Monthly Archives: December 2017

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 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)

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 (1) implies y_*(t), a solution of (5) is also a solution of (1)

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Solving y’ + a(x) y = b(x), Part 3

Yet another way to find the solution of

y' + a(x) y = b(x)\quad\quad\quad(1)

is to seek a function f(x) > 0 \; \forall x such that the result of multiply (1) by f(x), namely

f(x)y' + f(x) a(x) y =f(x) b(x)\quad\quad\quad(2)

can be written as

(f(x) y)' =  b(x) f(x),

i.e.,

f(x) y =\int b(x) f(x)\; dx + c_{3}

where c_{3} is a constant.

Or,

y = {1 \over f(x)} (\int b(x) f(x)\; dx + c_{3})\quad\quad\quad(3)

since f(x) > 0.

Let us proceed to find such f(x).

From (2) we see that if

f'(x)=a(x) f(x)

then the left side of (2) is certainly (f(x) y)' and consequently  for f(x) > 0 \;\forall x,

{1 \over f(x)} f'(x) = a(x)\quad\quad\quad(4)

or,

log(f(x)) = \int a(x) dx + c_{1}

where c_{1} is a constant.

Therefore, a solution to (4) is

f(x) = c_{2}e^{\int a(x)\;dx}\quad\quad\quad(5)

where c_{2} = exp(c_{1}). This is a positive function \forall x indeed.

With (5), (3) becomes

y = e^{-\int a(x)\;dx} (\int b(x) e^{ \int b(x) dx} dx + c)\quad\quad\quad(6)

where c = {c_{3} \over c_{2}}.

In fact, for any constant c,

 e^{-\int a(x) dx} (\int b(x) e^{\int a(x) dx} dx + c)

is a solution of (1):

(e^{-\int a(x) dx} (\int b(x) e^{\int a(x) dx} dx+ c))' +

a(x)\dot(e^{-\int a(x)\;dx} (\int b(x) e^{\int a(x) dx} dx + c))

= -a(x) (e^{-\int a(x) dx} (\int b(x) e^{\int a(x) dx} dx + c))

+ e^{-\int a(x) dx} b(x) e^{\int a(x) dx}

+ a(x)(e^{-\int a(x) dx} (\int b(x) e^{\int a(x) dx} dx + c))

=b(x)

Therefore, the solution of (1) is

 y = e^{-\int a(x) dx} (\int b(x) e^{\int a(x) dx} dx + c)

where c is any constant.

Exercise: prove that e^x  is a positive function.

Solving y’ + a(x) y = b(x), Part 2

In my previous post “Solving y’ + a(x) y = b(x), Part 1“, it was revealed that every solution of

y' + a(x) y = b(x)\quad\quad\quad(1)

is the sum of a particular solution,  e^{-\int a(x) dx}\int b(x) e^{\int a(x) dx}dx, with c e^{-\int a(x) dx}, a solution of the homogeneous equation y '+ a(x) y =0.

This structure suggests that (1) can be solved in two steps: solve y' +a(x) y= 0 first to get c e^{-\int a(x) dx}, then change the constant c in it to a function c(x) and solve for c(x) after submitting the variation into (1).

There is an alternative:

Let

y = u v

We have

(u v)' + a(x) (u v) = v u' + u (v' + a(x) v)=b(x)

From v' + a(x) v = 0, we obtain

v = c_1 e^{-\int a(x) dx}.

Solving

v u' = c_1 e^{-\int a(x) dx} u'= b(x)

subsequently for u yields

u = \int b(x) e^{\int a(x) dx} dx + c.

Hence,

y = u v  = e^{-\int a(x)dx} (\int b(x) e^{\int a(x) dx} dx + c).

Solving y’ + a(x) y = b(x), Part 1

Given a linear differential equation

y' + a(x) y = b(x)\quad\quad\quad(1)

It is clear that it has solution y(x) \equiv 0 only if b(x) \equiv 0. For if b(x) \not \equiv 0, substituting y(x) \equiv 0 into (1) results 0 = b(x), \forall x, i.e., b(x) \equiv 0, a contradiction. Hence, when b(x) \not \equiv 0, solution to (1) must be a function y(x) \not \equiv  0. If the solution is not everywhere zero, then for y(x) \neq 0, we have

y' = ({b(x) \over y} -a(x)) y

or equivalently

{1 \over y} y' = {b(x) \over y} -a(x),

from which we conclude

log(|y|) = \int {{b(x) \over y} - a(x)} dx + c_1

where c_1 is a constant.

It follows that

y = c_2 e^{\int {{b(x) \over y} - a(x)} dx}\quad\quad\quad(2)

where c_2 is either +e^{c_1} if y >0 or -e^{c_1} if y < 0.

We can express (2) differently as

y = c_2 e^{\int {b(x) \over y} dx} e^{- \int a(x) dx}\quad\quad\quad(3),

where c_2 e^{\int { b(x) \over y} dx} is clearly a function of x.

Let

y = \phi(x) e^{-\int a(x)  dx}\quad\quad\quad(4)

we have

(\phi(x) e^{-\int a(x) dx})' + a(x) \phi(x) e^{-\int a(x) dx}

= \phi'(x)e^{-{\int a(x) dx}}-\phi(x) a(x) e^{-\int a(x) dx} + a(x) \phi(x) e^{-\int a(x) dx}

= \phi'(x) e^{-\int a(x) dx}=b(x).

Hence,

\phi(x) = \int {b(x) e^{\int a(x) dx} dx} + c\quad\quad\quad(5)

where c is a constant.

With (5), (4) becomes

y = e^{-\int a(x) dx} (\int{b(x) e^{\int a(x) dx}} dx + c)\quad\quad\quad(6)

As a matter of fact, for any constant c,  (6) is a solution of (1):

(e^{-\int a(x) dx} (\int{b(x) e^{\int a(x) dx}} dx + c)' + a(x)  e^{-\int a(x) dx} (\int{b(x) e^{\int a(x) dx}} dx + c)

=-a(x) e^{-\int a(x) dx}(\int {b(x) e^{\int a(x) dx}} dx + c)+e^{-\int a(x) dx}b(x)e^{\int a(x) dx}

+ a(x) e^{-\int a(x) dx} (\int b(x)e^{\int a(x) dx} +c)

= b(x)

Notice when b(x)=0, c=0, (6) yields y(x) \equiv 0.

Consequently, (6) defines a family of (1)’s solution.

But are there any other solutions ?

To answer this question, let us first rewrite (6) as

y = e^{-\int a(x) dx} \int{b(x) e^{\int a(x) dx}} dx + ce^{-\int a(x) dx}\quad\quad\quad(7)

We see that e^{-\int a(x)dx}\int{b(x) e^{\int a(x) dx} }dx is a solution of (1) since it is the y(x) of (7) when c =0.

Let y(x) denotes any solution of (1) and,

y_*(x) = e^{-\int a(x) dx} \int{b(x) e^{\int a(x) dx}} dx,

z(x) = y(x) - y_*(x)

Since y'(x)+a(x)y=b(x), y_*'(x)+a(x)y_*(x)=b(x),

z'(t) + a(x) z(t)

= (y(x)-y_*(x))' + a(x)(y(x)-y_*(x))

= y'(x)-y_*'(x)+a(x)y(x)-a(x)y_*(x)

= y'(x) + a(x)y(x) - (y_*'(x)+a(x) y_*(x))

= b(x) - b(x)

= 0.

It means z(x) is a solution of y'+a(x)y=0.

We know c e^{-\int a(x) dx} is the solution of y'+a(x) y=0 (see “Solving y’+a(x) y =0“)

Therefore,

y(x) - y_*(x) = c e^{-\int a(x) dx},

i.e.,

y(x) = y_*(x) + c e^{-\int a(x) dx}

or,

y(x) =  e^{-\int a(x) dx}( \int{b(x) e^{\int a(x) dx}} dx + c).

Thus, we have shown that all solutions of (1) are of the form (6)!

Stated differently,

The collection of functions expressed by

y = e^{-\int a(x) dx} (\int{b(x) e^{\int a(x) dx}} dx + c)

where c is any constant constitutes all possible solutions of y' + a(x) y =b(x).

 

Solving y’ + a(x) y = 0

Let’s consider the 1st-Order linear differential equation

y'(x)+ a(x) y(x) =0\quad\quad\quad(1)

We see y(x) \equiv 0 is a solution.

However, if (1) has a solution whose value at x is not zero, it must be true that

{1 \over y(x)} y(x)' = -a(x)

i.e.,

log(|y(x)|) = -\int a(x) dx + c_1\quad\quad\quad(2)

where c_1 is a constant.

From (2), we obtain

y(x) = c\;e^{-\int a(x) dx}\quad\quad\quad(3)

where c is a constant. It is either +e^{c_1} or -e^{c_1}.

We assert and prove that for any constant c, a function defined by (3)  is a solution of (1):

(c\;e^{-\int a(x)dx)})' +a(x)c\;e^{-\int a(x) dx}=-c\;a(x)e^{-\int a(x) dx} +a(x) c\;e^{-\int a(x) dx} =0.

Notice when c=0, (3) yields y(x) \equiv 0, the zero solution of (1).

Moreover, to see there are no other solutions, let  u(x) be any solution of (1), we have

(u(x) e^{\int a(x)dx})' = u'(x) e^{\int a(x) dx} + u(x) a(x) e^{\int a(x) dx}

= e^{\int a(x)dx } (u'(x) + a(x) u(x))

= 0.

Therefore,

(u(x) e^{\int a(x) dx})' = 0.

That is

u(x) e^{\int a(x)dx} = c

or,

u(x) = c\;e^{-\int a(x) dx}

where c is a constant. Hence, any solution of (1) belongs to the family of functions defined by (3)