Asked to find first three terms in the local behavior as $x\to 0^+$ of the solutions of $\displaystyle{y'+ xy = \frac{1}{x^3}}$
This was taken by bender and orszag book
Working :
I tried to use method of dominance but later realised that we can find using the series expansion. But I am not sure how to proceed. Please advise how to proceed with it.
Since this equation is linear the simplest way to find asymptotic behaviour is to try to estimate the asymptotic behaviour of the constant while applying the variation of constant method.
So first you solve the homegeneous equation : $y'+xy=0\iff y=A\,e^{-x^2/2}$.
Then applying constant variation : $\require{cancel}\displaystyle{\frac{1}{x^3}=y'+xy=\cancel{-Ax\,e^{-x^2/2}}+A'e^{-x^2/2}+\cancel{Ax\,e^{-x^2/2}}}$
We get $\displaystyle{A'(x)=\frac{e^{\frac{x^2}{2}}}{x^3}}=\left(1+\frac{x^2}{2}+\frac{x^4}{8}+o(x^4)\right)/x^3=\frac{1}{x^3}+\frac{1}{2x}+\frac x8+o(x)$
We integrate this $\displaystyle{A(x)=B-\frac{1}{2x^2}+\frac{\ln(x)}{2}+\frac{x^2}{16}+o(x^2)}$
And multiply the by $\displaystyle{e^{-\frac{x^2}{2}}=1-\frac{x^2}{2}+o(x^2)}$
Finally $\displaystyle{y(x)=B-\frac{1}{2x^2}+\frac{\ln(x)}{2}+\frac{x^2}{16}-\frac{Bx^2}{2}+\frac{1}{4}-\frac{x^2\ln(x)}{4}+o(x^2)}$
If we limit ourselves to $3$ terms as asked then $\bbox[5px,border:2px solid ]{\displaystyle{y(x)=-\frac{1}{2x^2}+C+\frac{\ln(x)}{2}+o(x)}}$ with $C\in\mathbb R$ near $0^+$.
There are no negligible term in our equation.
But let's set $\displaystyle{g(x)=y(x)+\frac{1}{2x^2}-\frac12\ln(x)}\qquad(*)$
According to previous study we have to show that $g(x)\to C$
Rem: to find $(*)$ just do it incrementaly, negliglate $xy$, solve, and so on, until rhs do not diverge to infinity anymore.
We have $\displaystyle{g'=y'-\frac{1}{x^3}-\frac{1}{2x}=-xy-\frac{1}{2x}=\left(-xg+\frac{1}{2x}-\frac x2\ln(x)\right)-\frac{1}{2x}}$
So $\displaystyle{g'+xg=-\frac{x\ln(x)}{2}}\to 0$ when $x\to 0^+\quad$ now this equation has something negligible.
The homegeneous equation $g'+xg=0$ gives $g=A.e^{-\frac{x^2}{2}}\to A$ when $x\to 0$.
This is working but I feel that this method is full of fluff compared to first method.