Calculus of Variations - Lagrange Equation

Question

My instructor has given the question in the image and the sample answer has also been given by him.

Image of the question and the instructor's answer: drive(DOT)google(DOT)com/file/d/1XkjhzrN1MKNc0XDrLlAR_pu6l39P9AY7/view?usp=share_link

However, I don't understand how step 1 simplifies into step 2, as in his calculation. Instead, I get below as step 2.

Step 2 according to my calculations :

( y*(y'') + ((y')^2) )/(sqrt(1+((y')^2)))) - (y*((y')^2)/(1+((y')^2))^(3/2)) - (sqrt((1+((y')^2))))

Any help in understanding this will be highly appreciated!

Answer 1

F(x,y,y') = y√[1+(y')²]

∂F/∂y = √[1+(y')²]

∂F/∂y' = yy'/√[1+(y')²]

As the function F is independent of the variable x, then by the Beltrami identity, the Euler-Lagrange equation is equivalent to y'F_y' - F = C:

y'F_y' - F = C

y'(yy'/√[1+(y')²]) - y√[1+(y')²] = C

y(y')²/√[1+(y')²]) - y√[1+(y')²] = C

y(y')²/√[1+(y')²] - y[1+(y')²]/√[1+(y')²] = C

(y(y')²-y[1+(y')²])/√[1+(y')²] = C

[y(y')²-y-y(y')²]/√[1+(y')²] = C

-y/√[1+(y')²] = C

-y = C√[1+(y')²]

y² = C[1+(y')²]

y² = C + C(y')²

y² - C = C(y')²

(y²-C)/C = (y')²

y' = ±√[(y²-C)/C]

Hope this helped!

Answer 2

for the derivative d(yy'(1+y'²)^(-1/2))/dx

=yy''(1+y'²)^(-1/2)-yy'²y''(1+y'²)^(-3/2)+y'²(1+y'²)^(-1/2)

subtracting (1+y'²)^(1/2), set equal to 0, multiply by ((1+y'²)^(1/2)

have

yy''-[yy'²y''/(1+y'²)]+y'²-1-y'²=0

same as

[yy''-1+y'²]/(1+y'²)=0

the denominator is different than image, suppose could multiply by (1+y'²)^(1/2) if needed