## Download A Textbook of Engineering Mathematics-I, 2nd Edition by H.S. Gangwar, Dr. Prabhakar Gupta PDF

By H.S. Gangwar, Dr. Prabhakar Gupta

Written for the scholars of BTech I 12 months of UP Technical collage, Lucknow and different states, this ebook discusses intimately the ideas and strategies in Engineering arithmetic.

**Read or Download A Textbook of Engineering Mathematics-I, 2nd Edition PDF**

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**Additional info for A Textbook of Engineering Mathematics-I, 2nd Edition**

**Example text**

N = 5 By Euler’s theorem, we get x ∂u ∂u + y ∂y = 5u. Proved. ∂x Next, we know that (from Corollary 1, on page 36) or x2 ∂ 2u ∂ 2u ∂ 2u 2 + 2xy +y ∂y2 = n (n – 1) u = 5 (5 – 1) u ∂x∂y ∂x2 x2 ∂ 2u ∂ 2u ∂ 2u 2 + 2xy + y = 20 u. Hence proved. ∂x∂y ∂y2 ∂x2 45 DIFFERENTIAL CALCULUS-I FG y IJ H xK Example 12. If u = x f1 x2 + f2 FG y IJ , prove that H xK 2 ∂ 2u ∂ 2u 2 ∂ u = 0. + 2 xy + y ∂x∂y ∂x 2 ∂y 2 u1 = Sol. Let FG y IJ H xK x f1 and u2 = x0f2 FG y IJ , then u = u H xK 1 + u2 Since u1 is a homogeneous function of degree one.

I), (ii) and (iii), we get ∂u ∂u ∂u + ∂y + ∂x ∂z = e 3 x 2 + y 2 + z 2 – xy − yz − zx x + y + z − 3xyz 3 3 3 j e 3 x 2 + y 2 + z 2 – xy − yz − zx = bx + y + zgex j + y + z − xy − yz − zx 2 2 2 a j f As a 3 + b 3 + c 3 – 3abc = a + b + c ( a 2 + b 2 + c 2 − ab − bc − ca) ∂u ∂u ∂u + + ∂x ∂y ∂z or = 3 . (iv) F ∂ + ∂ + ∂ I u = FG ∂ + ∂ + ∂ IJ FG ∂ + ∂ + ∂ IJ u GH ∂x ∂y ∂z JK H ∂x ∂y ∂z K H ∂x ∂y ∂z K F ∂ ∂ ∂ I F ∂u ∂u ∂u I F ∂ ∂ ∂ I F 3 I = G ∂x + ∂y + ∂z J G ∂x + ∂y + ∂z J = G ∂x + ∂y + ∂z J G x + y + z J , H KH K H KH K L ∂ F 1 I ∂ F 1 I ∂ F 1 IO = 3 M ∂x G x + y + z J + ∂y G x + y + z J + ∂z G x + y + z J P K H K H K PQ MN H L 1 OP 1 1 −9 − − = 3 M− MN bx + y + zg bx + y + zg bx + y + zg PQ = bx + y + zg .

E1 + y j e1 + x + y j e1 + x + y j ∂ ∂x 2 2 2 = Example 10. If 2 2 2 2 2 −1/2 2 2 2 2 2 2 3/2 2 2 3/2 y2 x2 z2 + + = 1, show that a2 + u c2 + u b2 + u FG ∂u IJ + FG ∂u IJ + FG ∂u IJ H ∂x K H ∂y K H ∂z K 2 2 2 2 2 = 2 2 2 =2 F x ∂u + y ∂u + z ∂u I . , 2002) Sol. (ii) 2 2 LM 2x O 2y 2z P + + ∑ LMNx / e a + uj OPQ MN e a + uj eb + uj e c + uj PQ 2 2 2 2 ∑ LMNx 2 / e a 2 + uj 2 2 OP Q 2 2 1 2 2 2 [1], from (i) F x ∂u + y ∂u + z ∂u I GH ∂x ∂y ∂z JK . (iii) Hence proved. Example 11. If xx yy zz = c, show that at x = y = z, ∂ 2z = – (x log ex)–1.