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UPSC 2020 Maths Optional Paper 1 Q8a-ii — Step-by-Step Solution

10 marks · Section B

First-order higher-degree ODEs · ODEs · asked 5× in 14 yrs · Read the full method →

Question

Find the general and singular solutions of the differential equation $9p^2(2-y)^2=4(3-y)$, where $p=\frac{dy}{dx}$.

Technique

Equation absent in $x$: solve for $p$, separate, integrate; general solution $(x-C)^2=y^2(3-y)$; singular solution from the $p$-/$C$-discriminant, retaining only the branch that satisfies the ODE.

Solution

Here $x$ does not appear explicitly; the equation is solvable for $p$ and separable.

Step 1 — Solve for $p$ and separate variables

Separate (take the $+$ branch; the $-$ branch gives the same implicit relation):

Step 2 — Integrate

Put $u=3-y$, so $2-y=u-1$ and $dy=-du$:

Thus $x+C'=u^{1/2}(3-u)$. Restoring $u=3-y$ gives $3-u=y$, so

Step 3 — General solution

Check (general). Differentiating $(x-C)^2=y^2(3-y)$ implicitly: $2(x-C)=\big(6y-3y^2\big)p$, and substituting $x-C=y\sqrt{3-y}$ back into $9p^2(2-y)^2-4(3-y)$ gives $0$ identically (CAS-confirmed).

Step 4 — Singular solution

The singular solution is the envelope, obtained from the $p$-discriminant (or, equivalently, the $C$-discriminant). Writing the equation as a quadratic in $p$,

there is no linear $p$ term, so eliminating $p$ requires

Test each constant ($p=0$) against the ODE $9p^2(2-y)^2=4(3-y)$:

• $y=3$: LHS $=0$, RHS $=4(3-3)=0$ — satisfied.
• $y=2$: LHS $=0$, RHS $=4(3-2)=4\ne0$ — rejected.

The $C$-discriminant of the general solution ($\partial_C[(x-C)^2-y^2(3-y)]=0\Rightarrow x=C$) gives $y^2(3-y)=0$, i.e. $y=3$ (the envelope) together with the spurious double factor $y=0$ (a node/tac-locus, not a solution — it fails the ODE since $4(3-0)=12\ne0$).

Only $y=3$ both satisfies the ODE and is tangent to every member of the family (each member meets $y=3$ at $x=C$ with slope $p=0$). Hence the singular solution is

Answer