Let rewrite the differential equation in the form

( ddx−λ1)( ddx−λ2) y=f (x )by letting

h=( ddx−λ2) y= y '−λ2 y ,

we see that y solves the problem if and only if

h solves {h'− λ1h=f (x) ,h (0 )=0 ,

and y solves {y '−λ2 y=h(x) ,y (0 )=0

From the above we get

h ( x )=∫0

x

e λ1 ( x−t ) f (t )dt ,

y ( x )=∫0

x

eλ2 (x−t )h (t )dt ,

by substituting hfrom the first formula into the second, we obtain y (x ) as a repeated integral ( for any x∈ I ¿ :

y ( x )=∫0

x

eλ 2 (x−t )(∫0

t

e λ1 (t−s) f ( s)ds)dtLet us suppose that x>0. Then in the integral with respect to t we have 0≤ t ≤ x, whereas in the integral with respect to s we have 0≤ s≤ t . We can then rewrite y (x ) as a double integral:

y ( x )=eλ 2x∫T x

❑

e(λ¿¿1− λ2) t e−λ 1s f ( s)dt ,¿

where T x is the triangle in the (s , t) plane defined by 0≤ s≤ t ≤ x , with vertices at the points (0,0), (0,x) and (x,x). We first integrate with respect to s and then with respect to t . Since the triangle T x is convex both horizontally and vertically, and since the integrand function:

F ( s , t )=e( λ¿¿ 1−λ2)t e−λ1 sf ( s )¿

is continuous in T x, we can interchange the order of integration and integrate with respect to t first.

We thus obtain

y ( x )=∫0

x

¿¿

by substituting t with t+s in the integral with respect to t we finally get

y ( x )= y (x )=∫0

x

¿¿

for x<0 we can reason in a similar way and we get the same result.

The integral in formula (3.3) can be computed exactly as in the real field. We obtain the following expression of the function g:

1) If λ1≠ λ2 (↔∆=a2−4b≠0 ) then

g ( x )= 1λ1−λ2

(eλ1 x−eλ2x)

2) Ifλ1=λ2(↔∆=0) then

g ( x )=x eλ1 x

Note that g is always a real function. Letting α=−a/2 and

B={√−∆ /2 if ∆<0√∆ /2 if ∆>0

so that λ1 ,2={α ± iBif ∆<0α ±B if ∆>0

we have

g ( x )=¿

we can check that g solves the following homogeneous initial value problem:

{ y ' '+a y '+by=0y (0 )=0 , y ' (0 )=1

The function g is called the impulsive response of the differential operator L. First, let us prove the following formula for the derivative y ' :

y ' ( x )= ddx∫0

x

g ( x−t ) f (t )dt=g (0 ) f ( x )+∫0

x

g' (x−t ) f ( t )dt ( x∈ I )

Indeed, given h such that x+h∈ I , we have

y ( x+h )− y (x)h =

1h ¿

as g∈C2(R), we can apply Taylor’s formula with the Lagrange remainder

g (x0+h )=g (x0 )+g ' (x0 )h+ 12 g' ' (E)h2

at the point x0=x−t , where E is some point between x0 and x0+h . Substituting this equation and using

∫0

x+h

g ( x−t ) f (t )dt=∫0

x

g ( x−t ) f (t )dt+∫x

x+h

g ( x−t ) f (t )dt

it will give

1h

( y ( x+h )− y ( x ) )=1h ∫x

x+h

g ( x−t ) f (t )dt+∫0

x +h

g' ( x−t ) f (t ) dt+ 12h∫

0

x +h

g' '(E) f (t)dt

For some E between x−t∧x−t+h. When h tends to zero, the first term in the right-hand side of the equation tends to g (0 ) f ( x ) , by the Fundamental Theorem of

Calculus. The second term tends to ∫0

x

g' ( x−t ) f (t )dt , by the continuity of the integral

function. Finally, the third term tends to zero, since the integral that occurs in it is a bounded function of h in a neighbourhood of h=0.vGiven that g (0 )=0, we finally get

y ' ( x )=∫0

x

g ' ( x−t ) f ( t )dt (x∈ I )

In the same way we compute the second derivative

y ' ' ( x )=( ddx

)2

∫0

x

g ( x−t ) f ( t )dt

¿ ddx∫0

x

g' ( x−t ) f ( t )dt

¿ g' (0 ) f ( x )+∫0

x

g' ' (x−t ) f (t )dt

¿ f ( x )+∫0

x

g' '( x−t) f (t )dt

where we used g' (0 )=1. It follows that

y ' ' ( x )+a y ' ( x )+by ( x )=f ( x )+∫0

x

(g¿¿ ' '+a g'+bg) (x−t ) f ( t ) dt ¿

¿ f ( x )∀ x∈ I

g being a solution of the homogeneous equation. Therefore the function y given solves (3.1) in the interval I . The initial conditions y (0 )=0= y ' (0) are immediately verified. We now come to the solution of the initial value problem with an arbitrary initial data at the point x=0.

Theorem 3.2 (Camporesi, 2011) Let f ∈ c0 ( I ) ,0∈ I and let y0 , y0' be two arbitrary real numbers. Then the initial value problem

{ y ' '+a y '+by= f (x)y (0 )= y0 , y

' (0 )= y0'

The equations has a unique solution, defined on the whole of I, and given by

y ( x )=∫0

x

g (x−t ) f ( t )dt+( y0' ¿+a y0)g ( x )+ y0g

' ( x )(x∈R)¿

The uniqueness of equation follows from the fact that if y1∧ y2 both solve Theorem 3.2, then their difference y= y1− y2 solves the problem in Theorem 3.1. Function g' satisfies the homogeneous equation. Indeed, since L has constant coefficients, we have

Lg '=L ddx

g=[( ddx )2

+a ddx

+b] ddx g= ddx

g= ddx

Lg=0

by the linearity of L and by Theorem 3.1, the equation satisfies (Ly)(x )=f (x)∀ x∈ I . It immediate that y (0 )= y0. Finally, since

y ' ( x )=∫0

x

g ' ( x−t ) f (t ) dt+( y0' ¿+a y0)g

' ( x )+ y0g' ' (x )¿

we have

y ' (0 )= y0' +a y0+ y0 g

' ' (0)¿ y0

' +a y0+ y0(−ag¿¿ ' (0 )−bg (0))¿¿ y0

'

by proceeding as in the proof of this theorem, we find that y solves the problem of Theorem 3.2 if and only if y is given by

y ( x )=∫0

x

g (x−s ) f (s ) ds+( y0' −¿ λ2 y0)g ( x )+ y0 e

λ2 x¿

This formula show the equality e λ2x=g ' ( x )−λ1g(x ), which follows from by interchanging λ1∧λ2. By imposing the initial conditions at an arbitrary point of the interval I , we get the solution.

Theorem 3.3 (Camporesi, 2011) Let f ∈ c0 ( I ) , x0∈ I , y0 , y0' ∈R. The solution of the initial value problem

{ y ' '+a y '+by=f ( x)y (x0 )= y0 , y

' (x0 )= y0'

The solution is unique and is defined on the whole of I , and is given by

y ( x )=∫x0

x

g (x−t ) f ( t )dt+(¿ y0' +a y0)g (x−x0

❑)+ y0 g' ( x−x0 )(x∈ I )¿

In particular (taking f=0), the solution of the homogeneous problem

{ y ' '+a y '+by=0y (x0 )= y0 , y

' (x0 )= y0'

with x0∈ R arbitrary, of class C∞ on the whole of R, is given by

yh ( x )=( y0' +a y 0) g (x−x0 )+ y0g' (x−x0 ) (x∈R )

Let y be a solution of Theorem 3.3, and let τ x0 denote the translation map defined by τ x0 ( x )=x+x0 Then

y ( x )= y oτ x0 ( x )= y (x+x0)

As the differential operator L has constant coefficients, it is invariant under translations, that is

L ( y o τ x0 )=(Ly )o τx0

follows that

(Ly ) ( x )=(Ly ) (x+x0 )=f (x+x0)

Since y (0 )= y (x0 )= y0 , y' (0 )= y ' (x0 )= y0

' , we see that y solves the problem in the interval I if and only if y solves the initial value problem

{y ' '+a y '+by=0 f (x )y (0 )= y0, y' (0 )= y0

'

In the translated interval I−x0={t−x0 : t ϵ I } and with translated forcing term f ( x )=f (x+x0). By theorem 3.2 we have that y is unique and is given by

y ( x )= y ( x−x0 )= ∫0

x−x0

g (x−x0−s) f ( s+x0 )ds+( y0' +a y0) g (x−x0 )+ y0g' (x−x0)

The equation follows immediately from this by making the change of variable s+x0=t in the integral with respect to s.

Corollary 3.4 (Camporesi, 2011) The set V of real solutions of the homogeneous equation (3.2) is a vector space of dimension 2 over ℝ, and the two functions g , g ' form a basic of V.

Theorem 3.5 (Camporesi, 2011) Every complex solution of the homogeneous equation Ly=0 can be written in the form y=c1 y1+c2 y2 , where y1 and y2are given by:

( y1 ( x ) , y2 (x ) )={ (eλ1 x , eλ 2x ) if λ1≠ λ2(eλ 1x , x eλ1 x )if λ1=λ2

Conversely, any function of this form is a solution of Ly=0.