Advanced book on Mathematics Olympiad

596 Real Analysis

562.The relation from the statement implies right away thatfis differentiable. Differ-
entiating

f(x)+x

∫x

0

f(t)dt−

∫x

0

tf (t)dt= 1 ,

we obtain

f′(x)+

∫x

0

f(t)dt+xf (x)−xf (x)= 0 ,

that is,f′(x)+

∫x
0 f(t)dt=0. Again we conclude thatfis twice differentiable, and so
we can transform this equality into the differential equationf′′+f=0. The general
solution isf(x)=Acosx+Bsinx. Substituting in the relation from the statement, we
obtainA=1,B=0, that is,f(x)=cosx.
(E. Popa,Analiza Matematica, Culegere de Probleme ̆ (Mathematical Analysis, Col-
lection of Problems), Editura GIL, 2005)

563.The equation is of Laplace type, but we can bypass the standard method once we
make the following observation. The associated homogeneous equation can be written as

x(y′′+ 4 y′+ 4 y)−(y′′+ 5 y′+ 6 y)= 0 ,

and the equationsy′′+ 4 y′+ 4 y=0 andy′′+ 5 y′+ 6 y=0 have the common solution
y(x)=e−^2 x. This will therefore be a solution to the homogeneous equation, as well.
To find a solution to the inhomogeneous equation, we use the method of variation of the
constant. Sety(x)=C(x)e−^2 x. The equation becomes

(x− 1 )C′′−C′=x,

with the solution

C′(x)=λ(x− 1 )+(x− 1 )ln|x− 1 |− 1.

Integrating, we obtain

C(x)=

1

2

(x− 1 )^2 ln|x− 1 |+

(

λ

2

−

1

4

)

(x− 1 )^2 −x+C 1.

If we setc 2 =λ 2 −^14 , then the general solution to the equation is

y(x)=e−^2 x

[

C 1 +C 2 (x− 1 )^2 +

1

2

(x− 1 )^2 ln|x− 1 |−x

]

.

(D. Flondor, N. Donciu,Algebra ̧ ̆si Analiz ̆a Matematica ̆(Algebra and Mathematical
Analysis), Editura Didactica ̧ ̆si Pedagogica, Bucharest, 1965) ̆