1. ## second order ODE

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?

2. If all roots of the polynomial...

$a\cdot t^{2} + b\cdot t + c$

... have real part $<0$, then if $y(x)$ is solution of the second order DE you have prosed is...

$\lim_{x \rightarrow \infty} y(x) =0$

Kind regards

$\chi$ $\sigma$

3. but how do i show that all the roots are negative

4. A polynomial whose roots have all real part $<0$ is called 'Hurwitz polynomial'. If we write a polynomial of degree n in t as $p(t)= \sum_{k=0}^{n} a_{k} \cdot t^{k}$ and it is $a_{0}=1$, necessary [but not sufficient...] condition for $p(t)$ to be Hurwitz is that $\forall k\ge 1$ $a_{k}>0$. For a suffcient condition see...

Hurwitz polynomial - Wikipedia, the free encyclopedia

Kind regards

$\chi$ $\sigma$

5. Originally Posted by sssouljah
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?

Descartes rule of signs tell you that $a\lambda^2+b \lambda +c=0$ has no positive roots and one negative root when $a,\ b,\ c >0$. Hence it has one negative root of multiplicity 2 and no complex roots.