← 2023 Paper 1

UPSC 2023 Maths Optional Paper 1 Q7a-ii — Step-by-Step Solution

10 marks · Section B

Clairaut's equation · ODEs · asked 5× in 14 yrs · Read the full method →

Question

Reduce the equation $x^2 p^2+y(2x+y)p+y^2=0$ to Clairaut’s form by the substitution $y=u$ and $xy=v$. Hence solve the equation and show that $y+4x=0$ is a singular solution of the differential equation.

Technique

Substitution $y=u,\,xy=v$ converts the original equation to Clairaut’s form. The general solution is the one-parameter family with $P=c$; singular solution from differentiating the Clairaut form w.r.t. $P$ to get the envelope.

Solution

Here $p=dy/dx$.

Step 1 — Substitute $u=y,\,v=xy$

Compute $du/dv$. With $u,v$ as functions of $x$: $\dfrac{du}{dx}=p,\quad\dfrac{dv}{dx}=y+x\dfrac{dy}{dx}=u+\dfrac{v}{u}p.$ (Using $x=v/u$.)

So $\dfrac{du}{dv}=\dfrac{p}{u+(v/u)p}=\dfrac{up}{u^2+vp}$.

Let $P=du/dv$. Then $up=Pu^2+Pvp\Rightarrow p(u-Pv)=Pu^2\Rightarrow p=\dfrac{Pu^2}{u-Pv}.$

Step 2 — Substitute into the original equation

Original: $x^2p^2+y(2x+y)p+y^2=0$. With $x=v/u,\,y=u$:

Multiply by $(u-Pv)^2/u^2$:

Let me simplify more carefully. The first term: $\dfrac{v^2}{u^2}\cdot\dfrac{P^2 u^4}{(u-Pv)^2}=\dfrac{v^2 P^2 u^2}{(u-Pv)^2}.$

Second term: $u(2v/u+u)\cdot\dfrac{Pu^2}{u-Pv}=(2v+u^2)\cdot\dfrac{Pu^2}{u-Pv}$.

Third term: $u^2$.

Multiply all three terms by $(u-Pv)^2/u^2$:

First → $v^2 P^2$. Second → $(2v+u^2)P(u-Pv)$. Third → $(u-Pv)^2$.

Sum:

Expand: $(2v+u^2)P(u-Pv)=2vPu-2v^2P^2+u^3P-u^2P^2v.$ $(u-Pv)^2=u^2-2uPv+P^2v^2.$

Sum: $v^2P^2+2vPu-2v^2P^2+u^3P-u^2P^2v+u^2-2uPv+P^2v^2$ Combine $P^2$-terms: $v^2-2v^2+v^2=0$ (cancel) but $-u^2P^2v$ remains. Combine $P$-terms: $2vPu+u^3P-2uPv=u^3P$ (the $2vPu$ and $-2uPv$ cancel). Constant term: $u^2$.

So the equation reduces to:

Divide by $u^2$:

Solve for $u$:

This is Clairaut’s form $u=vP+f(P)$ with $f(P)=-1/P$.

Step 3 — General solution

Clairaut’s general solution: $P=c$ (constant), so $u=vc-1/c$, i.e., $y=cv-1/c=c(xy)-1/c$.

Rearrange: $y(1-cx)=-1/c\Rightarrow y=\dfrac{1}{c(cx-1)}$, equivalently

(Family of curves parameterised by $c$.)

Step 4 — Singular solution

Differentiate $u=vP-1/P$ w.r.t. $P$ (treating $u,v$ as functions of $P$ in the singular envelope):

Combined with $u=vP-1/P=-1/P-1/P=-2/P$, eliminate $P$: $P=-2/u$, so $v=-1/P^2=-u^2/4$.

In original variables: $xy=v=-u^2/4=-y^2/4\Rightarrow 4xy+y^2=0\Rightarrow y(4x+y)=0$.

The trivial solution $y=0$ is included; the non-trivial singular solution is $y+4x=0$, i.e., $y=-4x$.

Answer