Calculus of Variations solved problems pdf - Web Education

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June 4, 2012 ( public domain )

Acknowledgement. The following problems were solved using my own procedure in a program Maple V, release 5. All possible errors are my faults.

1

Solving the Euler equation

Theorem.(Euler) Suppose f (x, y, y0_{) has continuous partial derivatives of the}

second order on the interval [a, b]. If a functional F (y) =Rb

af (x, y, y

0_{)dx attains}

a weak relative extrema at y0, then y0 is a solution of the following equation

∂f ∂y − d dx ∂f ∂y0  = 0 .

Z b a 12 x y(x) + ( ∂ ∂xy(x)) 2_dx Hint: elementary Solution.

We denote auxiliary function f f(x, y(x), ∂ ∂xy(x)) = 12 x y(x) + ( ∂ ∂xy(x)) 2 in the form f(x, y, z) = 12 x y + z2

Substituting we x, y(x) and y0(x) for x, y and z we obtain the integrand in the given functional. We compute partial derivatives

∂ ∂yf(x, y(x), ∂ ∂xy(x)) = 12 x and ∂ ∂zf(x, y(x), ∂ ∂xy(x)) = 2 ( ∂ ∂xy) and ∂2 ∂x ∂zf(x, y(x), ∂ ∂xy(x)) = 2 ( ∂2 ∂x2y(x))

and finally obtain the Euler equation for our functional 12 x − 2 ( ∂

2

∂x2y(x)) = 0

We solve it using various tools and obtain the solution y(x) = x3+ C1 x + C2

Info.

cubic polynomial

Comment.

Z b a 3 x + r ∂ ∂xy(x) dx Hint: elementary Solution.

We denote auxiliary function f

f(x, y(x), ∂ ∂xy(x)) = 3 x + r ∂ ∂xy(x) in the form f(x, y, z) = 3 x +√z

Substituting we x, y(x) and y0(x) for x, y and z we obtain the integrand in the given functional. We compute partial derivatives

∂ ∂yf(x, y(x), ∂ ∂xy(x)) = 0 and ∂ ∂zf(x, y(x), ∂ ∂xy(x)) = 1 2 1 q ∂ ∂xy and ∂2 ∂x ∂zf(x, y(x), ∂ ∂xy(x)) = − 1 4 ∂2 ∂x2y(x) (_∂x∂ y(x))(3/2)

and finally obtain the Euler equation for our functional 1 4 ∂2 ∂x2y(x) (∂ ∂xy(x)) (3/2) = 0

We solve it using various tools and obtain the solution y(x) = C1 x + C2

Info.

linear function

Comment.

Z b a x + y(x)2+ 3 ( ∂ ∂xy(x)) dx Hint: elementary Solution.

We denote auxiliary function f f(x, y(x), ∂ ∂xy(x)) = x + y(x) 2_{+ 3 (} ∂ ∂xy(x)) in the form f(x, y, z) = x + y2+ 3 z

Substituting we x, y(x) and y0(x) for x, y and z we obtain the integrand in the given functional. We compute partial derivatives

∂ ∂yf(x, y(x), ∂ ∂xy(x)) = 2 y and ∂ ∂zf(x, y(x), ∂ ∂xy(x)) = 3 and ∂2 ∂x ∂zf(x, y(x), ∂ ∂xy(x)) = 0 and finally obtain the Euler equation for our functional

2 y(x) = 0

We solve it using various tools and obtain the solution y(x) = 0

Info.

constant solution

Comment.

Z b a y(x)2dx Hint: elementary Solution.

We denote auxiliary function f f(x, y(x), ∂

∂xy(x)) = y(x)

2

in the form

f(x, y, z) = y2

Substituting we x, y(x) and y0(x) for x, y and z we obtain the integrand in the given functional. We compute partial derivatives

∂ ∂yf(x, y(x), ∂ ∂xy(x)) = 2 y and ∂ ∂zf(x, y(x), ∂ ∂xy(x)) = 0 and ∂2 ∂x ∂zf(x, y(x), ∂ ∂xy(x)) = 0 and finally obtain the Euler equation for our functional

2 y(x) = 0

We solve it using various tools and obtain the solution y(x) = 0

Info.

zero

Comment.

Z b a y(x)2+ x2( ∂ ∂xy(x)) dx Hint: elementary Solution.

We denote auxiliary function f f(x, y(x), ∂ ∂xy(x)) = y(x) 2_{+ x}2₍ ∂ ∂xy(x)) in the form f(x, y, z) = y2+ x2z

Substituting we x, y(x) and y0(x) for x, y and z we obtain the integrand in the given functional. We compute partial derivatives

∂ ∂yf(x, y(x), ∂ ∂xy(x)) = 2 y and ∂ ∂zf(x, y(x), ∂ ∂xy(x)) = x 2 and ∂2 ∂x ∂zf(x, y(x), ∂ ∂xy(x)) = 2 x and finally obtain the Euler equation for our functional

2 y(x) − 2 x = 0 We solve it using various tools and obtain the solution

y(x) = x

Info.

linear

Comment.

Z b a y(x) + x ( ∂ ∂xy(x)) dx Hint: elementary Solution.

We denote auxiliary function f f(x, y(x), ∂ ∂xy(x)) = y(x) + x ( ∂ ∂xy(x)) in the form f(x, y, z) = y + x z

Substituting we x, y(x) and y0(x) for x, y and z we obtain the integrand in the given functional. We compute partial derivatives

∂ ∂yf(x, y(x), ∂ ∂xy(x)) = 1 and ∂ ∂zf(x, y(x), ∂ ∂xy(x)) = x and ∂2 ∂x ∂zf(x, y(x), ∂ ∂xy(x)) = 1 and finally obtain the Euler equation for our functional

0 = 0

We solve it using various tools and obtain the solution y(x) = Y(x)

Info.

all functions

Comment.

Z b a ∂ ∂xy(x) dx Hint: elementary Solution.

We denote auxiliary function f f(x, y(x), ∂ ∂xy(x)) = ∂ ∂xy(x) in the form f(x, y, z) = z

Substituting we x, y(x) and y0(x) for x, y and z we obtain the integrand in the given functional. We compute partial derivatives

∂ ∂yf(x, y(x), ∂ ∂xy(x)) = 0 and ∂ ∂zf(x, y(x), ∂ ∂xy(x)) = 1 and ∂2 ∂x ∂zf(x, y(x), ∂ ∂xy(x)) = 0 and finally obtain the Euler equation for our functional

0 = 0

We solve it using various tools and obtain the solution y(x) = Y(x)

Info.

all functions

Comment.

Z b a y(x)2− ( ∂ ∂xy(x)) 2_dx Hint: missing x Solution.

We denote auxiliary function f f(x, y(x), ∂ ∂xy(x)) = y(x) 2_{− (} ∂ ∂xy(x)) 2 in the form f(x, y, z) = y2− z2

Substituting we x, y(x) and y0(x) for x, y and z we obtain the integrand in the given functional. We compute partial derivatives

∂ ∂yf(x, y(x), ∂ ∂xy(x)) = 2 y and ∂ ∂zf(x, y(x), ∂ ∂xy(x)) = −2 ( ∂ ∂xy) and ∂2 ∂x ∂zf(x, y(x), ∂ ∂xy(x)) = −2 ( ∂2 ∂x2y(x))

and finally obtain the Euler equation for our functional 2 y(x) + 2 ( ∂

2

∂x2y(x)) = 0

We solve it using various tools and obtain the solution y(x) = C1 cos(x) + C2 sin(x)

Info.

trig functions

Comment.

Z b a r 1 + ( ∂ ∂xy(x)) 2_dx Hint: missing x y Solution.

We denote auxiliary function f

f(x, y(x), ∂ ∂xy(x)) = r 1 + ( ∂ ∂xy(x)) 2 in the form f(x, y, z) =p1 + z2

Substituting we x, y(x) and y0(x) for x, y and z we obtain the integrand in the given functional. We compute partial derivatives

∂ ∂yf(x, y(x), ∂ ∂xy(x)) = 0 and ∂ ∂zf(x, y(x), ∂ ∂xy(x)) = ∂ ∂xy q 1 + (_∂x∂ y)2 and ∂2 ∂x ∂zf(x, y(x), ∂ ∂xy(x)) = − (_∂x∂ y(x))2₍∂2 ∂x2y(x)) (1 + ( ∂ ∂xy(x)) 2₎(3/2) + ∂2 ∂x2y(x) q 1 + (_∂x∂ y(x))2

and finally obtain the Euler equation for our functional (_∂x∂ y(x))2(_∂x∂22 y(x)) (1 + ( ∂ ∂xy(x)) 2₎(3/2) − ∂2 ∂x2 y(x) q 1 + (_∂x∂ y(x))2 = 0 We solve it using various tools and obtain the solution

y(x) = C1 x + C2

Info.

linear function

Comment.

Z b a x ( ∂ ∂xy(x)) + ( ∂ ∂xy(x)) 2_dx Hint: missing y Solution.

We denote auxiliary function f f(x, y(x), ∂ ∂xy(x)) = x ( ∂ ∂xy(x)) + ( ∂ ∂xy(x)) 2 in the form f(x, y, z) = x z + z2

Substituting we x, y(x) and y0(x) for x, y and z we obtain the integrand in the given functional. We compute partial derivatives

∂ ∂yf(x, y(x), ∂ ∂xy(x)) = 0 and ∂ ∂zf(x, y(x), ∂ ∂xy(x)) = x + 2 ( ∂ ∂xy) and ∂2 ∂x ∂zf(x, y(x), ∂ ∂xy(x)) = 1 + 2 ( ∂2 ∂x2y(x))

and finally obtain the Euler equation for our functional −1 − 2 ( ∂

2

∂x2y(x)) = 0

We solve it using various tools and obtain the solution y(x) = −1 4x 2_{+ C1 x + C2} Info. quadratic function Comment. no comment

Z b a (1 + x) ( ∂ ∂xy(x)) 2_dx Hint: no hint Solution.

We denote auxiliary function f f(x, y(x), ∂ ∂xy(x)) = (1 + x) ( ∂ ∂xy(x)) 2 in the form f(x, y, z) = (1 + x) z2

Substituting we x, y(x) and y0(x) for x, y and z we obtain the integrand in the given functional. We compute partial derivatives

∂ ∂yf(x, y(x), ∂ ∂xy(x)) = 0 and ∂ ∂zf(x, y(x), ∂ ∂xy(x)) = 2 (1 + x) ( ∂ ∂xy) and ∂2 ∂x ∂zf(x, y(x), ∂ ∂xy(x)) = 2 ( ∂ ∂xy(x)) + 2 (1 + x) ( ∂2 ∂x2y(x))

and finally obtain the Euler equation for our functional −2 ( ∂

∂xy(x)) − 2 (1 + x) ( ∂2

∂x2y(x)) = 0

We solve it using various tools and obtain the solution y(x) = C1 + C2 ln(1 + x)

Info.

not supplied

Comment.

Z b a 2 y(x) ex+ y(x)2+ ( ∂ ∂xy(x)) 2_dx Hint: no hint Solution.

We denote auxiliary function f f(x, y(x), ∂ ∂xy(x)) = 2 y(x) e x_{+ y(x)}2_{+ (} ∂ ∂xy(x)) 2 in the form f(x, y, z) = 2 y ex+ y2+ z2

Substituting we x, y(x) and y0(x) for x, y and z we obtain the integrand in the given functional. We compute partial derivatives

∂ ∂yf(x, y(x), ∂ ∂xy(x)) = 2 e x_{+ 2 y} and ∂ ∂zf(x, y(x), ∂ ∂xy(x)) = 2 ( ∂ ∂xy) and ∂2 ∂x ∂zf(x, y(x), ∂ ∂xy(x)) = 2 ( ∂2 ∂x2y(x))

and finally obtain the Euler equation for our functional 2 ex+ 2 y(x) − 2 ( ∂

2

∂x2y(x)) = 0

We solve it using various tools and obtain the solution y(x) = (1 2cosh(x) 2₊1 2cosh(x) sinh(x) + 1 2x) sinh(x) + (−1 2cosh(x) sinh(x) + 1 2x − 1 2cosh(x)

2_{) cosh(x) + C1 sinh(x) + C2 cosh(x)} Info.

not supplied

Comment.

Z b a 2 y(x) + ( ∂ ∂xy(x)) 2_dx Hint: no hint Solution.

We denote auxiliary function f f(x, y(x), ∂ ∂xy(x)) = 2 y(x) + ( ∂ ∂xy(x)) 2 in the form f(x, y, z) = 2 y + z2

Substituting we x, y(x) and y0(x) for x, y and z we obtain the integrand in the given functional. We compute partial derivatives

∂ ∂yf(x, y(x), ∂ ∂xy(x)) = 2 and ∂ ∂zf(x, y(x), ∂ ∂xy(x)) = 2 ( ∂ ∂xy) and ∂2 ∂x ∂zf(x, y(x), ∂ ∂xy(x)) = 2 ( ∂2 ∂x2y(x))

and finally obtain the Euler equation for our functional 2 − 2 ( ∂

2

∂x2y(x)) = 0

We solve it using various tools and obtain the solution y(x) =1 2x 2_{+ C1 x + C2} Info. not supplied Comment. no comment

Z b a 2 (∂ ∂xy(x)) 3_dx Hint: no hint Solution.

We denote auxiliary function f f(x, y(x), ∂ ∂xy(x)) = 2 ( ∂ ∂xy(x)) 3 in the form f(x, y, z) = 2 z3

Substituting we x, y(x) and y0(x) for x, y and z we obtain the integrand in the given functional. We compute partial derivatives

∂ ∂yf(x, y(x), ∂ ∂xy(x)) = 0 and ∂ ∂zf(x, y(x), ∂ ∂xy(x)) = 6 ( ∂ ∂xy) 2 and ∂2 ∂x ∂zf(x, y(x), ∂ ∂xy(x)) = 12 ( ∂ ∂xy(x)) ( ∂2 ∂x2y(x))

and finally obtain the Euler equation for our functional −12 (∂

∂xy(x)) ( ∂2

∂x2y(x)) = 0

We solve it using various tools and obtain the solution y(x) = C1

Info.

not supplied

Comment.

Z b a 4 x ( ∂ ∂xy(x)) + ( ∂ ∂xy(x)) 2_dx Hint: no hint Solution.

We denote auxiliary function f f(x, y(x), ∂ ∂xy(x)) = 4 x ( ∂ ∂xy(x)) + ( ∂ ∂xy(x)) 2 in the form f(x, y, z) = 4 x z + z2

Substituting we x, y(x) and y0(x) for x, y and z we obtain the integrand in the given functional. We compute partial derivatives

∂ ∂yf(x, y(x), ∂ ∂xy(x)) = 0 and ∂ ∂zf(x, y(x), ∂ ∂xy(x)) = 4 x + 2 ( ∂ ∂xy) and ∂2 ∂x ∂zf(x, y(x), ∂ ∂xy(x)) = 4 + 2 ( ∂2 ∂x2y(x))

and finally obtain the Euler equation for our functional −4 − 2 ( ∂

2

∂x2y(x)) = 0

We solve it using various tools and obtain the solution y(x) = −x2+ C1 x + C2

Info.

not supplied

Comment.

Z b a 7 (∂ ∂xy(x)) 3_dx Hint: no hint Solution.

We denote auxiliary function f f(x, y(x), ∂ ∂xy(x)) = 7 ( ∂ ∂xy(x)) 3 in the form f(x, y, z) = 7 z3

Substituting we x, y(x) and y0(x) for x, y and z we obtain the integrand in the given functional. We compute partial derivatives

∂ ∂yf(x, y(x), ∂ ∂xy(x)) = 0 and ∂ ∂zf(x, y(x), ∂ ∂xy(x)) = 21 ( ∂ ∂xy) 2 and ∂2 ∂x ∂zf(x, y(x), ∂ ∂xy(x)) = 42 ( ∂ ∂xy(x)) ( ∂2 ∂x2y(x))

and finally obtain the Euler equation for our functional −42 (∂

∂xy(x)) ( ∂2

∂x2y(x)) = 0

We solve it using various tools and obtain the solution y(x) = C1

Info.

not supplied

Comment.

Z b a r y(x) (1 + ( ∂ ∂xy(x)) 2_{) dx} Hint: no hint Solution.

We denote auxiliary function f

f(x, y(x), ∂ ∂xy(x)) = r y(x) (1 + ( ∂ ∂xy(x)) 2₎ in the form f(x, y, z) =py (1 + z2₎

Substituting we x, y(x) and y0(x) for x, y and z we obtain the integrand in the given functional. We compute partial derivatives

∂ ∂yf(x, y(x), ∂ ∂xy(x)) = 1 2 1 + (_∂x∂ y)2 q y (1 + (_∂x∂ y)2₎ and ∂ ∂zf(x, y(x), ∂ ∂xy(x)) = y (_∂x∂ y) q y (1 + (_∂x∂ y)2₎ and ∂2 ∂x ∂zf(x, y(x), ∂ ∂xy(x)) = −1 2

y(x) (_∂x∂ y(x)) ((_∂x∂ y(x)) (1 + (_∂x∂ y(x))2) + 2 y(x) (_∂x∂ y(x)) (_∂x∂22y(x)))

(y(x) (1 + (_∂x∂ y(x))2₎₎(3/2) + ( ∂ ∂xy(x)) 2 q y(x) (1 + (∂ ∂xy(x)) 2₎ + y(x) ( ∂2 ∂x2y(x)) q y(x) (1 + (∂ ∂xy(x)) 2₎

and finally obtain the Euler equation for our functional 1 2 1 py(x)+ 1 2 y(x) ( ∂ ∂xy(x)) (( ∂ ∂xy(x)) (1 + ( ∂ ∂xy(x)) 2_{) + 2 y(x) (}∂ ∂xy(x)) ( ∂2 ∂x2y(x))) (y(x) (1 + (_∂x∂ y(x))2₎₎(3/2) − ( ∂ ∂xy(x)) 2 q y(x) (1 + (∂ ∂xy(x)) 2₎ − y(x) ( ∂2 ∂x2y(x)) q y(x) (1 + ( ∂ ∂xy(x)) 2₎ = 0

y(x) = 4 C1 Info. quadratic function Comment. no comment

Z b a x y(x) ( ∂ ∂xy(x)) dx Hint: no hint Solution.

We denote auxiliary function f f(x, y(x), ∂ ∂xy(x)) = x y(x) ( ∂ ∂xy(x)) in the form f(x, y, z) = x y z

Substituting we x, y(x) and y0(x) for x, y and z we obtain the integrand in the given functional. We compute partial derivatives

∂ ∂yf(x, y(x), ∂ ∂xy(x)) = x ( ∂ ∂xy) and ∂ ∂zf(x, y(x), ∂ ∂xy(x)) = x y and ∂2 ∂x ∂zf(x, y(x), ∂ ∂xy(x)) = y(x) + x ( ∂ ∂xy(x)) and finally obtain the Euler equation for our functional

−y(x) − x ( ∂

∂xy(x)) = 0 We solve it using various tools and obtain the solution

y(x) = 0

Info.

not supplied

Comment.

Z b a x y(x) + 2 ( ∂ ∂xy(x)) 2_dx Hint: no hint Solution.

We denote auxiliary function f f(x, y(x), ∂ ∂xy(x)) = x y(x) + 2 ( ∂ ∂xy(x)) 2 in the form f(x, y, z) = x y + 2 z2

Substituting we x, y(x) and y0(x) for x, y and z we obtain the integrand in the given functional. We compute partial derivatives

∂ ∂yf(x, y(x), ∂ ∂xy(x)) = x and ∂ ∂zf(x, y(x), ∂ ∂xy(x)) = 4 ( ∂ ∂xy) and ∂2 ∂x ∂zf(x, y(x), ∂ ∂xy(x)) = 4 ( ∂2 ∂x2y(x))

and finally obtain the Euler equation for our functional x − 4 ( ∂

2

∂x2y(x)) = 0

We solve it using various tools and obtain the solution y(x) = 1 24x 3_{+ C1 x + C2} Info. not supplied Comment. no comment

Z b

a

x y(x) + y(x)2− 2 y(x)2₍ ∂

∂xy(x)) dx

Hint:

no hint

Solution.

We denote auxiliary function f f(x, y(x), ∂

∂xy(x)) = x y(x) + y(x)

2_{− 2 y(x)}2₍ ∂

∂xy(x)) in the form

f(x, y, z) = x y + y2− 2 y2_z

Substituting we x, y(x) and y0(x) for x, y and z we obtain the integrand in the given functional. We compute partial derivatives

∂ ∂yf(x, y(x), ∂ ∂xy(x)) = x + 2 y − 4 y ( ∂ ∂xy) and ∂ ∂zf(x, y(x), ∂ ∂xy(x)) = −2 y 2 and ∂2 ∂x ∂zf(x, y(x), ∂ ∂xy(x)) = −4 y(x) ( ∂ ∂xy(x)) and finally obtain the Euler equation for our functional

x + 2 y(x) + 4 y(x) ( ∂

∂xy(x)) = 0 We solve it using various tools and obtain the solution

y(x) = −1 2x Info. no extremal Comment. no comment

Z b a y(x) r 1 + ( ∂ ∂xy(x)) 2_dx Hint: no hint Solution.

We denote auxiliary function f f(x, y(x), ∂ ∂xy(x)) = y(x) r 1 + (∂ ∂xy(x)) 2 in the form f(x, y, z) = yp1 + z2

Substituting we x, y(x) and y0(x) for x, y and z we obtain the integrand in the given functional. We compute partial derivatives

∂ ∂yf(x, y(x), ∂ ∂xy(x)) = r 1 + (∂ ∂xy) 2 and ∂ ∂zf(x, y(x), ∂ ∂xy(x)) = y (_∂x∂ y) q 1 + (_∂x∂ y)2 and ∂2 ∂x ∂zf(x, y(x), ∂ ∂xy(x)) = (_∂x∂ y(x))2 q 1 + (_∂x∂ y(x))2 −y(x) ( ∂ ∂xy(x)) 2₍∂2 ∂x2y(x)) (1 + (_∂x∂ y(x))2₎(3/2) + y(x) (_∂x∂22y(x)) q 1 + (_∂x∂ y(x))2

and finally obtain the Euler equation for our functional 1 − ( ∂ ∂xy(x)) 2 q 1 + (_∂x∂ y(x))2 +y(x) ( ∂ ∂xy(x)) 2₍∂2 ∂x2y(x)) (1 + (_∂x∂ y(x))2₎(3/2) − y(x) (_∂x∂22y(x)) q 1 + (_∂x∂ y(x))2 = 0 We solve it using various tools and obtain the solution

y(x) = 1 2 1 + (e( √ C1 (x+ C2 ))₎2 e(√C1 (x+ C2 ))√ _C1 Info. not supplied Comment.

Z b a y(x) ( ∂ ∂xy(x)) dx Hint: no hint Solution.

We denote auxiliary function f f(x, y(x), ∂ ∂xy(x)) = y(x) ( ∂ ∂xy(x)) in the form f(x, y, z) = y z

Substituting we x, y(x) and y0(x) for x, y and z we obtain the integrand in the given functional. We compute partial derivatives

∂ ∂yf(x, y(x), ∂ ∂xy(x)) = ∂ ∂xy and ∂ ∂zf(x, y(x), ∂ ∂xy(x)) = y and ∂2 ∂x ∂zf(x, y(x), ∂ ∂xy(x)) = ∂ ∂xy(x) and finally obtain the Euler equation for our functional

−( ∂

∂xy(x)) = 0 We solve it using various tools and obtain the solution

y(x) = Y(x)

Info.

not supplied

Comment.

Z b a y(x) ( ∂ ∂xy(x)) 2_dx Hint: no hint Solution.

We denote auxiliary function f f(x, y(x), ∂ ∂xy(x)) = y(x) ( ∂ ∂xy(x)) 2 in the form f(x, y, z) = y z2

Substituting we x, y(x) and y0(x) for x, y and z we obtain the integrand in the given functional. We compute partial derivatives

∂ ∂yf(x, y(x), ∂ ∂xy(x)) = ( ∂ ∂xy) 2 and ∂ ∂zf(x, y(x), ∂ ∂xy(x)) = 2 y ( ∂ ∂xy) and ∂2 ∂x ∂zf(x, y(x), ∂ ∂xy(x)) = 2 ( ∂ ∂xy(x)) 2_{+ 2 y(x) (} ∂ 2 ∂x2y(x))

and finally obtain the Euler equation for our functional −2 ( ∂

∂xy(x))

2_{− 2 y(x) (} ∂2

∂x2y(x)) = 0

We solve it using various tools and obtain the solution y(x) = 0

Info.

not supplied

Comment.

Z b a y(x)2+ 2 x y(x) ( ∂ ∂xy(x)) dx Hint: no hint Solution.

We denote auxiliary function f f(x, y(x), ∂ ∂xy(x)) = y(x) 2_{+ 2 x y(x) (} ∂ ∂xy(x)) in the form f(x, y, z) = y2+ 2 x y z

Substituting we x, y(x) and y0(x) for x, y and z we obtain the integrand in the given functional. We compute partial derivatives

∂ ∂yf(x, y(x), ∂ ∂xy(x)) = 2 y + 2 x ( ∂ ∂xy) and ∂ ∂zf(x, y(x), ∂ ∂xy(x)) = 2 x y and ∂2 ∂x ∂zf(x, y(x), ∂ ∂xy(x)) = 2 y(x) + 2 x ( ∂ ∂xy(x)) and finally obtain the Euler equation for our functional

−2 x ( ∂

∂xy(x)) = 0 We solve it using various tools and obtain the solution

y(x) = Y(x)

Info.

all functions

Comment.

Z b a y(x)2+ 4 y(x) ( ∂ ∂xy(x)) + 4 ( ∂ ∂xy(x)) 2_dx Hint: no hint Solution.

We denote auxiliary function f f(x, y(x), ∂ ∂xy(x)) = y(x) 2_{+ 4 y(x) (} ∂ ∂xy(x)) + 4 ( ∂ ∂xy(x)) 2 in the form f(x, y, z) = y2+ 4 y z + 4 z2

Substituting we x, y(x) and y0(x) for x, y and z we obtain the integrand in the given functional. We compute partial derivatives

∂ ∂yf(x, y(x), ∂ ∂xy(x)) = 2 y + 4 ( ∂ ∂xy) and ∂ ∂zf(x, y(x), ∂ ∂xy(x)) = 4 y + 8 ( ∂ ∂xy) and ∂2 ∂x ∂zf(x, y(x), ∂ ∂xy(x)) = 4 ( ∂ ∂xy(x)) + 8 ( ∂2 ∂x2y(x))

and finally obtain the Euler equation for our functional 2 y(x) − 4 ( ∂

∂xy(x)) − 8 ( ∂2

∂x2y(x)) = 0

We solve it using various tools and obtain the solution y(x) = C1 cosh(1 2x) + C2 sinh( 1 2x) Info. two exponentials Comment. no comment

Z b a y(x)2+ y(x) ( ∂ ∂xy(x)) + ( ∂ ∂xy(x)) 2_dx Hint: no hint Solution.

We denote auxiliary function f f(x, y(x), ∂ ∂xy(x)) = y(x) 2_{+ y(x) (} ∂ ∂xy(x)) + ( ∂ ∂xy(x)) 2 in the form f(x, y, z) = y2+ y z + z2

Substituting we x, y(x) and y0(x) for x, y and z we obtain the integrand in the given functional. We compute partial derivatives

∂ ∂yf(x, y(x), ∂ ∂xy(x)) = 2 y + ( ∂ ∂xy) and ∂ ∂zf(x, y(x), ∂ ∂xy(x)) = y + 2 ( ∂ ∂xy) and ∂2 ∂x ∂zf(x, y(x), ∂ ∂xy(x)) = ( ∂ ∂xy(x)) + 2 ( ∂2 ∂x2y(x))

and finally obtain the Euler equation for our functional 2 y(x) − ( ∂

∂xy(x)) − 2 ( ∂2

∂x2y(x)) = 0

We solve it using various tools and obtain the solution y(x) = C1 sinh(x) + C2 cosh(x)

Info.

not supplied

Comment.

Z b a 2 y(x) ex+ y(x)2+ ( ∂ ∂xy(x)) 2_dx Hint: no hint Solution.

We denote auxiliary function f f(x, y(x), ∂ ∂xy(x)) = 2 y(x) e x_{+ y(x)}2_{+ (} ∂ ∂xy(x)) 2 in the form f(x, y, z) = 2 y ex+ y2+ z2

Substituting we x, y(x) and y0(x) for x, y and z we obtain the integrand in the given functional. We compute partial derivatives

∂ ∂yf(x, y(x), ∂ ∂xy(x)) = 2 e x_{+ 2 y} and ∂ ∂zf(x, y(x), ∂ ∂xy(x)) = 2 ( ∂ ∂xy) and ∂2 ∂x ∂zf(x, y(x), ∂ ∂xy(x)) = 2 ( ∂2 ∂x2y(x))

and finally obtain the Euler equation for our functional 2 ex+ 2 y(x) − 2 ( ∂

2

∂x2y(x)) = 0

We solve it using various tools and obtain the solution y(x) = (1 2cosh(x) 2₊1 2cosh(x) sinh(x) + 1 2x) sinh(x) + (−1 2cosh(x) sinh(x) + 1 2x − 1 2cosh(x)

2_{) cosh(x) + C1 sinh(x) + C2 cosh(x)} Info.

not supplied

Comment.

Z b a y(x)2+ (∂ ∂xy(x)) 2_{− 2 y(x) sin(x) dx} Hint: no hint Solution.

We denote auxiliary function f f(x, y(x), ∂ ∂xy(x)) = y(x) 2_{+ (} ∂ ∂xy(x)) 2_{− 2 y(x) sin(x)} in the form f(x, y, z) = y2+ z2− 2 y sin(x)

Substituting we x, y(x) and y0(x) for x, y and z we obtain the integrand in the given functional. We compute partial derivatives

∂ ∂yf(x, y(x), ∂ ∂xy(x)) = 2 y − 2 sin(x) and ∂ ∂zf(x, y(x), ∂ ∂xy(x)) = 2 ( ∂ ∂xy) and ∂2 ∂x ∂zf(x, y(x), ∂ ∂xy(x)) = 2 ( ∂2 ∂x2y(x))

and finally obtain the Euler equation for our functional 2 y(x) − 2 sin(x) − 2 ( ∂

2

∂x2y(x)) = 0

We solve it using various tools and obtain the solution y(x) = (1 4e x_{cos(x) −}1 4sin(x) e x₊1 4e (−x)_{cos(x) +}1 4e (−x)_{sin(x)) sinh(x)} + (−1 4e x_{cos(x) +}1 4sin(x) e x₊1 4e (−x)_{cos(x) +} 1 4e (−x)_{sin(x)) cosh(x)} + C1 sinh(x) + C2 cosh(x) Info.

exponentials and sin

Comment.

Z b a y(x)2− 4 y(x) ( ∂ ∂xy(x)) + 4 ( ∂ ∂xy(x)) 2_dx Hint: no hint Solution.

We denote auxiliary function f f(x, y(x), ∂ ∂xy(x)) = y(x) 2_{− 4 y(x) (} ∂ ∂xy(x)) + 4 ( ∂ ∂xy(x)) 2 in the form f(x, y, z) = y2− 4 y z + 4 z2

Substituting we x, y(x) and y0(x) for x, y and z we obtain the integrand in the given functional. We compute partial derivatives

∂ ∂yf(x, y(x), ∂ ∂xy(x)) = 2 y − 4 ( ∂ ∂xy) and ∂ ∂zf(x, y(x), ∂ ∂xy(x)) = −4 y + 8 ( ∂ ∂xy) and ∂2 ∂x ∂zf(x, y(x), ∂ ∂xy(x)) = −4 ( ∂ ∂xy(x)) + 8 ( ∂2 ∂x2y(x))

and finally obtain the Euler equation for our functional 2 y(x) + 4 ( ∂

∂xy(x)) − 8 ( ∂2

∂x2y(x)) = 0

We solve it using various tools and obtain the solution y(x) = C1 cosh(1 2x) + C2 sinh( 1 2x) Info. not supplied Comment. no comment

Z b a y(x) + y(x) ( ∂ ∂xy(x)) + ( ∂ ∂xy(x)) + 1 2( ∂ ∂xy(x)) 2_dx Hint: no hint Solution.

We denote auxiliary function f f(x, y(x), ∂

∂xy(x)) = y(x) + y(x) ( ∂ ∂xy(x)) + ( ∂ ∂xy(x)) + 1 2( ∂ ∂xy(x)) 2 in the form f(x, y, z) = y + y z + z + 1 2z 2

Substituting we x, y(x) and y0(x) for x, y and z we obtain the integrand in the given functional. We compute partial derivatives

∂ ∂yf(x, y(x), ∂ ∂xy(x)) = 1 + ( ∂ ∂xy) and ∂ ∂zf(x, y(x), ∂ ∂xy(x)) = y + 1 + ( ∂ ∂xy) and ∂2 ∂x ∂zf(x, y(x), ∂ ∂xy(x)) = ( ∂ ∂xy(x)) + ( ∂2 ∂x2y(x))

and finally obtain the Euler equation for our functional 1 − ( ∂

∂xy(x)) − ( ∂2

∂x2y(x)) = 0

We solve it using various tools and obtain the solution y(x) =1 2x 2_{+ C1 x + C2} Info. quadratic function Comment. no comment

Z b a y(x) − y(x) ( ∂ ∂xy(x)) + x ( ∂ ∂xy(x)) 2_dx Hint: no hint Solution.

We denote auxiliary function f f(x, y(x), ∂

∂xy(x)) = y(x) − y(x) ( ∂ ∂xy(x)) + x ( ∂ ∂xy(x)) 2 in the form f(x, y, z) = y − y z + x z2

Substituting we x, y(x) and y0(x) for x, y and z we obtain the integrand in the given functional. We compute partial derivatives

∂ ∂yf(x, y(x), ∂ ∂xy(x)) = 1 − ( ∂ ∂xy) and ∂ ∂zf(x, y(x), ∂ ∂xy(x)) = −y + 2 x ( ∂ ∂xy) and ∂2 ∂x ∂zf(x, y(x), ∂ ∂xy(x)) = ( ∂ ∂xy(x)) + 2 x ( ∂2 ∂x2y(x))

and finally obtain the Euler equation for our functional 1 − (∂

∂xy(x)) − 2 x ( ∂2

∂x2y(x)) = 0

We solve it using various tools and obtain the solution y(x) = 1 2x + C1 + C2 ln(x) Info. not supplied Comment. no comment

Z b a ( ∂ ∂xy(x)) (1 + x 2₍ ∂ ∂xy(x))) dx Hint: no hint Solution.

We denote auxiliary function f f(x, y(x), ∂ ∂xy(x)) = ( ∂ ∂xy(x)) (1 + x 2₍∂ ∂xy(x))) in the form f(x, y, z) = z (1 + x2z)

Substituting we x, y(x) and y0(x) for x, y and z we obtain the integrand in the given functional. We compute partial derivatives

∂ ∂yf(x, y(x), ∂ ∂xy(x)) = 0 and ∂ ∂zf(x, y(x), ∂ ∂xy(x)) = 1 + 2 x 2₍ ∂ ∂xy) and ∂2 ∂x ∂zf(x, y(x), ∂ ∂xy(x)) = 4 x ( ∂ ∂xy(x)) + 2 x 2₍ ∂ 2 ∂x2y(x))

and finally obtain the Euler equation for our functional −4 x ( ∂

∂xy(x)) − 2 x

2₍ ∂2

∂x2y(x)) = 0

We solve it using various tools and obtain the solution y(x) = C1 + C2 x Info. hyperbolic function Comment. no comment

Z b a (_∂x∂ y(x))2 x3 dx Hint: no hint Solution.

We denote auxiliary function f f(x, y(x), ∂ ∂xy(x)) = (_∂x∂ y(x))2 x3 in the form f(x, y, z) = z 2 x3

Substituting we x, y(x) and y0(x) for x, y and z we obtain the integrand in the given functional. We compute partial derivatives

∂ ∂yf(x, y(x), ∂ ∂xy(x)) = 0 and ∂ ∂zf(x, y(x), ∂ ∂xy(x)) = 2 ∂ ∂xy x3 and ∂2 ∂x ∂zf(x, y(x), ∂ ∂xy(x)) = 2 ∂2 ∂x2y(x) x3 − 6 ∂ ∂xy(x) x4

and finally obtain the Euler equation for our functional

−2 ∂2 ∂x2y(x) x3 + 6 ∂ ∂xy(x) x4 = 0

We solve it using various tools and obtain the solution y(x) = C1 + C2 x4

Info.

not supplied

Comment.

Z b a ( ∂ ∂xy(x)) 2_{+ 2 y(x) (} ∂ ∂xy(x)) − 16 y(x) 2_dx Hint: no hint Solution.

We denote auxiliary function f f(x, y(x), ∂ ∂xy(x)) = ( ∂ ∂xy(x)) 2_{+ 2 y(x) (} ∂ ∂xy(x)) − 16 y(x) 2 in the form f(x, y, z) = z2+ 2 y z − 16 y2

Substituting we x, y(x) and y0(x) for x, y and z we obtain the integrand in the given functional. We compute partial derivatives

∂ ∂yf(x, y(x), ∂ ∂xy(x)) = 2 ( ∂ ∂xy) − 32 y and ∂ ∂zf(x, y(x), ∂ ∂xy(x)) = 2 ( ∂ ∂xy) + 2 y and ∂2 ∂x ∂zf(x, y(x), ∂ ∂xy(x)) = 2 ( ∂2 ∂x2y(x)) + 2 ( ∂ ∂xy(x)) and finally obtain the Euler equation for our functional

−32 y(x) − 2 ( ∂

2

∂x2y(x)) − 2 (

∂

∂xy(x)) = 0 We solve it using various tools and obtain the solution

y(x) = C1 cos(4 x) + C2 sin(4 x)

Info.

trig functions

Comment.

Z b a ( ∂ ∂xy(x)) 2_{+ cos(y(x)) dx} Hint: no hint Solution.

We denote auxiliary function f f(x, y(x), ∂ ∂xy(x)) = ( ∂ ∂xy(x)) 2_{+ cos(y(x))} in the form f(x, y, z) = z2+ cos(y)

Substituting we x, y(x) and y0(x) for x, y and z we obtain the integrand in the given functional. We compute partial derivatives

∂ ∂yf(x, y(x), ∂ ∂xy(x)) = −sin(y) and ∂ ∂zf(x, y(x), ∂ ∂xy(x)) = 2 ( ∂ ∂xy) and ∂2 ∂x ∂zf(x, y(x), ∂ ∂xy(x)) = 2 ( ∂2 ∂x2y(x))

and finally obtain the Euler equation for our functional −sin(y(x)) − 2 ( ∂

2

∂x2y(x)) = 0

We solve it using various tools and obtain the solution y(x) = 0

Info.

not supplied

Comment.