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If U = Sin − 1 ( X 1 3 + Y 1 3 X 1 2 − Y 1 2 ) , Prove that X 2 ∂ 2 U ∂ X 2 + 2 X Y ∂ 2 U ∂ X ∂ Y + Y 2 ∂ 2 U ∂ Y 2 = Tan U 144 ( Tan 2 U + 13 ) - Applied Mathematics 1

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Sum

If u `=sin^(-1)((x^(1/3)+y^(1/3))/(x^(1/2)-y^(1/2)))`, Prove that 

`x^2(del^2u)/(delx^2)+2xy(del^2u)/(delxdely)+y^2(del^2u)/(dely^2)=tanu/144(tan^2u+13)`

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Solution

`z=sinu=sqrt((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)`=sqrt((x^(1/3)+y^(1/3))/(x^(1/2)-y^(1/2)))= sqrt((xt^(1/3)+yt^(1/3))/(xt^(1/2)-yt^(1/2))) = sqrt((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 homogeneous function of x,y of degrees 1/12 Hence, by the above corollary.

`x^2(del^2u)/(delx^2)+2xy(del^2u)/(delxdely)+y^2(del^2u)/(dely^2)=g(u)[g'(u)-1]`
Where, `g(u)=n(f(u))/(f'(u))=(-1)/12.(sinu)/(cosu)=(-1)/12tanu`

`g'(u)-1=(-1)/12sec^2u-1=(-1)/12(1-tan^2u)-1 =(-1)/12tan^2u-13/12`

`=(-1)/12(tan^2u+13)`

`therefore g(u)[g'(u)-1]=((-1)/12tanu)((-1)/12(tan^2u+13))`

`x^2(del^2u)/(delx^2)+2xy(del^2u)/(delxdely)+y^2(del^2u)/(dely^2)=(tanu)/144(tan^2u+13)`

Concept: System of Homogeneous and Non – Homogeneous Equations
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