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}}.


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}}


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)


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

This is also easy to prove. From (2) we haver_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)


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


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


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),


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)

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