JEE Advanced 2024 Paper 1 · Mathematics · Differential Equations · medium

A curve y=y(x) for x>0 passes through (1,3). For every point P(x,y) on it, the tangent at P meets the y-axis at Q, and the midpoint of PQ lies on the…

Numerical answer

A curve

y=y(x)

for

x>0

passes through

(1,3)

. For every point

P(x,y)

on it, the tangent at

P

meets the

y

-axis at

Q

, and the midpoint of

PQ

lies on the line

y=x

. The value of

y(3)

is ______.

Show answer and solution

Answer: 21

The tangent at

P(x,y)

Y-y=y'(X-x)

. Putting

X=0

gives

Q=(0,\,y-xy')

. The midpoint of

PQ

\left(\tfrac x2,\ \tfrac{2y-xy'}{2}\right)

. Lying on

y=x

requires

2y-xy'=x

, i.e.

xy'-2y=-x

, or

y'-\tfrac{2}{x}y=-1

. This is linear with integrating factor

x^{-2}

, so

\dfrac{d}{dx}\!\left(\dfrac{y}{x^2}\right)=-\dfrac1{x^2}

and

\dfrac{y}{x^2}=\dfrac1x+C

. Hence

y=x+Cx^2

. The point

(1,3)

gives

C=2

, so

y=x+2x^2

. Finally

y(3)=3+18=21

. Symbolic solution of the ODE with the initial condition gives the same function.

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