Thread: 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?

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

$\displaystyle \chi$ $\displaystyle \sigma$

but how do i show that all the roots are negative

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

$\displaystyle \chi$ $\displaystyle \sigma$

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

CB