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प्रश्न
If U `=sin^(-1)[(x^(1/3)+y^(1/3))/(x^(1/2)+y^(1/2))]`prove that `x^2(del^2u)/(del^2x)+2xy(del^2u)/(delxdely)+y^2(del^2u)/(del^2y)=(tanu)/144[tan^2"U"+13].`
उत्तर १
Given `"U"=sin^(-1)[(x^(1/3)+y^(1/3))/(x^(1/2)+y^(1/2))]`
z = sin u =`[(x^(1/3)+y^(1/3))/(x^(1/2)+y^(1/2))]^(1/2)`is homogeneous function in x and y with degree `-1/12`
∴ We have the result,
If z = f (u) is homogeneous function of degree x and y then
`x^2(del^2u)/(del^2x)+2xy(del^2u)/(delxdely)+y^2(del^2u)/(del^2y)`= g (u) [g' (u) – 1] where g(u) = n `9(f(u))/(f'(u)).`
`n = -1/12`, f (u) = sin u, f'(u) = cos u
`therefore g(U)=-1/12(sinu)/(cosu)`
`therefore g(U) =-1/21tan"U"`
`thereforeg'(U)=-1/12sec^2"U"`
By above result,
`x^2(del^2u)/(del^2x)+2xy(del^2u)/(delxdely)+y^2(del^2u)/(del^2y)=-1/12[-1/12sec^2"U"-1]`
`=1/12[1/12sec^2"U"-1]`
`=1/12[1/12sec^2"U"+1]=1/12[(1+tan^2u)/12+1]`
`=1/144tan"U"[tan^2"U"+13]`
`therefore x^2(del^2u)/(del^2x) +2xy(del^2u)/(delxdely) +y^2(del^2u)/(del^2y)=tanu/144[tan^2"U"+13].`
Hence proved
उत्तर २
Given `"U"=sin^(-1)[(x^(1/3)+y^(1/3))/(x^(1/2)+y^(1/2))]`
z = sin u =`[(x^(1/3)+y^(1/3))/(x^(1/2)+y^(1/2))]^(1/2)`is homogeneous function in x and y with degree `-1/12`
∴ We have the result,
If z = f (u) is homogeneous function of degree x and y then
`x^2(del^2u)/(del^2x)+2xy(del^2u)/(delxdely)+y^2(del^2u)/(del^2y)`= g (u) [g' (u) – 1] where g(u) = n `9(f(u))/(f'(u)).`
`n = -1/12`, f (u) = sin u, f'(u) = cos u
`therefore g(U)=-1/12(sinu)/(cosu)`
`therefore g(U) =-1/21tan"U"`
`thereforeg'(U)=-1/12sec^2"U"`
By above result,
`x^2(del^2u)/(del^2x)+2xy(del^2u)/(delxdely)+y^2(del^2u)/(del^2y)=-1/12[-1/12sec^2"U"-1]`
`=1/12[1/12sec^2"U"-1]`
`=1/12[1/12sec^2"U"+1]=1/12[(1+tan^2u)/12+1]`
`=1/144tan"U"[tan^2"U"+13]`
`therefore x^2(del^2u)/(del^2x) +2xy(del^2u)/(delxdely) +y^2(del^2u)/(del^2y)=tanu/144[tan^2"U"+13].`
Hence proved
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