Finding $L^2$ norm of solution of ODE I have a linear differential equation with real constant coefficients
$$
\sum\limits_{i=0}^3 a_i y^{(i)}(x)=0
$$
with initial conditions $y^{(i)}(0)=y_i\in\mathbb{R}$ where $i=0,1,2$. I need to find $L^2(\mathbb{R}_+)$ norm of $y(x)$ assuming that $\lim\limits_{x\to+\infty}y(x)=0$. 
How can I solve it? The chracteristic equation is of the third order with arbitrary coefficients!
 A: Presumably the $a_j$ are real.  Let $p(z) = \sum_{j=0}^3 a_j z^j$.    Let $r_j$ be the roots of this polynomial (counted by multiplicity).  
If all $r_j$ have real part $\ge 0$, the only solution $y$ with $y \to 0$ at $+\infty$ is $0$.  
If two $r_j$ have real part $\ge 0$ and one (say $r_3$) is negative, the only solution with $y(0) = y_0$ and 
$y \to 0$ at $\infty$ is $y = y_0 e^{r_3 t}$, and the $L^2({\mathbb R}_+)$ norm of this is
$\|y_0|/\sqrt{-2 r_3}$.
If a complex conjugate pair of $r_j$  have real part $< 0$  (say $\alpha \pm \beta i$ with $\beta > 0$ and $\alpha < 0$)and the other root is nonnegative, the real solutions with $y \to 0$ at $\infty$ are $e^{\alpha t} \left(y_0 \cos(\beta t) + \dfrac{y_1 - \alpha y_0}{\beta} \sin(\beta t)\right)$, and their $L^2({\mathbb R}_+)$ norms are  $$\sqrt{\frac {5\,{{\it y_0}}^{2}{\alpha}^{2}+{{\it y_0}}^{2}{\beta}^{2}-4\,\alpha\,{\it y_0}\,{\it y_1}+{{\it y_1}}^{2}}{ -4\left( {\alpha}^{2}+{
\beta}^{2} \right) \alpha}}$$ 
Other cases may be more complicated.
EDIT: if all $r_j$ have real part $< 0$, so all solutions go to $0$ as $t \to \infty$, 
then (with help from Maple) I get 
$$\|y\|^2 = (-2a_{{1}}{a_{{2}}}^{2}y_{{0}}y_{{1}}-2a_{{3}}y_{{1}}y_{{2}}{a_{{2}
}}^{2}+ \left( -a_{{0}}{a_{{2}}}^{2}-{a_{{1}}}^{2}a_{{2}}+a_{{0}}a_{{1
}}a_{{3}} \right) {y_{{0}}}^{2}- \left( {a_{{2}}}^{3}+{a_{{3}}}^{2}a_{
{0}} \right) {y_{{1}}}^{2}-{a_{{3}}}^{2}{y_{{2}}}^{2}a_{{2}}+2\,a_{{3}
} \left( -a_{{2}}a_{{1}}+a_{{0}}a_{{3}} \right) y_{{0}}y_{{2}})/
 (2 a_{{0}} \left( -a_{{2}}a_{{1}}+a_{{0}}a_{{3}} \right) 
)
$$
EDIT: This was assuming real $y_i$.  
The general solution of the differential equation is
$ y(t) = \sum_{r} c_r e^{rt}$, the sum being over the roots of $p(z)$.  The initial conditions are $y(0) = \sum_{r} c_r = y_0$, $y'(0) = \sum_{r} r c_r = y_1$, $y''(0) = \sum_r r^2 c_r = y_2$.  Write these three equations as $Y = V C$ where  $$V = \pmatrix{1 & 1 & 1\cr r_1 & r_2 & r_3\cr r_1^2 & r_2^2 & r_3^2\cr}$$ 
If $y_i$ are real, so is $y(t)$, and the square of its $L^2$ norm is
$$ \int_0^\infty y(t)^2\ dt = \sum_r \sum_s c_r c_s \int_0^\infty e^{(r+s)t}\ dt
= \sum_r \sum_s \dfrac{-c_r c_s}{r+s} = - C^T M C$$
where $$M = \pmatrix{\dfrac{1}{2r_1} & \dfrac{1}{r_1+r_2} & \dfrac{1}{r_1+r_3}\cr
\dfrac{1}{r_2+r_1} & \dfrac{1}{2r_2} & \dfrac{1}{r_2+r_3}\cr
\dfrac{1}{r_3+r_1} & \dfrac{1}{r_3+r_2} & \dfrac{1}{2r_3}\cr}$$
Now $-C^T M C = -Y^T (V^{-1})^T M V^{-1} Y$, so what we need to do is express each entry of $(V^{-1})^T M V^{-1}$ in terms
of the coefficients $a_j$.  These entries are symmetric rational functions in $r_1$, $r_2$, $r_3$, so that should be possible: every symmetric polynomial in $k$ variables can be expressed as a polynomial in the elementary symmetric polynomials.  But doing it by hand looks rather daunting.    
