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 e-book discusses intimately the strategies and methods in Engineering arithmetic.

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

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2006) Proof. t. (ii) 2 FG y IJ H xK + xn−1 y F′ FG y IJ H xK (from (i)). , xn then x1 ∂f ∂f + x2 ∂x1 ∂x 2 ∂f + ... + xn ∂x = nf. n Corollary 1. If f is a homogeneous function of degree n, then 2 x ∂2 f ∂x 2 ∂2 f ∂2 f 2 + 2xy +y = n (n –1)f. ∂x∂y ∂y 2 Proof. t. (iii) Multiplying (ii) by x and (iii) by y and adding, we have x2 ∂2 f ∂x 2 + 2xy ∂2 f ∂2 f ∂f ∂f + y2 +y =n 2 + x ∂x∂y xy ∂y ∂x F x ∂f + y ∂t I GH ∂x ∂y JK 37 DIFFERENTIAL CALCULUS-I ⇒ x 2 ∂2 f ∂x 2 ∂2 f ∂2 f 2 + 2xy +y ∂x∂y ∂y 2 = n2f − nf = n (n – 1) f Example 1.

U. t. x, we get y1 = cos(a sin–1 x). (v) Replace n by (n – 2) in relation (v), we get yn (0) = {(n – 2)2 – a2}yn–2 (0) Again replace n by (n – 4) in equation (v) and putting yn–2 (0) in above relation, we get yn (0) = {(n – 2)2 – a2} {(n – 4)2 – a2}yn–4 (0) Case I. When n is odd: yn (0) = {(n – 2)2 – a2} {(n – 4)2 – a2} ..... (vi) [The last term in (vi) obtain by putting n = 1 in equation (v)] Putting x = 0, in equation (i), we get y1 (0) = cos(a sin–1 0). a = cos 0 . a ⇒ y1 (0) = a yn (0) = {(n – 2)2 – a2} {(n – 4)2 – a2} .....

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 . 2 Now, 2 2 2 from (iv) 2 Hence proved. Example 9. If u = tan–1 ∂ 2u ∂x∂y Sol. 1 − y 1 e1 + x + y j 2 ×x M + 1 x e +y j MN e1 + x + y j − y x .

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