Q.3 (b) For u=sin⁻¹(x²+y²)^(1/5) deduce x²∂²u/∂ x²+2xy∂²u/∂ x∂ y+y²∂²u/∂ y²=frac225tan u (2tan²u-3) .

z=sin u=(x²+y²)^(1/5) is homogeneous of degree n=2/5 . Euler: xz_x+yz_y=nz⇒cos u (xu_x+yu_y)=2/5sin u⇒ xu_x+yu_y=2/5tan u=g(u) . Deduction: x²u_xx+2xyu_xy+y²u_yy=g(u)[g'(u)-1] , with g'(u)=2/5sec²u . =2/5tan u(2/5sec²u-1)=frac225tan u (2sec²u-5)=frac225tan u (2+2tan²u-5) =frac225tan u (2tan²u-3) (proved).

For \(u=\sin^{-1}(x^2+y^2)^{1/5}\) deduce \(x^2\frac{\partial^2u}{\partial x^2}+2xy\frac{\partial^2u}{\partial x\partial y}+y^2\frac{\partial^2u}{\partial y^2}=\frac2{25}\tan u\,(2\tan^2u-3)\).

Q.3 (b) For \(u=\sin^{-1}(x^2+y^2)^{1/5}\) deduce \(x^2\frac{\partial^2u}{\partial x^2}+2xy\frac{\partial^2u}{\partial x\partial y}+y^2\frac{\partial^2u}{\partial y^2}=\frac2{25}\tan u\,(2\tan^2u-3)\).

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\(z=\sin u=(x^2+y^2)^{1/5}\) is homogeneous of degree \(n=\frac25\).

Euler: \(xz_x+yz_y=nz\Rightarrow\cos u\,(xu_x+yu_y)=\frac25\sin u\Rightarrow xu_x+yu_y=\frac25\tan u=g(u)\).

Deduction: \(x^2u_{xx}+2xyu_{xy}+y^2u_{yy}=g(u)[g'(u)-1]\), with \(g'(u)=\frac25\sec^2u\).

\(=\frac25\tan u\left(\frac25\sec^2u-1\right)=\frac2{25}\tan u\,(2\sec^2u-5)=\frac2{25}\tan u\,(2+2\tan^2u-5)\)

\(=\frac2{25}\tan u\,(2\tan^2u-3)\) (proved).

Doubtly AI · 25 Oct 2026 AI-verified · Oct 2026 Report

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For \(u=\sin^{-1}(x^2+y^2)^{1/5}\) deduce \(x^2\frac{\partial^2u}{\partial x^2}+2xy\frac{\partial^2u}{\partial x\partial y}+y^2\frac{\partial^2u}{\partial y^2}=\frac2{25}\tan u\,(2\tan^2u-3)\).

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