Advanced book on Mathematics Olympiad

196 3 Real Analysis

z(t)=(a+ib)exp

(

− 1 +

√

5

2

it

)

+(c+id)exp

(

− 1 −

√

5

2

it

)

,

for arbitrary real numbersa, b, c, d. Sincexandyare, respectively, the real and complex

parts of the solution, they have the general form

x(t)=acos

− 1 +

√

5

2

t−bsin

− 1 +

√

5

2

t+ccos

− 1 −

√

5

2

t−dsin

− 1 −

√

5

2

t,

y(t)=asin

− 1 +

√

5

2

t+bcos

− 1 +

√

5

2

t+csin

− 1 −

√

5

2

t+dcos

− 1 −

√

5

2

t.

The problem is solved. 

Our second example is an equation published by M. Ghermanescu in the ̆ Mathematics

Gazette, Bucharest. Its solution combines several useful techniques.

Example.Solve the differential equation

2 (y′)^3 −yy′y′′−y^2 y′′′= 0.

Solution.In a situation like this, where the variablexdoes not appear explicitly, one can

reduce the order of the equation by takingyas the variable andp=y′as the function.

The higher-order derivatives ofy′′are

y′′=

d

dx

y′=

d

dy

p

dy

dx

=p′p,

y′′′=

d

dx

y′′=

(

d

dy

pp′

)

dy

dx

=

(

(p′)^2 +pp′′

)

p.

We end up with a second-order differential equation

2 p^3 −yp^2 p′−y^2 pp′′−y^2 p(p′)^2 = 0.

A family of solutions isp=0, that is,y′=0. This family consists of the constant
functionsy=C. Dividing the equation by−p, we obtain

y^2 p′′+y^2 (p′)^2 +ypp′− 2 p^2 = 0.

The distribution of the powers ofyreminds us of the Euler–Cauchy equation, while the

last terms suggests the substitutionu=p^2. And indeed, we obtain the Euler–Cauchy

equation

y^2 u′′+yu′− 4 u= 0 ,