788
Appendix: All Solutions to the Problems
Integrating two times, we get
y
−
1
2 = c 1 x + c 2
or
y
=
1
(c 1 x + c 2 ) 2
Integrating two times, we get
c
2
1 y = ln(c 1 x + c 2 ) + c 3 x + c 4
When k = 1, the value of the functional has nothing to do with y.
2.64 If the functional is J [y] =
1
2
x 1
x 0
(Dy
2
+ ky
2
− 2qy)dx, where, k is a constant,
q is a given function of x, find the Euler equation of the functional.
Solution The Euler equation of the functional is
ky − q +
d
2
dx 2 (Dy
) = 0
If D is a constant, then
ky − q + Dy
(4)
= 0
or
y
(4)
+
k
D
y =
k
D
q
2.65 Find the extremal curve of the functional J [y] =
x 1
x 0
(y
2
+ 2y
2
+ y
2
)dx.
Solution Let the integrand F = y
2
+2y
2
+y
2 , substituting it into the Euler–Poisson
equation, there is
2y − 4y + 2y
(4)
= 0
or
y
(4)
− 2y
+ y = 0
This is a fourth order differential equation with constant coefficients, the general
solution is
Appendix: All Solutions to the Problems
Integrating two times, we get
y
−
1
2 = c 1 x + c 2
or
y
=
1
(c 1 x + c 2 ) 2
Integrating two times, we get
c
2
1 y = ln(c 1 x + c 2 ) + c 3 x + c 4
When k = 1, the value of the functional has nothing to do with y.
2.64 If the functional is J [y] =
1
2
x 1
x 0
(Dy
2
+ ky
2
− 2qy)dx, where, k is a constant,
q is a given function of x, find the Euler equation of the functional.
Solution The Euler equation of the functional is
ky − q +
d
2
dx 2 (Dy
) = 0
If D is a constant, then
ky − q + Dy
(4)
= 0
or
y
(4)
+
k
D
y =
k
D
q
2.65 Find the extremal curve of the functional J [y] =
x 1
x 0
(y
2
+ 2y
2
+ y
2
)dx.
Solution Let the integrand F = y
2
+2y
2
+y
2 , substituting it into the Euler–Poisson
equation, there is
2y − 4y + 2y
(4)
= 0
or
y
(4)
− 2y
+ y = 0
This is a fourth order differential equation with constant coefficients, the general
solution is