← 2020 Paper 2

UPSC 2020 Maths Optional Paper 2 Q3c — Step-by-Step Solution

20 marks · Section A

Partial derivatives; equality of mixed partials (Schwarz) · Real Analysis · asked 2× in 13 yrs · Read the full method →

Question

If $u=\tan^{-1}\frac{x^3+y^3}{x-y}$, $x\ne y$, then show that $x^2 u_{xx}+2xy\,u_{xy}+y^2 u_{yy}=(1-4\sin^2 u)\sin 2u$.

Technique

Euler’s theorem on the homogeneous function $z=\tan u$ (degree $2$); first-order relation $x u_x+y u_y=\sin2u$; differentiate again to get the second-order identity $x^2u_{xx}+2xyu_{xy}+y^2u_{yy}=g(g'-1)$ with $g=\sin2u$.

Solution

The function $u$ itself is not homogeneous, but $\tan u$ is. The trick is to apply Euler’s homogeneous-function theorem to $z:=\tan u$ and translate back.

Step 1 — $z=\tan u$ is homogeneous of degree $2$

Replacing $(x,y)\to(tx,ty)$: numerator scales by $t^3$, denominator by $t^1$, so $z(tx,ty)=t^{2}z(x,y)$. Hence $z$ is homogeneous of degree $n=2$.

Step 2 — Euler’s theorem (first order)

For a homogeneous function $z$ of degree $n$:

Since $z=\tan u$, $z_x=\sec^2u\,\cdot u_x$, $z_y=\sec^2u\,\cdot u_y$. Substituting into (1):

Denote $g(u):=\sin 2u=x u_x+y u_y$.

Step 3 — Second-order Euler relation

Differentiate (2) partially. Apply $\dfrac{\partial}{\partial x}$ to $x u_x+y u_y=g(u)$:

Apply $\dfrac{\partial}{\partial y}$:

Multiply the first by $x$, the second by $y$, and add (using $u_{xy}=u_{yx}$):

Using (2), $x u_x+y u_y=g(u)$:

so

Step 4 — Substitute $g(u)=\sin 2u$

$g(u)=\sin2u$, $g'(u)=2\cos2u$. Therefore

Now reduce to the target form using double-angle identities. Note $\cos2u=1-2\sin^2u$, so

Hence

Combining with (3):

$\blacksquare$

Verification

Algebraic. $(1-4\sin^2u)\sin2u = \sin2u-4\sin^2u\sin2u = \sin2u-2\sin2u(2\sin^2u)=\sin2u-2\sin2u(1-\cos2u)=2\sin2u\cos2u-\sin2u$, identical to $g(g'-1)$ above ✓.

Symbolic (SymPy). Computing $x^2u_{xx}+2xyu_{xy}+y^2u_{yy}$ directly from $u=\arctan\frac{x^3+y^3}{x-y}$ and subtracting $(1-4\sin^2u)\sin2u$ simplifies identically to $0$; numeric spot-checks at $(x,y)=(2,1)$ and $(3,0.5)$ give residual $\sim10^{-16}$ ✓.