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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**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 n2 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 yx y y x 2 z y 2 z z = (n 1) 2 xy y y Multiplying (2) by x, we have 2 z z = ( n 1) x 2 xy x x Multiplying (3) by y, we have z 2 z 2 z xy. y 2 . 2 = (n 1) y y yx 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 xy 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.