# Advanced Engineering Mathematics by H.K. Dass

By H.K. Dass

Bargains with partial differentiation, a number of integrals, functionality of a fancy variable, specific features, laplace transformation, advanced numbers, and facts.

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Extra info for Advanced Engineering Mathematics

Example text

2006) Statement. If z is a homogeneous function of x, y of order n, then x. z z  y. = nz x y Created with Print2PDF. com/ Partial Differentiation 17 Proof. Since z is a homogeneous function of x, y of order n. t. ‘x’, we have  y  y y  n    x . f     2  x    x x  z  y n 1  y  n2 y. f     = nx . f    x x x x Multiplying both sides by x, we have z  y  y n n 1 x = n x . f    x y. t. ‘y’, we have z y 1 n = x f  . y x x Multiplying both sides by y, we get z  y n 1 y.

Com/ Partial Differentiation 21 Solution. By Euler’s Theorem x. t. ‘x’, we get z z 2 z 2 z  x. t. ‘y’, we have x.  x. (3) z  2 z z 2 z  y 2 = n  y yx y y x 2 z y 2 z z  = (n  1) 2 xy y y Multiplying (2) by x, we have  2 z z = ( n  1) x 2 xy x x Multiplying (3) by y, we have z 2 z 2 z xy.  y 2 . 2 = (n  1) y y yx y Adding (4) and (5), we get x2 . (5) 2  z z  2 z 2  z  y . = (n  1)  x  y  2 2  x  y xy  x y = (n – 1) n z [From (1)] = n(n – 1) z Proved (iii) Example 26.

X y x y Solution. t. x (treating y as constant), we obtain  f u  f v .  . 1 = u  x v  x 0 =  u  v .  . u  x v  x v u and , we obtain x x  v  f   f  .  . u v v u   u  f   f  .  . t. y, we get  f u  f v .  . (4) Solving the equations (3) and (4) for u = x v = x Similarly, differentiating (1) and 0 = 1 =  u  v .  . u  y v  y Solving the equations (5) and (6) for Ans. Ans. (6) u v and , we obtain y y Created with Print2PDF.