By H.K. Dass

ISBN-10: 8121903459

ISBN-13: 9788121903455

Bargains with partial differentiation, a number of integrals, functionality of a posh variable, exact services, laplace transformation, complicated numbers, and facts.

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Show that z z z ( x y) 0 is equivalent to = 0 x y v 7. If u = f (x2 + 2 y z, y2 + 2 z x), prove that u du du ( y 2 z x) ( x 2 y z) ( z 2 xy ) 0 x dy dz (x + y) Created with Print2PDF. com/ Partial Differentiation 36 8. By changing the independent variables x and t to u and v by means of the relationships u = x – at, v = x + at 2 y a2 Show that x2 2 y t2 4 a2 2 y u v du dx 1 v y 9. d x d u y v 2 y x v u uv z z z , show that 2 x u v .

2 2u 2 u y and = n(n – 1)u xy x 2 y 2 As v is a homogeneous function of x, y of degree –n. (4) Created with Print2PDF. (7) [From (1)] xy x y On adding (6) and (7), we have 2 2 z z z 2 z y x y 2 2 x y x y = n(n – 1)u + n (n + 1) v + nu – nv x y = nu (n – 1 + 1) + nv (n + 1 – 1) = n2u + n2v = n2 (u + v) = n2z II. Deduction: Prove that x2 2 z 2 xy x2 2u x 2 2 xy Proved. 2u 2u y 2 2 = g(u) [g(u) – 1] (Nagpur University, Winter 2003) xy y f (u ) g(u) = n f (u ) Proof.

2 z 2 z y 2 . 2 n(n 1) z. xy y (Uttarakhand Ist Semester, Dec. 2006) Created with Print2PDF. 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 .

### Advanced Engineering Mathematics by H.K. Dass

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