$x^2 \frac{(∂^2 u)}{(∂x^2 )} + 2xy(∂^2 u)/∂x∂y + y^2 (∂^2 u)/(∂y^2 ) = \frac{tanu}{144}((tan)^2 u + 13) $

Mumbai Unversity>FE>Sem1>Applied Maths1

$Z =sinu = √(\frac{(x^{1/3} + y^{1/3})}{(x^{1/2} - y^{1/2} )}) = f(u) = F(X,Y) say .$ Putting X =xt, Y = yt $F(X,Y) = √(\frac{(x^{1/3} + y^{1/3})}{(x^{1/2} - y^{1/2} )}) = √(\frac{((xt)^{1/3} + (yt)^{1/3})}{((xt)^{1/2} - (yt)^{1/2} )}) = √\frac{t^{1/3}{t^{1/2}} ((x^{1/3} + y^{1/3})}{(x^{1/2} - y^{1/2} ))} = t^{-1/12}f(x,y)$ Thus Z = f(u) = sinu is a homogenous function of x, y of degrees 1/12 Hence , by the above corollary. $x^2 (∂^2 u)/(∂x^2 ) + 2xy(∂^2 u)/∂x∂y + y^2 (∂^2 u)/(∂y^2 ) = g(u)[g’(u)-1]$ $Where , g(u) = n(f(u))/(f'(u)) = (-1)/12.sinu/cosu = (-1)/12 tanu g’(u)-1 =(-1)/12 (sec)^2 u – 1 = (-1)/12 ((1-tan)^2 u)– 1 = (-1)/12 (tan)^2 u -13/12$ $= (-1)/12((tan)^2 u +13)$ $∴ g(u)[g’(u)-1] = ( (-1)/12 tanu)((-1)/12((tan)^2 u +13))$ $x^2 (∂^2 u)/(∂x^2 ) + 2xy(∂^2 u)/∂x∂y + y^2 (∂^2 u)/(∂y^2 ) = tanu/144((tan)^2 u + 13)$