# On Viete Theorem and Second-Order Linear Differential Equations with Constant Coefficients

Viete theorem, named after French mathematician Franciscus Viete relates the coefficients of a polynomial equation to sums and products of its roots. It states:

For quadratic equation $a x^2+b x + c =0$, with roots $r_1, r_2$,

$r_1+r_2 = {{-b} \over {a}}$,

$r_1 r_2 = {{c} \over {a}}$.

This is easy to prove. We know that the roots of $a x^2 + b x +c =0$ are

$r_1 = {{-b + \sqrt{b^2-4 a c}} \over {2 a}}$,       $r_2 = {{-b - \sqrt{b^2-4 a c}} \over {2 a}}$.

Therefore,

$r_1 + r_2 ={{-b + \sqrt{b^2-4 a c}} \over {2 a}} + {{-b - \sqrt{b^2-4 a c}} \over {2 a}}={{-b}\over{a}}$

and

$r_1 r_2 = ({{-b + \sqrt{b^2-4 a c}} \over {2 a}})({{-b - \sqrt{b^2-4 a c}} \over {2 a}})={{(-b)^2-(b^2-4 a c)}\over{4 a}}={{c}\over {a}}$.

In fact, the converse is also true. If two given numbers $r_1, r_2$ are such that

$r_1 + r_2 = {{-b}\over{a}}\qquad\qquad(1)$

$r_1 r_2 = {{c}\over {a}}\qquad\qquad\qquad(2)$

then

$r_1, r_2$ are the roots of $a x^2 + b x + c=0$.

This is also easy to prove. From (2) we have$r_1 = {{-b}\over{a}}-r_2$. Hence, (2) implies that $({{-b} \over{a}} -r_2) r_2 = {{c} \over a}$, or

$a r_2^2 + b r_2 +c =0\qquad\qquad(3)$

Since $r_1, r_2$ are symmetric in both (1) and (2), (3) implies that $r_1$ is also the root of $a x^2 + b x +c = 0$.

Let us consider the second-order linear ordinary differential equation with constant coefficients:

$y''(t)+b y'(t) + c y = f(t)\qquad\qquad(4)$

Let $\lambda_1, \lambda_2$ be the roots of quadratic equation $x^2+b x + c =0$ with unknown $x$.

By Viete’s theorem,

$b = -(\lambda_1 + \lambda_2)$

$c =\lambda_1 \lambda_2$.

Therefore, (4) can be written as

$y''(t)-(\lambda_1+\lambda_2)y'(t) +\lambda_1 \lambda_2 y(t) = f(t)$.

Rearrange the terms, we have

$y''(t)-\lambda_1 y'(t) - \lambda_2 y'(t) + \lambda_1 \lambda_2 y(t) = f(t)$

i.e.,

$y''(t)-\lambda_1 y'(t) -\lambda_2 (y'(t)-\lambda_1 y(t) )=f(t)$

or,

$(y'(t)-\lambda_1 y(t))' - \lambda_2 (y'(t)-\lambda_1 y(t)) =f(t)\qquad(5)$

Let

$z(t) = y'(t) - \lambda_1 y(t)\qquad\qquad(6)$

(5), a second-order equation is reduced to a first-order equation

$z'(t) - \lambda_2 z(t) = f(t)\qquad\qquad(7)$

To obtain $y(t)$,  we solve two first-order equations, (7) for $z(t)$ first, then (6) for $y(t)$.

We are now ready to show that any solution obtained as described above is also a solution of (1):

Let $y_*(t)$ be the result of solving (7) for $z(t)$ then (6) for $y(t)$,

then

$z(t) = y'_*(t)-\lambda_1 y_*(t)$.

By (7),

$(y'_*(t)-\lambda_1 y_*(t) )'- \lambda_2(y_*(t)-\lambda_1 y_*(t)) = f(t)\quad\quad\quad(8)$

(8) tells that $y_*(t)$ is a solution of (5).

The fact that (5) is equivalent to (4) implies $y_*(t)$, a solution of (5) is also a solution of (4)