By H.S. Gangwar, Dr. Prabhakar Gupta
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Additional info for A Textbook of Engineering Mathematics: v. 1
Consider a homogeneous polynomial in x and y f (x, y) = a0xn + a1 xn−1 y + a2 xn-2y2 + ..... + a n Hence every homogeneous function of x and y of degree n can be written in above form. NOTE: Degree of Homogeneous function = degree of numerator – degree of denominator. Remark 1: If f (x, y) = a0xn + a1xn+1 . y–1 + a2xn+2 . y–2 + ...... + anxn+n y –n R| F xI F xI x + a G J + .... + a G J S| y H yK H yK T F xI f (x, y) = x F G J ; degree = n H yK f (x, y) = a y + a y . x + ...... + a y . x R| F x I F x I U| = y Sa + a G J + ....
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. (ii) = 0 ∂2 ∂2 (u1 + u2) + y2 (u1 + u2) = 0 ∂x∂y ∂y 2 ∂ 2u ∂ 2u ∂ 2u 2 + 2xy +y ∂x∂y ∂x 2 ∂y 2 = 0. Hence proved. Example 13. If u = sin–1 (x3 + y3)2/5, evaluate 2 ∂ 2u x 2∂ 2u 2 ∂ u . + + xy y 2 ∂x∂y ∂x 2 ∂y 2 u = sin–1 (x3 + y3)2/5 Sol.
B) 2 Thus, from (A) and (B), we find ∂ 2u ∂ 2u = . ∂y∂x ∂x∂y Hence proved. FG y IJ , verify that ∂ f H xK ∂y∂x F yI f = tan G J H xK 2 Example 2. If f = tan–1 Sol. We have = ∂2 f . (iv) 2 Differentiating (iii) partially with respect to x, we get F GH ∂2 f x ∂ = 2 ∂x∂y ∂x x + y 2 = Example 3. If u(x + y) = x Sol. (v) F ∂u ∂u I + y , prove that G − J H ∂x ∂y K F1 − ∂u − ∂u I . GH ∂x ∂y JK 2 =4 x2 + y2 x+y bx + yga2xf − ex + y ja1f = x + 2xy − y b x + yg bx + y g bx + ygb2yg − ex + y ja1f = y + 2xy − x bx + yg bx + yg 2 = ∂u ∂y = ∂u ∂u + ∂x ∂y = b x + yg = 1− = ex 2 2 2 2 2 2 = 2 2 2 2 2 2 2 2 4xy 2 4xy ( x + y) 2 = (x − y) 2 (x + y) 2 j e bx + y g 2 ex − y j 2 bx − yg = bx + yg bx + y g 2 + 2xy − y 2 − y 2 + 2xy − x 2 2 2 ∴ 3 ∂2 f ∂2 f = .
A Textbook of Engineering Mathematics: v. 1 by H.S. Gangwar, Dr. Prabhakar Gupta