UPSC 2021 Maths Optional Paper 1 Q7b — Step-by-Step Solution

15 marks · Section B

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

Question

Find all possible solutions of $y^2\log y=xy\dfrac{dy}{dx}+\!\left(\dfrac{dy}{dx}\right)^2$ .

Substitute $u=\log y$ to convert to Clairaut form $u=xu'+(u')^2$ ; general = lines $u=cx+c^2$ ; singular = envelope.

Setup. Let $p=y'$ . Equation: $y^2\log y=xyp+p^2$ , i.e., $p^2+xyp-y^2\log y=0$ .

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

Let $u=\log y$ , so $du/dx=y'/y=p/y$ .

Multiply original by $1/y^2$ : $\log y=x(p/y)+(p/y)^2$ , i.e., $u=xu'+(u')^2$ .

This is Clairaut’s equation: $u=xu'+(u')^2$ with $f(u')=(u')^2$ .

General solution: $u'=c$ (constant), so $u=cx+c^2$ .

In terms of $y$ : $\log y=cx+c^2$ , i.e., $y=e^{cx+c^2}$ .

Singular solution: Differentiate Clairaut w.r.t. $u'$ : $0=x+2u'$ , so $u'=-x/2$ .

Substitute back: $u=x(-x/2)+(-x/2)^2=-x^2/2+x^2/4=-x^2/4$ .

So singular: $\log y=-x^2/4$ , i.e., $y=e^{-x^2/4}$ .

\boxed{\;\text{General: }y=e^{cx+c^2}\text{ for arbitrary }c.\quad\text{Singular: }y=e^{-x^2/4}.\;}