UPSC 2026 Maths Optional Paper 2 Q5a — Step-by-Step Solution

10 marks · Section B

Quasilinear first-order PDEs (Lagrange's method) · PDEs · asked 10× in 14 yrs · Read the full method →

Question

Find the integral surface of the equations $\dfrac{dx}{xz - y} = \dfrac{dy}{yz - x} = \dfrac{dz}{1 - z^2}$ .

Technique

These are Lagrange’s auxiliary (characteristic) equations. The trigger is the antisymmetry between the first two denominators: adding them gives $(x+y)(z-1)$ and subtracting gives $(x-y)(z+1)$ — both factorise, and each factor pairs with one of the two factors of $1-z^2=(1-z)(1+z)$ . So the multipliers $(1,1,0)$ and $(1,-1,0)$ split the system into two separated first-order ODEs. Two independent integrals $u,v$ then give the general integral surface $F(u,v)=0$ .

Solution

Step 0 — What the question is asking.

The system

\frac{dx}{P}=\frac{dy}{Q}=\frac{dz}{R},\qquad P=xz-y,\quad Q=yz-x,\quad R=1-z^{2},

is the family of characteristic curves of the quasi-linear PDE

(xz-y)\,p+(yz-x)\,q=1-z^{2},\qquad p=z_x,\ q=z_y .

An integral surface is a surface generated by a one-parameter family of these characteristics. By Lagrange’s theorem, if $u(x,y,z)=c_1$ and $v(x,y,z)=c_2$ are two functionally independent first integrals of the system, then the general integral surface is

F(u,v)=0,\qquad F\ \text{an arbitrary } C^1 \text{ function with } (F_u,F_v)\neq(0,0).

So the whole task is to produce two independent integrals. No initial (Cauchy) curve is prescribed, so the definite article in “find the integral surface” is loose wording: what is determinable is the general integral $F(u,v)=0$ , a family parametrised by one arbitrary function, and that family is the complete answer. A candidate who invents a side condition in order to force a single surface is answering a different question.

Step 1 — First multiplier: $(1,\,1,\,0)$ .

By the rule of equal ratios (add numerators, add denominators with the chosen multipliers $\ell,m,n$ ; the common ratio equals $\dfrac{\ell\,dx+m\,dy+n\,dz}{\ell P+m Q+n R}$ ):

\frac{dx+dy}{(xz-y)+(yz-x)}=\frac{d(x+y)}{z(x+y)-(x+y)}=\frac{d(x+y)}{(x+y)(z-1)} .

Pair this with the third ratio, writing $1-z^{2}=(1-z)(1+z)$ :

\frac{d(x+y)}{(x+y)(z-1)}=\frac{dz}{(1-z)(1+z)} .

Multiply through by $(z-1)$ and note $\dfrac{z-1}{(1-z)(1+z)}=-\dfrac{1}{1+z}$ :

\frac{d(x+y)}{x+y}=-\frac{dz}{1+z}.

Integrating, $\ \ln|x+y|+\ln|1+z|=\text{const}$ , i.e.

\boxed{\,u\equiv (x+y)(1+z)=c_1\,}

Step 2 — Second multiplier: $(1,\,-1,\,0)$ .

\frac{dx-dy}{(xz-y)-(yz-x)}=\frac{d(x-y)}{z(x-y)+(x-y)}=\frac{d(x-y)}{(x-y)(z+1)} .

Pairing again with $\dfrac{dz}{(1-z)(1+z)}$ and multiplying by $(z+1)$ , with $\dfrac{z+1}{(1-z)(1+z)}=\dfrac{1}{1-z}$ :

\frac{d(x-y)}{x-y}=\frac{dz}{1-z}\quad\Longrightarrow\quad \ln|x-y|=-\ln|1-z|+\text{const},

\boxed{\,v\equiv (x-y)(1-z)=c_2\,}

Step 3 — Verify both are genuine integrals (this is a mark, and it costs two lines).

Along a characteristic, $\dfrac{dx}{dt}=P,\ \dfrac{dy}{dt}=Q,\ \dfrac{dz}{dt}=R$ , so a function $w$ is a first integral iff the characteristic derivative $P w_x+Qw_y+Rw_z$ vanishes identically.

For $u=(x+y)(1+z)$ :

P u_x+Qu_y+Ru_z=(1+z)\bigl[(xz-y)+(yz-x)\bigr]+(1-z^{2})(x+y)

=(1+z)(x+y)(z-1)+(1-z^{2})(x+y)=(x+y)(z^{2}-1)+(x+y)(1-z^{2})=0.\ \checkmark

For $v=(x-y)(1-z)$ :

P v_x+Qv_y+Rv_z=(1-z)\bigl[(xz-y)-(yz-x)\bigr]-(1-z^{2})(x-y)

=(1-z)(x-y)(z+1)-(1-z^{2})(x-y)=(x-y)(1-z^{2})-(x-y)(1-z^{2})=0.\ \checkmark

Step 4 — Independence of $u$ and $v$ .

\frac{\partial(u,v)}{\partial(x,y,z)}=\begin{pmatrix}1+z & 1+z & x+y\\[2pt] 1-z & -(1-z) & -(x-y)\end{pmatrix}.

The $2\times2$ minor from the first two columns is $-(1+z)(1-z)-(1+z)(1-z)=-2(1-z^{2})$ , which is non-zero wherever $z\neq\pm1$ . Hence $u,v$ are functionally independent on the open set $z\neq\pm1$ , and Lagrange’s theorem applies there.

Step 5 — The integral surface.

F\bigl((x+y)(1+z),\ (x-y)(1-z)\bigr)=0,

equivalently, solving for one integral in terms of the other,

(x+y)(1+z)=f\bigl((x-y)(1-z)\bigr),\qquad f\ \text{arbitrary}. \qquad\blacksquare

Step 6 — What the arbitrary function means, and the degenerate members.

Geometry. The two families $u=c_1$ and $v=c_2$ are each a one-parameter family of surfaces; their intersections $\{u=c_1\}\cap\{v=c_2\}$ are exactly the characteristic curves. Choosing $F$ is choosing a curve $F(c_1,c_2)=0$ in the $(c_1,c_2)$ -plane, i.e. selecting a one-parameter subfamily of characteristics; the integral surface is the union (the ruled-by-characteristics surface) they sweep out. Every integral surface arises this way, and through every point with $z\neq\pm1$ there passes exactly one characteristic — which is why an integral surface is pinned down only after a transversal initial curve is prescribed.

Degenerate members. The planes $z=1$ and $z=-1$ (where $R=0$ ) and the planes $x=y$ , $x=-y$ are themselves integral surfaces; they are not lost, since

v=0\iff (x-y)(1-z)=0\iff \{x=y\}\cup\{z=1\},\qquad u=0\iff\{x=-y\}\cup\{z=-1\},

so they occur as the members $F=v$ and $F=u$ of the family.

Consistency check on the pair. Multiplying the two integrals,

uv=(x^{2}-y^{2})(1-z^{2})=\text{const},

which is exactly what the multipliers $(x,-y,0)$ give directly: $\dfrac{x\,dx-y\,dy}{z(x^{2}-y^{2})}=\dfrac{dz}{1-z^{2}}\Rightarrow\dfrac{d(x^{2}-y^{2})}{x^{2}-y^{2}}=\dfrac{2z\,dz}{1-z^{2}}=-\dfrac{d(1-z^{2})}{1-z^{2}}$ . So that third, “obvious” integral is not independent of $u,v$ — it is their product. Worth one line: it shows the candidate knows two integrals must be checked for independence, not just counted.

Answer

\boxed{\;u=(x+y)(1+z)=c_1,\qquad v=(x-y)(1-z)=c_2,\qquad\text{integral surface: } F\bigl((x+y)(1+z),\,(x-y)(1-z)\bigr)=0\;}

equivalently $(x+y)(1+z)=f\bigl((x-y)(1-z)\bigr)$ with $f$ arbitrary.