← 2019 Paper 1

UPSC 2019 Maths Optional Paper 1 Q8a — Step-by-Step Solution

15 marks · Section B

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

Question

Obtain the singular solution of the differential equation

Also find the complete primitive of the given differential equation. Give the geometrical interpretations of the complete primitive and singular solution.

Technique

Treat as quadratic in $p$; singular solution $=$ $p$-discriminant $B^2-4AC=0$. For the complete primitive, the substitution $X=x^2,Y=y^2$ linearises it to $Y=X\phi(P)$ (Lagrange form), which integrates to $X(P+1)^2=c^2$; eliminate $P$ to get the circle family.

Solution

Write $p=\dfrac{dy}{dx}$ and $v=\dfrac yx$. The equation is a quadratic in $p$:

Step 1 — Singular solution from the $p$-discriminant

The singular solution (envelope) satisfies $B^2-4AC=0$:

Divide by $4v^2$ (the case $v=0$ gives $y=0$, a trivial line):

Since $1+\cot^2\alpha=\csc^2\alpha$, this reduces to $\csc^2\alpha=v^2\cot^2\alpha\csc^2\alpha$, i.e. $v^2\cot^2\alpha=1$, so $v=\pm\tan\alpha$. Restoring $v=y/x$:

Step 2 — Reduce to a solvable form (substitution $X=x^2,\ Y=y^2$)

Multiply the equation by $x^2$ and use $py=\tfrac12(y^2)'$ in the new variables. Put $X=x^2,\ Y=y^2$ and $P=\dfrac{dY}{dX}=\dfrac{2y\,dy}{2x\,dx}=\dfrac yx\,p=vp$, so $vp=P$ and $v^2p^2=P^2$, while $v^2=Y/X$. The $x^2\times$ equation

becomes (divide by $x^2$ and substitute):

Solve for $Y$:

Step 3 — Lagrange-type integration

This has the form $Y=X\,\phi(P)$ with $P=dY/dX$. Differentiate w.r.t. $X$:

Now

Separate:

The integral of the right side is (sympy below):

Hence $\ln X=-2\ln(P+1)+\text{const}$, i.e.

Step 4 — Eliminate the parameter $P$

From $X(P+1)^2=c^2$ and $X=x^2$: $(P+1)=c/x$, so $P=\dfrac cx-1$. Substitute into $Y=X\phi(P)$ (algebra carried out symbolically):

Rearrange, writing $b=c$:

Answer