0
$\begingroup$

I have this non-constant coefficient homogeneous second ODE:

$$(1 + x^{2})y'' + 4xy' + 2y = 0.$$

I have found a power series solution for this equation but I am then asked to transform it to system of first order ODE and also fundamental matrix form for that. But i don’t know how to transform it. I can do it if that equation is originally with constant coefficient, but that is polynomial coefficient and i just don’t know how. I am also not sure if that equation is in Euler form or not since it has $1 + x^2$ instead of $x^{2}$

Any help would be appreciated. Thank you very much :)

$\endgroup$
1
  • $\begingroup$ Notice that $(uy)'' = u''y + 2u'y' + uy''$. So let $u=1+x^2$, $u'=2x$ and $u''=2$ and you have your equation back. Its a trivial example of this form of problem. So $uy=Cx+D$ where $u$ is now known. $\endgroup$ Apr 21, 2018 at 20:45

2 Answers 2

1
$\begingroup$

Making the substitution

$$ y = \frac{x^\lambda}{1+x^2} $$

we get

$$ \lambda(\lambda-1)x^{\lambda-2} = 0 \Rightarrow \left\{ \begin{array}{rcl} \lambda & = & 0\\ \lambda & = & 1 \end{array}\right. $$

hence

$$ y = \frac{C_1}{1+x^2}+\frac{C_2x}{1+x^2} $$

$\endgroup$
0
1
$\begingroup$

Let $u=x^2+1$. Then $u'=2x$ and $u''=2$. Clearly then $(1+x^2)y'' + 4xy' + 2y = 0$ can be rewritten $uy'' + 2u'y' + u''y=0$. This is equivalent to $(uy)''=0$. Now just integrate twice. $(uy)' = C$ and $uy = Cx + D$. We know $u=x^2+1$ so $(x^2+1)y = Cx+D$ or if you prefer $y=\frac{Cx + D}{x^2+1}$.

$\endgroup$
1
  • 1
    $\begingroup$ After such transformation you get $$\tag{$*$}\mathbf{x}'=\begin{bmatrix}0&1\\ \frac{2}{1+x^2}&\frac{4x}{1+x^2}\end{bmatrix}\mathbf{x},$$ and a fundamental matrix $$\begin{bmatrix}\frac{1}{1+x^2}&\frac{x}{1+x^2} \\ \frac{-2x}{(1+x^2)^2}&\frac{1-x^2}{(1+x^2)^2}\end{bmatrix}.$$ The latter equals $I$ for $x=0$, so is the principal fundamental matrix solution of $(*)$ at $x=0$. $\endgroup$
    – user539887
    Apr 22, 2018 at 19:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy