Partial Derivatives And Euler's Equation MCQ Level - 2

u = sin^-1(x + y/(√x + √y))

Put x = xt and y = yt 

u = sin^-1[(xt-yt)/(√xt + √yt)]

sinu = t^1/2[(x+y)/(√x + √y)]

= t^1/2 f(x,y)

This function sin u is homogeneous with degree 1/2.

By Euler's theorem

xu_x + yu_y = G(u) = nf(u)/f,(u) = 1/2 tan u

xu_x + yu_y = 1/2 tan u

x2u_xx + 2xyu_yy + y2u_yy = G(u)[G'(u) - 1]

= 1/2 tan u [(sec²u - 2)/2]

= 1/4 tan u [(tan²u - 1)/1]

= 1/4 * sin u/cos u [(sin²u - cos²u)/cos²u]

x²u_xx + 2xyu_yy + y2u_yy = -(sinu cos²u)/4ucos³u

We have 

v is homogeneous function of degree n then

The correct answer is: n(n - 1)

The correct answer is: II - f(x, y),  I - f_x(x, y), III - f_y(x, y)

We are given 

f(u) is homogeneous function degree 1, then


 
  [By dividing with cos³ u]

The correct answer is:  

The correct answer is: 

We have w = f(u, v)

The correct answer is: 

Let w = u²v²
then z = f(w) + u

So, 

The correct answer is: 

f(u) is homogeneous function of degree 2

Now, let g(u) = 2tan u

The correct answer is: 2tan u(2sec²u –1)

(1) Differentiate with respect to y, I get:

0+F2+F3∂z/∂y=0

So 

F3 ∂z/∂y = −F2

(2) Differentiate with respect to x, I get:

F1 + F2 ∂y/∂x + 0 = 0

So F2 ∂y/∂x = −F1

(3) Differentiate with respect to z, I get:

F1 ∂x/∂z +0 + F3 = 0

4) After some manipulations with the Fi, I get to the conclusion that 

∂z/∂y∗∂y/∂x∗∂x/∂z=−1, so when evaluated with x, z, y respectively, conclusion is still true

The correct answer is: