x\rightarrow 0^{+}$$, we can decide how many terms to keep in the RHS, If we keep one term, then we can say\begin{align*} S_{0} & \thicksim \pm \int e^{\frac{1}{x}}dx\\ & \thicksim \pm x^{2}e^{\frac{1}{x}} \end{align*}

For two terms$S_{0}\thicksim \pm \int e^{\frac{1}{x}}dx\thicksim \pm e^{\frac{1}{x}}\left ( x^{2}+2x^{3}\right )$ And so on. Let us use one term for now for the rest of the solution.$S_{0}\thicksim \pm x^{2}e^{\frac{1}{x}}$ To ﬁnd leading behavior, let $S\left ( x\right ) =S_{0}\left ( x\right ) +S_{1}\left ( x\right )$ Then $$y\left ( x\right ) =e^{S_{0}\left ( x\right ) +S_{1}\left ( x\right ) }$$ and hence now\begin{align*} y^{\prime }\left ( x\right ) & =\left ( S_{0}\left ( x\right ) +S_{1}\left ( x\right ) \right ) ^{\prime }e^{S_{0}+S_{1}}\\ y^{\prime \prime }\left ( x\right ) & =\left ( \left ( S_{0}+S_{1}\right ) ^{\prime }\right ) ^{2}e^{S_{0}+S_{1}}+\left ( S_{0}+S_{1}\right ) ^{\prime \prime }e^{S_{0}+S_{1}} \end{align*}

Substituting into the given ODE gives\begin{align*} x^{2}\left [ \left ( \left ( S_{0}+S_{1}\right ) ^{\prime }\right ) ^{2}+\left ( S_{0}+S_{1}\right ) ^{\prime \prime }\right ] +\left ( 2x+1\right ) \left ( S_{0}\left ( x\right ) +S_{1}\left ( x\right ) \right ) ^{\prime }-x^{2}\left ( e^{\frac{2}{x}}+1\right ) & =0\\ x^{2}\left [ \left ( S_{0}^{\prime }+S_{1}^{\prime }\right ) ^{2}+\left ( S_{0}+S_{1}\right ) ^{\prime \prime }\right ] +\left ( 2x+1\right ) \left ( S_{0}^{\prime }\left ( x\right ) +S_{1}^{\prime }\left ( x\right ) \right ) -x^{2}\left ( e^{\frac{2}{x}}+1\right ) & =0\\ x^{2}\left [ \left ( S_{0}^{\prime }\right ) ^{2}+\left ( S_{1}^{\prime }\right ) ^{2}+2S_{0}^{\prime }S_{1}^{\prime }+\left ( S_{0}^{\prime \prime }+S_{1}^{\prime \prime }\right ) \right ] +\left ( 2x+1\right ) S_{0}^{\prime }\left ( x\right ) +\left ( 2x+1\right ) S_{1}^{\prime }\left ( x\right ) -x^{2}\left ( e^{\frac{2}{x}}+1\right ) & =0\\ x^{2}\left ( S_{0}^{\prime }\right ) ^{2}+x^{2}\left ( S_{1}^{\prime }\right ) ^{2}+2x^{2}S_{0}^{\prime }S_{1}^{\prime }+x^{2}S_{0}^{\prime \prime }+x^{2}S_{1}^{\prime \prime }+\left ( 2x+1\right ) S_{0}^{\prime }\left ( x\right ) +\left ( 2x+1\right ) S_{1}^{\prime }\left ( x\right ) & =x^{2}\left ( e^{\frac{2}{x}}+1\right ) \end{align*}

But $$x^{2}\left ( S_{0}^{\prime }\right ) ^{2}\thicksim x^{2}\left ( e^{\frac{2}{x}}+1\right )$$ since we found that $$S_{0}^{\prime }\thicksim e^{\frac{1}{x}}$$. Hence the above simpliﬁes to\begin{align} x^{2}\left ( S_{1}^{\prime }\right ) ^{2}+2x^{2}S_{0}^{\prime }S_{1}^{\prime }+x^{2}S_{0}^{\prime \prime }+x^{2}S_{1}^{\prime \prime }+\left ( 2x+1\right ) S_{0}^{\prime }\left ( x\right ) +\left ( 2x+1\right ) S_{1}^{\prime }\left ( x\right ) & =0\nonumber \\ \left ( S_{1}^{\prime }\right ) ^{2}+2S_{0}^{\prime }S_{1}^{\prime }+S_{0}^{\prime \prime }+S_{1}^{\prime \prime }+\frac{\left ( 2x+1\right ) }{x^{2}}S_{0}^{\prime }\left ( x\right ) +\frac{\left ( 2x+1\right ) }{x^{2}}S_{1}^{\prime }\left ( x\right ) & =0 \tag{2} \end{align}

Now looking at $$S_{0}^{\prime \prime }+\frac{\left ( 2x+1\right ) }{x^{2}}S_{0}^{\prime }\left ( x\right )$$ terms in the above. We can simplify this since we know $$S_{0}^{\prime }=e^{\frac{1}{x}}$$,$$S_{0}^{\prime \prime }=-\frac{1}{x^{2}}e^{\frac{1}{x}}$$ This terms becomes $-\frac{1}{x^{2}}e^{\frac{1}{x}}+\frac{\left ( 2x+1\right ) }{x^{2}}e^{\frac{1}{x}}=\frac{-e^{\frac{1}{x}}+2xe^{\frac{1}{x}}+e^{\frac{1}{x}}}{x^{2}}=\frac{2xe^{\frac{1}{x}}}{x^{2}}=\frac{2e^{\frac{1}{x}}}{x}$ Therefore (2) becomes\begin{align*} \left ( S_{1}^{\prime }\right ) ^{2}+2S_{0}^{\prime }S_{1}^{\prime }+S_{1}^{\prime \prime }+\frac{\left ( 2x+1\right ) }{x^{2}}S_{1}^{\prime }\left ( x\right ) & \thicksim \frac{-2e^{\frac{1}{x}}}{x}\\ \frac{\left ( S_{1}^{\prime }\right ) ^{2}}{S_{0}^{\prime }}+2S_{1}^{\prime }+\frac{S_{1}^{\prime \prime }}{S_{0}^{\prime }}+\frac{\left ( 2x+1\right ) }{x^{2}S_{0}^{\prime }}S_{1}^{\prime }\left ( x\right ) & \thicksim \frac{-2e^{\frac{1}{x}}}{xS_{0}^{\prime }}\\ \frac{\left ( S_{1}^{\prime }\right ) ^{2}}{e^{\frac{1}{x}}}+2S_{1}^{\prime }+\frac{S_{1}^{\prime \prime }}{e^{\frac{1}{x}}}+\frac{\left ( 2x+1\right ) }{x^{2}e^{\frac{1}{x}}}S_{1}^{\prime }\left ( x\right ) & \thicksim \frac{-2}{x} \end{align*}

Assuming the balance is$S_{1}^{\prime }\thicksim \frac{-1}{x}$ Hence$S_{1}\left ( x\right ) \thicksim -\ln \left ( x\right ) +c$ Since $$c$$ subdominant as $$x\rightarrow 0^{+}$$ then$\fbox{S_1\left ( x\right ) \thicksim -\ln \left ( x\right ) }$ Veriﬁcation\begin{align*} S_{1}^{\prime } & \ggg \frac{\left ( S_{1}^{\prime }\right ) ^{2}}{e^{\frac{1}{x}}}\\ \frac{1}{x} & \ggg \frac{1}{x^{2}}\frac{1}{e^{\frac{1}{x}}}\\ e^{\frac{1}{x}} & \ggg \frac{1}{x} \end{align*}

Yes, for $$x\rightarrow 0^{+}$$

\begin{align*} S_{1}^{\prime } & \ggg \frac{S_{1}^{\prime \prime }}{e^{\frac{1}{x}}}\\ \frac{1}{x} & \ggg \frac{1}{x^{2}}\frac{1}{e^{\frac{1}{x}}} \end{align*}

Yes.

\begin{align*} S_{1}^{\prime } & \ggg \frac{\left ( 2x+1\right ) }{x^{2}e^{\frac{1}{x}}}S_{1}^{\prime }\left ( x\right ) \\ \frac{1}{x} & \ggg \frac{\left ( 2x+1\right ) }{x^{2}e^{\frac{1}{x}}}\frac{1}{x}\\ \frac{1}{x} & \ggg \frac{1}{x^{3}e^{\frac{1}{x}}} \end{align*}

Yes. All assumptions veriﬁed. Hence leading behavior is\begin{align*} y\left ( x\right ) & \thicksim \exp \left ( S_{0}\left ( x\right ) +S_{1}\left ( x\right ) \right ) \\ & \thicksim \exp \left ( \pm x^{2}e^{\frac{1}{x}}-\ln \left ( x\right ) \right ) \\ & \thicksim \frac{1}{x}\left ( \exp \left ( x^{2}e^{\frac{1}{x}}\right ) +\exp \left ( -x^{2}e^{\frac{1}{x}}\right ) \right ) \end{align*}

For small $$x$$, then we ignore $$\exp \left ( -x^{2}e^{\frac{1}{x}}\right )$$ since much smaller than $$\exp \left ( x^{2}e^{\frac{1}{x}}\right )$$. Therefore$y\left ( x\right ) \thicksim \frac{1}{x}\exp \left ( x^{2}e^{\frac{1}{x}}\right )$

3.2.5 problem 3.39(h)

problem Find leading asymptotic behavior as $$x\rightarrow \infty$$ for $$y^{\prime \prime }=e^{-\frac{3}{x}}y$$

solution Let $$y\left ( x\right ) =e^{S\left ( x\right ) }$$. Hence\begin{align*} y\left ( x\right ) & =e^{S_{0}\left ( x\right ) }\\ y^{\prime }\left ( x\right ) & =S_{0}^{\prime }e^{S_{0}}\\ y^{\prime \prime } & =S_{0}^{\prime \prime }e^{S_{0}}+\left ( S_{0}^{\prime }\right ) ^{2}e^{S_{0}}\\ & =\left ( S_{0}^{\prime \prime }+\left ( S_{0}^{\prime }\right ) ^{2}\right ) e^{S_{0}} \end{align*}

Substituting in the ODE gives\begin{align*} \left ( S_{0}^{\prime \prime }+\left ( S_{0}^{\prime }\right ) ^{2}\right ) e^{S_{0}} & =e^{-\frac{3}{x}}e^{S_{0}}\\ S_{0}^{\prime \prime }+\left ( S_{0}^{\prime }\right ) ^{2} & =e^{-\frac{3}{x}} \end{align*}

Assuming $$\left ( S_{0}^{\prime }\right ) ^{2}\ggg S_{0}^{\prime \prime }$$ the above becomes\begin{align*} \left ( S_{0}^{\prime }\right ) ^{2} & \thicksim e^{-\frac{3}{x}}\\ S_{0}^{\prime } & \thicksim \pm e^{-\frac{3}{2x}} \end{align*}

Hence$S_{0}\thicksim \pm \int e^{-\frac{3}{2x}}dx$ Integration by parts. Since $$\frac{d}{dx}e^{\frac{-3}{2x}}=\frac{3}{2x^{2}}e^{-\frac{3}{2x}}$$, then we rewrite the integral above as$\int e^{-\frac{3}{2x}}dx=\int \frac{3}{2x^{2}}e^{-\frac{3}{2x}}\left ( \frac{2x^{2}}{3}\right ) dx$ And now apply integration by parts. Let $$dv=\frac{3}{2x^{2}}e^{-\frac{3}{2x}}\rightarrow v=e^{-\frac{3}{2x}},u=\frac{2x^{2}}{3}\rightarrow du=\frac{4}{3}x$$, hence\begin{align*} \int e^{-\frac{3}{2x}}dx & =\left [ uv\right ] -\int vdu\\ & =\frac{2x^{2}}{3}e^{-\frac{3}{2x}}-\int \frac{4}{3}xe^{-\frac{3}{2x}}dx \end{align*}

Ignoring higher terms, then we use$S_{0}\thicksim \pm \frac{2x^{2}}{3}e^{-\frac{3}{2x}}$ Veriﬁcation\begin{align*} & \left ( S_{0}^{\prime }\right ) ^{2}\ggg S_{0}^{\prime \prime }\\ & \left ( e^{-\frac{3}{2x}}\right ) ^{2}\ggg \frac{3}{2x^{2}}e^{-\frac{3}{2x}}\\ & e^{-\frac{3}{x}}\ggg \frac{3}{2x^{2}}e^{-\frac{3}{2x}} \end{align*}

Yes, as $$x\rightarrow \infty .$$ To ﬁnd leading behavior, let $S\left ( x\right ) =S_{0}\left ( x\right ) +S_{1}\left ( x\right )$ Then $$y\left ( x\right ) =e^{S_{0}\left ( x\right ) +S_{1}\left ( x\right ) }$$ and hence now\begin{align*} y^{\prime }\left ( x\right ) & =\left ( S_{0}\left ( x\right ) +S_{1}\left ( x\right ) \right ) ^{\prime }e^{S_{0}+S_{1}}\\ y^{\prime \prime }\left ( x\right ) & =\left ( \left ( S_{0}+S_{1}\right ) ^{\prime }\right ) ^{2}e^{S_{0}+S_{1}}+\left ( S_{0}+S_{1}\right ) ^{\prime \prime }e^{S_{0}+S_{1}} \end{align*}

Using the above, the ODE $$y^{\prime \prime }=e^{-\frac{3}{x}}y$$ now becomes\begin{align*} \left ( \left ( S_{0}+S_{1}\right ) ^{\prime }\right ) ^{2}+\left ( S_{0}+S_{1}\right ) ^{\prime \prime } & =e^{-\frac{3}{x}}\\ \left ( S_{0}^{\prime }+S_{1}^{\prime }\right ) ^{2}+S_{0}^{\prime \prime }+S_{1}^{\prime \prime } & =e^{-\frac{3}{x}}\\ \left ( S_{0}^{\prime }\right ) ^{2}+\left ( S_{1}^{\prime }\right ) ^{2}+2S_{0}^{\prime }S_{1}^{\prime }+S_{0}^{\prime \prime }+S_{1}^{\prime \prime } & =e^{-\frac{3}{x}} \end{align*}

But $$\left ( S_{0}^{\prime }\right ) ^{2}\thicksim e^{-\frac{3}{x}}$$ hence the above simpliﬁes to$\left ( S_{1}^{\prime }\right ) ^{2}+2S_{0}^{\prime }S_{1}^{\prime }+S_{0}^{\prime \prime }+S_{1}^{\prime \prime }=0$ Assuming $$\left ( 2S_{0}^{\prime }S_{1}^{\prime }\right ) \ggg S_{1}^{\prime \prime }$$ the above becomes$\left ( S_{1}^{\prime }\right ) ^{2}+2S_{0}^{\prime }S_{1}^{\prime }+S_{0}^{\prime \prime }=0$ Assuming $$2S_{0}^{\prime }S_{1}^{\prime }\ggg \left ( S_{1}^{\prime }\right ) ^{2}$$\begin{align*} 2S_{0}^{\prime }S_{1}^{\prime }+S_{0}^{\prime \prime } & =0\\ S_{1}^{\prime } & \thicksim -\frac{S_{0}^{\prime \prime }}{2S_{0}^{\prime }}\\ S_{1} & \thicksim -\frac{1}{2}\ln \left ( S_{0}^{\prime }\right ) \end{align*}

But $$S_{0}^{\prime }\thicksim e^{-\frac{3}{2x}}$$, hence the above becomes\begin{align*} S_{1} & \thicksim -\frac{1}{2}\ln \left ( e^{-\frac{3}{2x}}\right ) +c\\ & \thicksim \frac{3}{4x}+c \end{align*}

Veriﬁcation\begin{align*} & \left ( 2S_{0}^{\prime }S_{1}^{\prime }\right ) \ggg S_{1}^{\prime \prime }\\ & \left ( 2e^{-\frac{3}{2x}}\frac{-3}{4x^{2}}\right ) \ggg \frac{3}{2x^{3}}\\ & \frac{3}{2}\frac{e^{-\frac{3}{2x}}}{x^{2}}\ggg \frac{3}{2x^{3}} \end{align*}

For large $$x$$ the above simpliﬁes to$\frac{1}{x^{2}}\ggg \frac{1}{x^{3}}$ Yes.\begin{align*} & \left ( 2S_{0}^{\prime }S_{1}^{\prime }\right ) \ggg \left ( S_{1}^{\prime }\right ) ^{2}\\ & \left ( 2e^{-\frac{3}{2x}}\frac{-3}{4x^{2}}\right ) \ggg \left ( \frac{-3}{4x^{2}}\right ) ^{2}\\ & \frac{3}{2}\frac{e^{-\frac{3}{2x}}}{x^{2}}\ggg \frac{9}{16x^{4}} \end{align*}

For large $$x$$ the above simpliﬁes to$\frac{1}{x^{2}}\ggg \frac{1}{x^{4}}$ Yes. All veriﬁed. Therefore, the leading behavior is\begin{align} y\left ( x\right ) & \sim \exp \left ( S_{0}\left ( x\right ) +S_{1}\left ( x\right ) \right ) \nonumber \\ & \sim \exp \left ( \pm \frac{2x^{2}}{3}e^{-\frac{3}{2x}}-\frac{1}{2}\ln \left ( e^{-\frac{3}{2x}}\right ) +c\right ) \nonumber \\ & \sim ce^{\frac{3}{4x}}\exp \left ( \pm \frac{2x^{2}}{3}e^{-\frac{3}{2x}}\right ) \tag{1} \end{align}

Check if we can use 3.4.28 to verify: $\lim _{x\rightarrow \infty }\left \vert x^{n}Q\left ( x\right ) \right \vert =\lim _{x\rightarrow \infty }\left \vert x^{2}e^{-\frac{3}{x}}\right \vert \rightarrow \infty$ We can use it. Lets verify using 3.4.28$y\left ( x\right ) \thicksim c\left [ Q\left ( x\right ) \right ] ^{\frac{1-n}{2n}}\exp \left ( \omega \int ^{x}Q\left [ t\right ] ^{\frac{1}{n}}dt\right )$ Where $$\omega ^{2}=1$$. For $$n=2,Q\left ( x\right ) =e^{-\frac{3}{x}}$$, the above gives\begin{align} y\left ( x\right ) & \thicksim c\left [ e^{-\frac{3}{x}}\right ] ^{\frac{1-2}{4}}\exp \left ( \omega \int ^{x}\left [ e^{-\frac{3}{t}}\right ] ^{\frac{1}{2}}dt\right ) \nonumber \\ & \thicksim c\left [ e^{-\frac{3}{x}}\right ] ^{\frac{-1}{4}}\exp \left ( \omega \int ^{x}e^{-\frac{3}{2t}}dt\right ) \nonumber \\ & \thicksim ce^{\frac{3}{4x}}\exp \left ( \omega \int ^{x}e^{-\frac{3}{2t}}dt\right ) \tag{2} \end{align}

We see that (1,2) are the same. Veriﬁed OK. Notice that in (1), we use the approximation for the $$e^{-\frac{3}{2x}}dx\approx \frac{2x^{2}}{3}e^{-\frac{3}{2x}}$$ we found earlier. This was done, since there is no closed form solution for the integral.

QED.

3.2.6 problem 3.42(a)

Problem: Extend investigation of example 1 of section 3.5 (a) Obtain the next few corrections to the leading behavior (3.5.5) then see how including these terms improves the numerical approximation of $$y\left ( x\right )$$ in 3.5.1.

Solution Example 1 at page 90 is  $$xy^{\prime \prime }+y^{\prime }=y$$. The leading behavior is given by 3.5.5 as ($$x\rightarrow \infty$$)$$y\left ( x\right ) \thicksim cx^{\frac{-1}{4}}e^{2x^{\frac{1}{2}}} \tag{3.5.5}$$ Where the book gives $$c=\frac{1}{2}\pi ^{\frac{-1}{2}}$$ on page 91. And 3.5.1 is$$y\left ( x\right ) =\sum _{n=0}^{\infty }\frac{x^{n}}{\left ( n!\right ) ^{2}} \tag{3.5.1}$$ To see the improvement, the book method is followed. This is described at end of page 91. This is done by plotting the leading behavior as ratio to $$y\left ( x\right )$$ as given in 3.5.1. Hence for the above leading behavior, we need to plot $\frac{\frac{1}{2}\pi ^{\frac{-1}{2}}x^{\frac{-1}{4}}e^{2x^{\frac{1}{2}}}}{y\left ( x\right ) }$ We are given $$S_{0}\left ( x\right ) ,S_{1}\left ( x\right )$$ in the problem. They are \begin{align*} S_{0}\left ( x\right ) & =2x^{\frac{1}{2}}\\ S_{1}\left ( x\right ) & =-\frac{1}{4}\ln x+c \end{align*}

Hence\begin{align} S_{0}^{\prime }\left ( x\right ) & =x^{\frac{-1}{2}}\nonumber \\ S_{0}^{\prime \prime } & =\frac{-1}{2}x^{-\frac{3}{2}}\nonumber \\ S_{1}^{\prime }\left ( x\right ) & =-\frac{1}{4}\frac{1}{x}\nonumber \\ S_{1}^{\prime \prime }\left ( x\right ) & =\frac{1}{4x^{2}} \tag{1} \end{align}

We need to ﬁnd $$S_{2}\left ( x\right ) ,S_{3}\left ( x\right )$$ $$,\cdots$$ to see that this will improve the solution $$y\left ( x\right ) \thicksim \exp \left ( S_{0}+S_{1}+S_{2}+\cdots \right )$$ as $$x\rightarrow x_{0}$$ compared to just using leading behavior $$y\left ( x\right ) \thicksim \exp \left ( S_{0}+S_{1}\right )$$. So now we need to ﬁnd $$S_{2}\left ( x\right )$$

Let $$y\left ( x\right ) =e^{S}$$, then the ODE becomes$x\left ( S^{\prime \prime }+\left ( S^{\prime }\right ) ^{2}\right ) +S^{\prime }=1$ Replacing $$S$$ by $$S_{0}\left ( x\right ) +S_{1}\left ( x\right ) +S_{2}\left ( x\right )$$ in the above gives\begin{align*} \left ( S_{0}+S_{1}+S_{2}\right ) ^{\prime \prime }+\left [ \left ( S_{0}+S_{1}+S_{2}\right ) ^{\prime }\right ] ^{2}+\frac{1}{x}\left ( S_{0}+S_{1}+S_{2}\right ) ^{\prime } & \thicksim \frac{1}{x}\\ \left \{ S_{0}^{\prime \prime }+S_{1}^{\prime \prime }+S_{2}^{\prime \prime }\right \} +\left [ \left ( S_{0}^{\prime }+S_{1}^{\prime }+S_{2}^{\prime }\right ) \right ] ^{2}+\frac{1}{x}\left ( S_{0}^{\prime }+S_{1}^{\prime }+S_{2}^{\prime }\right ) & \thicksim \frac{1}{x}\\ \left \{ S_{0}^{\prime \prime }+S_{1}^{\prime \prime }+S_{2}^{\prime \prime }\right \} +\left \{ \left [ S_{0}^{\prime }\right ] ^{2}+2S_{0}^{\prime }S_{1}^{\prime }+2S_{0}^{\prime }S_{2}^{\prime }+\left [ S_{1}^{\prime }\right ] ^{2}+2S_{1}^{\prime }S_{2}^{\prime }+\left [ S_{2}^{\prime }\right ] ^{2}\right \} +\frac{1}{x}\left \{ S_{0}^{\prime }+S_{1}^{\prime }+S_{2}^{\prime }\right \} & \thicksim \frac{1}{x} \end{align*}

Moving all known quantities to the RHS, these are $$S_{0}^{\prime \prime },S_{1}^{\prime \prime },\left [ S_{0}^{\prime }\right ] ^{2},2S_{0}^{\prime }S_{1}^{\prime },S_{0}^{\prime },S_{1}^{\prime },\left [ S_{1}^{\prime }\right ] ^{2}$$ then the above reduces to$\left \{ S_{2}^{\prime \prime }\right \} +\left \{ +2S_{0}^{\prime }S_{2}^{\prime }+2S_{1}^{\prime }S_{2}^{\prime }+\left [ S_{2}^{\prime }\right ] ^{2}\right \} +\frac{1}{x}\left \{ S_{2}^{\prime }\right \} \thicksim \frac{1}{x}-S_{0}^{\prime \prime }-S_{1}^{\prime \prime }-\left [ S_{0}^{\prime }\right ] ^{2}-2S_{0}^{\prime }S_{1}^{\prime }-\frac{1}{x}S_{0}^{\prime }-\frac{1}{x}S_{1}^{\prime }-\left [ S_{1}^{\prime }\right ] ^{2}$ Replacing known terms, by using (1) into the above gives\begin{multline*} \left \{ S_{2}^{\prime \prime }\right \} +\left \{ 2S_{0}^{\prime }S_{2}^{\prime }+2S_{1}^{\prime }S_{2}^{\prime }+\left [ S_{2}^{\prime }\right ] ^{2}\right \} +\frac{1}{x}\left \{ S_{2}^{\prime }\right \} \thicksim \\ \frac{1}{x}+\frac{1}{2}x^{-\frac{3}{2}}-\frac{1}{4x^{2}}-\left [ x^{\frac{-1}{2}}\right ] ^{2}-2\left ( x^{\frac{-1}{2}}\right ) \left ( -\frac{1}{4}\frac{1}{x}\right ) -\frac{1}{x}\left ( x^{\frac{-1}{2}}\right ) -\frac{1}{x}\left ( -\frac{1}{4}\frac{1}{x}\right ) -\left ( -\frac{1}{4}\frac{1}{x}\right ) ^{2} \end{multline*} Simplifying gives$\left \{ S_{2}^{\prime \prime }\right \} +\left \{ 2S_{0}^{\prime }S_{2}^{\prime }+2S_{1}^{\prime }S_{2}^{\prime }+\left [ S_{2}^{\prime }\right ] ^{2}\right \} +\frac{1}{x}\left \{ S_{2}^{\prime }\right \} \thicksim \frac{1}{x}+\frac{1}{2}x^{-\frac{3}{2}}-\frac{1}{4x^{2}}-x^{-1}+\frac{1}{2}x^{\frac{-3}{2}}-x^{\frac{-3}{2}}+\frac{1}{4}\frac{1}{x^{2}}-\frac{1}{16}\frac{1}{x^{2}}$ Hence$\left \{ S_{2}^{\prime \prime }\right \} +\left \{ 2S_{0}^{\prime }S_{2}^{\prime }+2S_{1}^{\prime }S_{2}^{\prime }+\left [ S_{2}^{\prime }\right ] ^{2}\right \} +\frac{1}{x}\left \{ S_{2}^{\prime }\right \} \thicksim -\frac{1}{16}\frac{1}{x^{2}}$ Lets assume now that $$2S_{0}^{\prime }S_{2}^{\prime }\thicksim -\frac{1}{16}\frac{1}{x^{2}} \tag{2}$$ Therefore \begin{align*} S_{2}^{\prime } & \thicksim -\frac{1}{32}\frac{1}{S_{0}^{\prime }x^{2}}\\ & \thicksim -\frac{1}{32}\frac{1}{\left ( x^{\frac{-1}{2}}\right ) x^{2}}\\ & \thicksim -\frac{1}{32}x^{\frac{-3}{2}} \end{align*}

We can now verify this before solving the ODE. We need to check that (as $$x\rightarrow \infty$$)\begin{align*} 2S_{0}^{\prime }S_{2}^{\prime } & \ggg S_{2}^{\prime \prime }\\ 2S_{0}^{\prime }S_{2}^{\prime } & \ggg 2S_{1}^{\prime }S_{2}^{\prime }\\ 2S_{0}^{\prime }S_{2}^{\prime } & \ggg \left [ S_{2}^{\prime }\right ] ^{2}\\ 2S_{0}^{\prime }S_{2}^{\prime } & \ggg \frac{1}{x}S_{2}^{\prime } \end{align*}

Where $$S_{2}^{\prime \prime }\thicksim x^{\frac{-5}{2}}$$, Hence\begin{align*} 2S_{0}^{\prime }S_{2}^{\prime } & \ggg S_{2}^{\prime \prime }\\ x^{\frac{-1}{2}}\left ( x^{\frac{-3}{2}}\right ) & \ggg x^{\frac{-5}{2}}\\ x^{-2} & \ggg x^{\frac{-5}{2}} \end{align*}

Yes.\begin{align*} 2S_{0}^{\prime }S_{2}^{\prime } & \ggg 2S_{1}^{\prime }S_{2}^{\prime }\\ x^{-2} & \ggg \left ( \frac{1}{x}\right ) \left ( x^{\frac{-3}{2}}\right ) \\ x^{-2} & \ggg x^{\frac{-5}{2}} \end{align*}

Yes\begin{align*} 2S_{0}^{\prime }S_{2}^{\prime } & \ggg \left [ S_{2}^{\prime }\right ] ^{2}\\ x^{-2} & \ggg \left ( x^{\frac{-3}{2}}\right ) ^{2}\\ x^{-2} & \ggg x^{-3} \end{align*}

Yes\begin{align*} 2S_{0}^{\prime }S_{2}^{\prime } & \ggg \frac{1}{x}S_{2}^{\prime }\\ x^{-2} & \ggg \frac{1}{x}x^{\frac{-3}{2}}\\ x^{-2} & \ggg x^{\frac{-5}{2}} \end{align*}

Yes. All assumptions are veriﬁed. Therefore we can g ahead and solve for $$S_{2}$$ using (2)\begin{align*} 2S_{0}^{\prime }S_{2}^{\prime } & \thicksim -\frac{1}{16}\frac{1}{x^{2}}\\ S_{2}^{\prime } & \thicksim -\frac{1}{32}\frac{1}{x^{2}}\frac{1}{S_{0}^{\prime }}\\ & \thicksim -\frac{1}{32}\frac{1}{x^{2}}\frac{1}{x^{\frac{-1}{2}}}\\ & \thicksim -\frac{1}{32}\frac{1}{x^{\frac{3}{2}}} \end{align*}

Hence$\fbox{S_2\thicksim \frac{1}{16}\frac{1}{\sqrt{x}}}$ The leading behavior now is \begin{align*} y\left ( x\right ) & \thicksim \exp \left ( S_{0}+S_{1}+S_{2}\right ) \\ & \thicksim \exp \left ( 2x^{\frac{1}{2}}-\frac{1}{4}\ln x+c+\frac{1}{16}\frac{1}{\sqrt{x}}\right ) \end{align*}

Now we will ﬁnd $$S_{3}$$. From$x\left ( S^{\prime \prime }+\left ( S^{\prime }\right ) ^{2}\right ) +S^{\prime }=1$ Replacing $$S$$ by $$S_{0}+S_{1}+S_{2}+S_{3}$$ in the above gives\begin{align*} \left ( S_{0}+S_{1}+S_{2}+S_{3}\right ) ^{\prime \prime }+\left [ \left ( S_{0}+S_{1}+S_{2}+S_{3}\right ) ^{\prime }\right ] ^{2}+\frac{1}{x}\left ( S_{0}+S_{1}+S_{2}+S_{3}\right ) ^{\prime } & \thicksim \frac{1}{x}\\ \left \{ S_{0}^{\prime \prime }+S_{1}^{\prime \prime }+S_{2}^{\prime \prime }+S_{3}^{\prime \prime }\right \} +\left [ \left ( S_{0}^{\prime }+S_{1}^{\prime }+S_{2}^{\prime }+S_{3}^{\prime }\right ) \right ] ^{2}+\frac{1}{x}\left ( S_{0}^{\prime }+S_{1}^{\prime }+S_{2}^{\prime }+S_{3}^{\prime }\right ) & \thicksim \frac{1}{x} \end{align*}

Hence\begin{multline*} \left \{ S_{0}^{\prime \prime }+S_{1}^{\prime \prime }+S_{2}^{\prime \prime }+S_{3}^{\prime \prime }\right \} +\\ \left \{ \left [ S_{0}^{\prime }\right ] ^{2}+2S_{0}^{\prime }S_{1}^{\prime }+2S_{0}^{\prime }S_{2}^{\prime }+2S_{1}^{\prime }S_{2}^{\prime }+\left [ S_{1}^{\prime }\right ] ^{2}+\left [ S_{2}^{\prime }\right ] ^{2}+2S_{0}^{\prime }S_{3}^{\prime }+2S_{1}^{\prime }S_{3}^{\prime }+2S_{2}^{\prime }S_{3}^{\prime }+\left [ S_{3}^{\prime }\right ] ^{2}\right \} \\ +\frac{1}{x}\left \{ S_{0}^{\prime }+S_{1}^{\prime }+S_{2}^{\prime }+S_{3}^{\prime }\right \} \thicksim \frac{1}{x} \end{multline*} Moving all known quantities to the RHS gives\begin{align} & \left \{ S_{3}^{\prime \prime }\right \} +\left \{ 2S_{0}^{\prime }S_{3}^{\prime }+2S_{1}^{\prime }S_{3}^{\prime }+2S_{2}^{\prime }S_{3}^{\prime }+\left [ S_{3}^{\prime }\right ] ^{2}\right \} +\frac{1}{x}\left \{ S_{3}^{\prime }\right \} \nonumber \\ & \thicksim \frac{1}{x}-S_{0}^{\prime \prime }-S_{1}^{\prime \prime }-S_{2}^{\prime \prime }-\left [ S_{0}^{\prime }\right ] ^{2}-2S_{0}^{\prime }S_{1}^{\prime }-2S_{0}^{\prime }S_{2}^{\prime }-2S_{1}^{\prime }S_{2}^{\prime }-\left [ S_{1}^{\prime }\right ] ^{2}-\left [ S_{2}^{\prime }\right ] ^{2}-\frac{1}{x}S_{0}^{\prime }-\frac{1}{x}S_{1}^{\prime }-\frac{1}{x}S_{2}^{\prime } \tag{3} \end{align}

Now we will simplify the RHS, since it is all known. Using\begin{align*} S_{0}^{\prime }\left ( x\right ) & =x^{\frac{-1}{2}}\\ \left [ S_{0}^{\prime }\right ] ^{2} & =x^{-1}\\ S_{0}^{\prime \prime } & =\frac{-1}{2}x^{-\frac{3}{2}}\\ S_{1}^{\prime }\left ( x\right ) & =-\frac{1}{4}\frac{1}{x}\\ \left [ S_{1}^{\prime }\left ( x\right ) \right ] ^{2} & =\frac{1}{16}\frac{1}{x^{2}}\\ S_{1}^{\prime \prime }\left ( x\right ) & =\frac{1}{4x^{2}}\\ S_{2}^{\prime }\left ( x\right ) & =-\frac{1}{32}\frac{1}{x^{\frac{3}{2}}}\\ \left [ S_{2}^{\prime }\left ( x\right ) \right ] ^{2} & =\frac{1}{1024x^{3}}\\ S_{2}^{\prime \prime }\left ( x\right ) & =\frac{3}{64}\frac{1}{x^{\frac{5}{2}}}\\ 2S_{0}^{\prime }S_{1}^{\prime } & =2\left ( x^{\frac{-1}{2}}\right ) \left ( -\frac{1}{4}\frac{1}{x}\right ) =-\frac{1}{2}\frac{1}{x^{\frac{3}{2}}}\\ 2S_{0}^{\prime }S_{2}^{\prime } & =2\left ( x^{\frac{-1}{2}}\right ) \left ( -\frac{1}{32}\frac{1}{x^{\frac{3}{2}}}\right ) =-\frac{1}{16x^{2}}\\ 2S_{1}^{\prime }S_{2}^{\prime } & =2\left ( -\frac{1}{4}\frac{1}{x}\right ) \left ( -\frac{1}{32}\frac{1}{x^{\frac{3}{2}}}\right ) =\frac{1}{64x^{\frac{5}{2}}} \end{align*}

Hence (3) becomes\begin{multline*} \left \{ S_{3}^{\prime \prime }\right \} +\left \{ 2S_{0}^{\prime }S_{3}^{\prime }+2S_{1}^{\prime }S_{3}^{\prime }+2S_{2}^{\prime }S_{3}^{\prime }+\left [ S_{3}^{\prime }\right ] ^{2}\right \} +\frac{1}{x}\left \{ S_{3}^{\prime }\right \} \thicksim \\ \frac{1}{x}+\frac{1}{2x^{\frac{3}{2}}}-\frac{1}{4x^{2}}-\frac{3}{64}\frac{1}{x^{\frac{5}{2}}}-\frac{1}{x}+\frac{1}{2}\frac{1}{x^{\frac{3}{2}}}+\frac{1}{16x^{2}}-\frac{1}{64x^{\frac{5}{2}}}-\frac{1}{16}\frac{1}{x^{2}}-\frac{1}{1024x^{3}}-\frac{1}{x^{\frac{3}{2}}}+\frac{1}{4}\frac{1}{x^{2}}+\frac{1}{32}\frac{1}{x^{\frac{5}{2}}} \end{multline*} Simplifying gives$\left \{ S_{3}^{\prime \prime }\right \} +\left \{ 2S_{0}^{\prime }S_{3}^{\prime }+2S_{1}^{\prime }S_{3}^{\prime }+2S_{2}^{\prime }S_{3}^{\prime }+\left [ S_{3}^{\prime }\right ] ^{2}\right \} +\frac{1}{x}\left \{ S_{3}^{\prime }\right \} \thicksim -\left ( \frac{32}{1024x^{\frac{5}{2}}}+\frac{1}{1024x^{3}}\right )$ Let us now assume that \begin{align*} S_{0}^{\prime }S_{3}^{\prime } & \ggg S_{1}^{\prime }S_{3}^{\prime }\\ S_{0}^{\prime }S_{3}^{\prime } & \ggg S_{2}^{\prime }S_{3}^{\prime }\\ S_{0}^{\prime }S_{3}^{\prime } & \ggg \left [ S_{3}^{\prime }\right ] ^{2}\\ S_{0}^{\prime }S_{3}^{\prime } & \ggg \frac{1}{x}\left \{ S_{3}^{\prime }\right \} \\ S_{0}^{\prime }S_{3}^{\prime } & \ggg S_{3}^{\prime \prime } \end{align*}

Therefore, we end up with the balance\begin{align*} 2S_{0}^{\prime }S_{3}^{\prime } & \thicksim -\left ( \frac{32}{1024x^{\frac{5}{2}}}+\frac{1}{1024x^{3}}\right ) \\ S_{3}^{\prime } & \thicksim -\left ( \frac{32}{1024x^{\frac{5}{2}}S_{0}^{\prime }}+\frac{1}{1024x^{3}S_{0}^{\prime }}\right ) \\ & \thicksim -\left ( \frac{32}{1024x^{\frac{5}{2}}\left ( x^{\frac{-1}{2}}\right ) }+\frac{1}{1024x^{3}\left ( x^{\frac{-1}{2}}\right ) }\right ) \\ & \thicksim -\left ( \frac{1}{32x^{2}}+\frac{1}{1024x^{\frac{5}{2}}}\right ) \end{align*}

Hence$\fbox{S_3\thicksim \frac{1}{1024}\left ( \frac{2}{32x^2}+\frac{32}{x}\right ) }$ Where constant of integration was ignored. Let us now verify the assumptions made\begin{align*} S_{0}^{\prime }S_{3}^{\prime } & \ggg S_{1}^{\prime }S_{3}^{\prime }\\ S_{0}^{\prime } & \ggg S_{1}^{\prime } \end{align*}

Yes. \begin{align*} S_{0}^{\prime }S_{3}^{\prime } & \ggg S_{2}^{\prime }S_{3}^{\prime }\\ S_{0}^{\prime } & \ggg S_{2}^{\prime } \end{align*}

Yes\begin{align*} S_{0}^{\prime }S_{3}^{\prime } & \ggg \left [ S_{3}^{\prime }\right ] ^{2}\\ S_{0}^{\prime } & \ggg S_{3}^{\prime } \end{align*}

Yes.\begin{align*} S_{0}^{\prime }S_{3}^{\prime } & \ggg \frac{1}{x}\left \{ S_{3}^{\prime }\right \} \\ S_{0}^{\prime } & \ggg \frac{1}{x}\\ x^{\frac{-1}{2}} & \ggg \frac{1}{x} \end{align*}

Yes, as $$x\rightarrow \infty$$, and ﬁnally\begin{align*} S_{0}^{\prime }S_{3}^{\prime } & \ggg S_{3}^{\prime \prime }\\ x^{\frac{-1}{2}}\left ( \frac{1}{32x^{2}}+\frac{1}{1024x^{\frac{5}{2}}}\right ) & \ggg \left ( \frac{5}{2048x^{\frac{7}{2}}}+\frac{1}{16x^{3}}\right ) \\ \frac{\left ( 32\sqrt{x}+1\right ) }{1024x^{3}} & \ggg \frac{128\sqrt{x}+5}{2048x^{\frac{7}{2}}} \end{align*}

Yes, as $$x\rightarrow \infty .$$ All assumptions veriﬁed. The leading behavior now is \begin{align*} y\left ( x\right ) & \thicksim \exp \left ( S_{0}+S_{1}+S_{2}+S_{3}\right ) \\ & \thicksim \exp \left ( 2x^{\frac{1}{2}}-\frac{1}{4}\ln x+c+\frac{1}{16}\frac{1}{\sqrt{x}}+\frac{1}{1024}\left ( \frac{2}{32x^{2}}+\frac{32}{x}\right ) \right ) \\ & \thicksim cx^{\frac{-1}{4}}\exp \left ( 2x^{\frac{1}{2}}+\frac{1}{16}\frac{1}{\sqrt{x}}+\frac{1}{1024}\left ( \frac{2}{32x^{2}}+\frac{32}{x}\right ) \right ) \end{align*}

Now we will show how adding more terms to leading behavior improved the $$y\left ( x\right )$$ solution for large $$x$$. When plotting the solutions, we see that $$\frac{\exp \left ( S_{0}+S_{1}+S_{2}+S_{3}\right ) }{y\left ( x\right ) }$$ approached the ratio $$1$$ sooner than $$\frac{\exp \left ( S_{0}+S_{1}+S_{2}\right ) }{y\left ( x\right ) }$$ and this in turn approached the ratio $$1$$ sooner than just using $$\frac{\exp \left ( S_{0}+S_{1}\right ) }{y\left ( x\right ) }$$. So the eﬀect of adding more terms, is that the solution becomes more accurate for larger range of $$x$$ values. Below is the code used and the plot generated.

3.2.7 problem 3.49(c)

Problem Find the leading behavior as $$x\rightarrow \infty$$ of the general solution of $$y^{\prime \prime }+xy=x^{5}$$

Solution This is non-homogenous ODE. We solve this by ﬁrst ﬁnding the homogenous solution (asymptotic solution) and then ﬁnding particular solution.  Hence we start with $y_{h}^{\prime \prime }+xy_{h}=0$ $$x=\infty$$ is ISP point. Therefore, we assume $$y_{h}\left ( x\right ) =e^{S\left ( x\right ) }$$ and obtain$$S^{\prime \prime }+\left ( S^{\prime }\right ) ^{2}+x=0 \tag{1}$$ Let $S\left ( x\right ) =S_{0}+S_{1}+\cdots$ Therefore (1) becomes\begin{align} \left ( S_{0}^{\prime \prime }+S_{1}^{\prime \prime }+\cdots \right ) +\left ( S_{0}^{\prime }+S_{1}^{\prime }+\cdots \right ) ^{2} & =-x\nonumber \\ \left ( S_{0}^{\prime \prime }+S_{1}^{\prime \prime }+\cdots \right ) +\left ( \left [ S_{0}^{\prime }\right ] ^{2}+2S_{0}^{\prime }S_{1}^{\prime }+\left [ S_{1}^{\prime }\right ] ^{2}+\cdots \right ) & =-x \tag{2} \end{align}

Assuming $$\left [ S_{0}^{\prime }\right ] ^{2}\ggg S_{0}^{\prime \prime }$$ we obtain\begin{align*} \left [ S_{0}^{\prime }\right ] ^{2} & \thicksim -x\\ S_{0}^{\prime } & \thicksim \omega \sqrt{x} \end{align*}

Where $$\omega =\pm i$$

Veriﬁcation\begin{align*} \left [ S_{0}^{\prime }\right ] ^{2} & \ggg S_{0}^{\prime \prime }\\ x & \ggg \frac{1}{2}\frac{1}{\sqrt{x}} \end{align*}

Yes, as $$x\rightarrow \infty$$. Hence$S_{0}\thicksim \frac{3}{2}\omega x^{\frac{3}{2}}$ Now we will ﬁnd $$S_{1}$$. From (2), and moving all known terms to RHS$$\left ( S_{1}^{\prime \prime }+\cdots \right ) +\left ( 2S_{0}^{\prime }S_{1}^{\prime }+\left [ S_{1}^{\prime }\right ] ^{2}+\cdots \right ) \thicksim -x-S_{0}^{\prime \prime }-\left [ S_{0}^{\prime }\right ] ^{2} \tag{3}$$ Assuming \begin{align*} 2S_{0}^{\prime }S_{1}^{\prime } & \ggg S_{1}^{\prime \prime }\\ 2S_{0}^{\prime }S_{1}^{\prime } & \ggg \left [ S_{1}^{\prime }\right ] ^{2} \end{align*}

Then (3) becomes (where $$S_{0}^{\prime }\thicksim \omega \sqrt{x},\left [ S_{0}^{\prime }\right ] ^{2}\thicksim -x,S_{0}^{\prime \prime }\thicksim \frac{1}{2}\omega \frac{1}{\sqrt{x}}$$)\begin{align*} 2S_{0}^{\prime }S_{1}^{\prime } & \thicksim -x-S_{0}^{\prime \prime }-\left [ S_{0}^{\prime }\right ] ^{2}\\ S_{1}^{\prime } & \thicksim \frac{-x-S_{0}^{\prime \prime }-\left [ S_{0}^{\prime }\right ] ^{2}}{2S_{0}^{\prime }}\\ & \thicksim \frac{-x-\frac{1}{2}\omega \frac{1}{\sqrt{x}}-\left ( -x\right ) }{2\omega \sqrt{x}}\\ & \thicksim -\frac{1}{4x} \end{align*}

Veriﬁcation (where $$S_{1}^{\prime \prime }\thicksim \frac{1}{4}\frac{1}{x^{2}}$$)\begin{align*} 2S_{0}^{\prime }S_{1}^{\prime } & \ggg S_{1}^{\prime \prime }\\ \sqrt{x}\left ( \frac{1}{4x}\right ) & \ggg \frac{1}{4}\frac{1}{x^{2}}\\ \frac{1}{x^{\frac{1}{2}}} & \ggg \frac{1}{x^{2}} \end{align*}

Yes, as $$x\rightarrow \infty$$\begin{align*} 2S_{0}^{\prime }S_{1}^{\prime } & \ggg \left [ S_{1}^{\prime }\right ] ^{2}\\ \frac{1}{x^{\frac{1}{2}}} & \ggg \left ( \frac{1}{4x}\right ) ^{2}\\ \frac{1}{x^{\frac{1}{2}}} & \ggg \frac{1}{16x^{2}} \end{align*}

Yes, as $$x\rightarrow \infty$$. All validated. We solve for $$S_{1}$$\begin{align*} S_{1}^{\prime } & \thicksim -\frac{1}{4x}\\ S_{1} & \thicksim -\frac{1}{4}\ln x+c \end{align*}

$$y_{h}$$ is found. It is given by\begin{align*} y_{h}\left ( x\right ) & \thicksim \exp \left ( S_{0}\left ( x\right ) +S_{1}\left ( x\right ) \right ) \\ & \thicksim \exp \left ( \frac{3}{2}\omega x^{\frac{3}{2}}-\frac{1}{4}\ln x+c\right ) \\ & \thicksim cx^{\frac{-1}{4}}\exp \left ( \frac{3}{2}\omega x^{\frac{3}{2}}\right ) \end{align*}

Now that we have found $$y_{h}$$, we go back and look at $y^{\prime \prime }+xy=x^{5}$ And consider two cases  (a) $$y^{\prime \prime }\thicksim x^{5}$$ (b) $$xy\thicksim x^{5}$$. The case of $$y^{\prime \prime }\thicksim xy$$ was covered above. This is what we did to ﬁnd $$y_{h}\left ( x\right )$$.

case (a)\begin{align*} y_{p}^{\prime \prime } & \thicksim x^{5}\\ y_{p}^{\prime } & \thicksim \frac{1}{5}x^{4}\\ y_{p} & \thicksim \frac{1}{20}x^{3} \end{align*}

Where constants of integration are ignored since subdominant for $$x\rightarrow \infty$$. Now we check if this case is valid. \begin{align*} xy_{p} & \lll x^{5}\\ x\frac{1}{20}x^{3} & \lll x^{5}\\ x^{4} & \lll x^{5} \end{align*}

No. Therefore case (a) did not work out. We try case (b) now\begin{align*} xy_{p} & \thicksim x^{5}\\ y_{p} & \thicksim x^{4} \end{align*}

Now we check is this case is valid. \begin{align*} y_{p}^{\prime \prime } & \lll x^{5}\\ 12x^{2} & \lll x^{5} \end{align*}

Yes. Therefore, we found $y_{p}\thicksim x^{4}$ Hence the complete asymptotic solution is \begin{align*} y\left ( x\right ) & \thicksim y_{h}\left ( x\right ) +y_{p}\left ( x\right ) \\ & \thicksim cx^{\frac{-1}{4}}\exp \left ( \frac{3}{2}\omega x^{\frac{3}{2}}\right ) +x^{4} \end{align*}

3.2.8 key solution of selected problems

3.2.8.1 section 3 problem 27
3.2.8.2 section 3 problem 33 b
3.2.8.3 section 3 problem 33 c
3.2.8.4 section 3 problem 35
3.2.8.5 section 3 problem 42a