second order ODE

**sssouljah** · Sep 4th 2009, 07:08 AM

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!

**chisigma** · Sep 4th 2009, 11:38 AM

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

$\chi$ $\sigma$

**sssouljah** · Sep 5th 2009, 09:52 AM

but how do i show that all the roots are negative

**chisigma** · Sep 5th 2009, 07:38 PM

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

$\chi$ $\sigma$

**CaptainBlack** · Sep 5th 2009, 10:28 PM

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?

Appreciate Your help!

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.

CB

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