Title Unit - 1 Typed Notes.pdf ✅

which is the required partial differential equation. ELIMINATION OF ARBITRARY FUNCTIONS Let us consider the relationf(uv) = 0, where u and v are functions of x, y, z and f is an arbitrary function to be eliminated. Differentiating, f(uv) = 0 partially with respect to x and y we get the PDE of the form Pp Qq = R. Problems 1. Form the partial differential equation by eliminating the arbitrary function ‘f ’ from z xy + yz + zx = f ( ) x + y Solution xy + yz + zx = f ( z x+y ) (1) Differentiating (1) partially with respect to x and y, we have ′ ( ) (x+y)p−z y + yp + xp + z = f u { (x+y) 2 } (2) and x + yq + xq + z = f′(u) { (x+y)q−z } (3) (x+y) 2 Dividing (2) by (3), we have (y+z)+(x+y)p = (x+y)p−z (z+x)+(x+y)q (x+y)q−z (x + y)(z + x)p − z(z + x) − z(x + y)q = (x + y)(y + z)q − z(y + z) − z(x + y)p i.e. (x + y)(x + 2z)p − (x + y)(y + 2z)q = z(x − y) which is the required partial differential equation. 2.Form the partial differential equation by eliminating the arbitrary function f and g from z = f(2x + y) + g(3x-y) Solution
Given z = f(2x + y) + g(3x-y) (1) Differentiating (1) partially with respect to x, P = f′(u)2 + g′(v)3 where u = 2x +y and v = 3x – y (2) Differentiating (1) partially with respect to y, q = f′(u)1 + g′(v)(−1) (3) Differentiating (2) partially with respect to x and y, r = f  (u)4 + g(v)9 (4) and s = f  (u)2 + g(v)(−3) (5) Eliminating f (u) and g(v) from (4), (5) and (6) using determinants, we have 4 9 r |2 −3 s| = 0 1 1 t i.e. 5r + 5s – 30t = 0 or ∂ 2 z + ∂ 2 z − 6 ∂ 2 z = 0 ∂x2 ∂x∂y ∂y2 3.Find the Partial differential equation of all planes cutting equal intercepts from x and y axis solution: Nov.,2009 Intercepts form of the plane equation is x a + y b + z = 1 c Given: a = b [ Equal intercepts on the x and y –axis] x y z + + = 1 a b c Here a and c are the two arbitrary constants. diff., (1) p.w.r to x , we get