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Appendix: All Solutions to the Problems
or
ln(y +
1 + y 2 )
Solving for y, we get y = sinh(c 1 x + c 2 ).
2.49 Find the extremal curve of the functional J [y] =
x 1
x 0
1+y
2
y dx through points (0,
0) and (1, 1).
Solution Because the integrand does not contain x, the first integral of the Euler
equation is
1 + y
2
y + y
1 + y
2
y = c
or
y
= c 1 (1 + y
2
)
The above equation is rewritten as
dy
1 + y 2 = c 1 dx
Integrating we get
arctan y = c 1 x + c 2
The constant in accordance with boundary conditions is c 1 =
π
4
, c 2 = 0, therefore
the extremal curve is y = tan
π x
4
.
2.50 Find the extremal curve of the functional J [y] =
x 1
x 0
y(1 + y 2 )dx.
Solution Because Let the integrand F =
y(1 + y 2 ) does not depend on x, and
F y =
yy
√
y(1+y 2 )
, the Euler equation has the first integral
F − y
F y =
y(1 + y 2 ) −
yy
2
y(1 + y 2 )
= c
or
y
1 + y 2 = c