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 publication discusses intimately the options and strategies in Engineering arithmetic.

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

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

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

**Example text**

Iv) yn(0) = (–1)2 {(n – 2)2 – m2} {(n – 4}2 – m2} yn–4 (0) Case I. If n is odd But or ⇒ Case II. If n is even yn(0) y1(0) y1(0) yn(0) = = = = – {(n – 2)2 – m2} {(n – 4)2 – m2} ... {12 – m2} y1 (0) my(0) m (As y (0) = 1) {m2 – (n – 2)2}{m2 – (n – 4)2} ... (m2 – 12)·m. yn (0) = {m2 – (n – 2)2} {m2 – (n – 4)2}... (m2 – 22) y2 (0) ⇒ yn (0) = {m2 – (n – 2)2} {m2 – (n – 4)2} ... (m2 – 22)·m2. (As y2 (0) = m2). –1 Example 4. Find the nth differential coefficient of the function on cos (2 cos x) at the point x = 0.

If u = x 2 + y 2 + z 2 j 1 2 , then prove that ∂2u ∂x 2 11. If z = f (x + ay) + φ (x – ay), prove that 13. If u = log (x2 + y2), show that ∂2u ∂x 2 16. If u = r , where r = x +y +z 2 17. If u = (x2 + y2 + z2)–1, prove that 18. If θ = t e − r2 4t + y2 = 0. ∂ 2u = 0. ∂y 2 + ∂2 f ∂x 2 = b2 ∂2 f ∂t 2 . ∂ 2u ∂ 2u ∂ 2u . find 2 + ∂y 2 + ∂x ∂z 2 2 [Ans. m (m + 1)rm−2] ∂ 2u ∂ 2u ∂2u + + = 2 (x2 + y2 + z2)–2. ∂x 2 ∂z 2 ∂y 2 , find the value of n, when 1 ∂ r 2 ∂r LMAns. n = − 3 OP 2Q N FG r ∂θ IJ = ∂θ . H ∂r K ∂t 2 19.

T. x, we get x or ∂u ∂ 2u ∂u ∂ 2u + + y = 2 cos 2u . t. (iii) Adding (ii) and (iii), we get x2 ∂ 2u ∂ 2u ∂ 2u 2 = ( 2cos 2u – 1) 2 + 2xy ∂x∂y + y ∂x ∂y2 F x ∂u + y ∂u I GH ∂x ∂y JK = (2cos 2u – 1) sin 2u, (from (i)) = (2sin 2u cos 2u – sin 2u) = sin 4u – sin 2u = 2 cos FG 4u + 2uIJ H 2 K FG 4u − 2uIJ . H 2 K . cos ∂ 2u ∂ 2u 2 Hence, x + 2xy +y ∂y2 = 2 cos 3u · cos u. ∂x∂y ∂x2 2 ∂ 2u Example 8. If z = xm f x2 Sol. (i) Now, u is homogeneous function of degree m. (iii) Adding (ii) and (iii), we get ∂ ∂2 ∂2 2 x 2 (u + v) + 2xy ∂x∂y (u + v) + y ∂y 2 (u + v) = m (m – 1) u + n (n – 1) v ∂x 2 ⇒ x2 ∂ 2z ∂2z ∂ 2z 2 + 2xy + y ∂y2 = m (m – 1) u + n (n – 1) v (As z = u + v).