a,b,c > 0
ay'' + by' +cy = 0
and y(0) = c and y'(0) = d
the equation has either one real root, two real roots or two complex roots, but how do i then show that in the limit x--> +infinity, y(x) will go to zero?
Appreciate Your help!
a,b,c > 0
ay'' + by' +cy = 0
and y(0) = c and y'(0) = d
the equation has either one real root, two real roots or two complex roots, but how do i then show that in the limit x--> +infinity, y(x) will go to zero?
Appreciate Your help!
If all roots of the polynomial...
$\displaystyle a\cdot t^{2} + b\cdot t + c$
... have real part $\displaystyle <0$, then if $\displaystyle y(x)$ is solution of the second order DE you have prosed is...
$\displaystyle \lim_{x \rightarrow \infty} y(x) =0$
Kind regards
$\displaystyle \chi$ $\displaystyle \sigma$
A polynomial whose roots have all real part $\displaystyle <0$ is called 'Hurwitz polynomial'. If we write a polynomial of degree n in t as $\displaystyle p(t)= \sum_{k=0}^{n} a_{k} \cdot t^{k}$ and it is $\displaystyle a_{0}=1$, necessary [but not sufficient...] condition for $\displaystyle p(t)$ to be Hurwitz is that $\displaystyle \forall k\ge 1$ $\displaystyle a_{k}>0$. For a suffcient condition see...
Hurwitz polynomial - Wikipedia, the free encyclopedia
Kind regards
$\displaystyle \chi$ $\displaystyle \sigma$