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) \begin{equation} S_{1}\left ( x\right ) \thicksim \frac{1-n}{2n}\ln \left ( Q\left ( x\right ) \right ) +c\tag{10.2.11} \end{equation} 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)
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\begin{equation} 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} \end{equation} 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
\begin{equation} 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} \end{equation}
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.
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\begin{equation} S_{0}^{\prime \prime }+\left ( S_{0}^{\prime }\right ) ^{2}=\sqrt{x}\tag{1} \end{equation} 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*}
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\[ y\left ( x\right ) \thicksim \frac{1}{x}\exp \left ( x^{2}e^{\frac{1}{x}}\right ) \]
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.
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 (\(x\rightarrow \infty \))\begin{equation} y\left ( x\right ) \thicksim cx^{\frac{-1}{4}}e^{2x^{\frac{1}{2}}} \tag{3.5.5} \end{equation} Where the book gives \(c=\frac{1}{2}\pi ^{\frac{-1}{2}}\) on page 91. And 3.5.1 is\begin{equation} y\left ( x\right ) =\sum _{n=0}^{\infty }\frac{x^{n}}{\left ( n!\right ) ^{2}} \tag{3.5.1} \end{equation} 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 \begin{equation} 2S_{0}^{\prime }S_{2}^{\prime }\thicksim -\frac{1}{16}\frac{1}{x^{2}} \tag{2} \end{equation} 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.
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\begin{equation} S^{\prime \prime }+\left ( S^{\prime }\right ) ^{2}+x=0 \tag{1} \end{equation} 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\begin{equation} \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} \end{equation} 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*}