Question

If u= `sin^-1 ((x+y)/(sqrtx+sqrty)), " prove that ""`i.xu_x+yu_y=1/2 tanu`

ii. `x^2uxx+2xyu_xy+y^2u_(y y)=(-sinu.cos2u)/(4cos^3u)`

Answer 1

`u= sin^1(x+y/(sqrtx+sqrty))`

Put x = xt and y = yt to find degree. 

∴ `u=sin_1((xt-yt)/(sqrt(xt)+sqrt(yt)))` 

∴ `sin u= t^(1/2).(x+y)/(sqrtx+sqrty)=t^(1/2). f(x,y)`

The function sin u is homogeneous with degree ½.
But sin u is the function of u and u is the function of x and y.
By Euler’s theorem , 

`xu_x+yu_y=G(u)=n. f(u)/(f,(u))=1/2 tanu` 

∴` xu_x+yu_y= 1/2 tanu`

∴` x^2u_(x x)+2xyu_(y y)+y^2u_(yy)=G(u)[G'(u)-1]`

                            =`1/2 tanu[(sec^2u-2)/2]`

                           =` 1/4 tan u [(tan^2u-1)/1]`

                           =`1/4xx sin u /cos u[(sin^2 u-cos^2 u)/(cos^2u)]` 

∴` x^2u_(x x)+2xyu_(x y)+y^2u_(y y)= -(sinu.cos2u)/(4cos^3u)`

Hence Proved.