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UPSC 2022 Maths Optional Paper 1 Q8a-i — Step-by-Step Solution

10 marks · Section B

Clairaut's equation · ODEs · asked 5× in 13 yrs · Read the full method →

Question

Find the general and singular solutions of $(x^2-a^2)p^2-2xyp+y^2+a^2=0$, where $p=dy/dx$. Also give the geometric relation between the general and singular solutions.

Technique

Recognise the algebraic form $(x^2-a^2)p^2-2xyp+y^2+a^2=0$ as $(xp-y)^2=a^2(p^2-1)$, then $y=xp\mp a\sqrt{p^2-1}$ — Clairaut form. General solution by replacing $p$ with constant $c$; singular solution by $\partial/\partial p=0$ (envelope condition).

Solution

Setup. First-order ODE quadratic in $p$. Try to solve for $p$ (or for $y$ in terms of $p$).

Step 1 — Solve as quadratic in $p$

$(x^2-a^2)p^2-2xy p+(y^2+a^2)=0$.

$p=\dfrac{2xy\pm\sqrt{4x^2 y^2-4(x^2-a^2)(y^2+a^2)}}{2(x^2-a^2)}$.

Discriminant: $4x^2 y^2-4(x^2-a^2)(y^2+a^2)=4[x^2 y^2-(x^2 y^2+a^2 x^2-a^2 y^2-a^4)]$ $=4[-a^2 x^2+a^2 y^2+a^4]=4a^2(y^2-x^2+a^2)$.

So $p=\dfrac{xy\pm a\sqrt{y^2-x^2+a^2}}{x^2-a^2}$.

Step 2 — Rewrite the original as Clairaut form?

Try rearranging $(x^2-a^2)p^2-2xyp+y^2+a^2=0$: $x^2 p^2-2xy p+y^2=a^2 p^2-a^2$ $(xp-y)^2=a^2(p^2-1)$ $xp-y=\pm a\sqrt{p^2-1}$ $y=xp\mp a\sqrt{p^2-1}$.

This is the Clairaut form $y=xp+f(p)$ with $f(p)=\mp a\sqrt{p^2-1}$.

Step 3 — General solution (Clairaut)

For Clairaut $y=xp+f(p)$, the general solution is $y=cx+f(c)$ for arbitrary constant $c$ (replace $p$ by $c$):

Squaring (and combining both signs into one): $(y-cx)^2=a^2(c^2-1)$, $y^2-2cxy+c^2 x^2=a^2 c^2-a^2$, $y^2-2cxy+c^2(x^2-a^2)=-a^2$, $(x^2-a^2)c^2-2xy c+(y^2+a^2)=0$.

So the general solution is the original equation viewed as a quadratic in $c$ with solutions $c=c_1,c_2$ — i.e., the family is parametrised by $c$, and the equation $(x^2-a^2)c^2-2xyc+(y^2+a^2)=0$ relates $(x,y)$ on each family member with parameter $c$.

Cleaner statement of general solution: $y=cx\pm a\sqrt{c^2-1}$ (family of straight lines parametrised by $c$, with $|c|\ge 1$).

Step 4 — Singular solution

Singular solution: differentiate Clairaut w.r.t. $p$ and set $\partial/\partial p=0$: $0=x+f'(p)$, $0=x\mp\dfrac{ap}{\sqrt{p^2-1}}$, $x=\pm\dfrac{ap}{\sqrt{p^2-1}}$.

Square: $x^2=\dfrac{a^2 p^2}{p^2-1}$, so $x^2(p^2-1)=a^2 p^2$, $p^2(x^2-a^2)=x^2$, $p=\dfrac{x}{\sqrt{x^2-a^2}}$.

Substitute into $y=xp\mp a\sqrt{p^2-1}$: $y=\dfrac{x^2}{\sqrt{x^2-a^2}}\mp a\sqrt{\dfrac{x^2}{x^2-a^2}-1}=\dfrac{x^2}{\sqrt{x^2-a^2}}\mp a\sqrt{\dfrac{a^2}{x^2-a^2}}$ $=\dfrac{x^2}{\sqrt{x^2-a^2}}\mp\dfrac{a^2}{\sqrt{x^2-a^2}}=\dfrac{x^2\mp a^2}{\sqrt{x^2-a^2}}$.

Two cases:

• "$-$": $y=\dfrac{x^2-a^2}{\sqrt{x^2-a^2}}=\sqrt{x^2-a^2}$. So $y^2=x^2-a^2$, i.e. $x^2-y^2=a^2$ (hyperbola).
• "$+$": $y=\dfrac{x^2+a^2}{\sqrt{x^2-a^2}}$. Squaring: $y^2(x^2-a^2)=(x^2+a^2)^2$ — different curve, less clean.

The standard singular solution is the hyperbola $x^2-y^2=a^2$.

Answer