JEE Advanced 2023 Paper 1 · Mathematics · Differential Equations · medium

Let y=y(x) be the solution of 2xy dy/dx=x^2+y^2 for x>0 with y(1)=2 and y>0. The slope of the tangent to the curve at the point where x=3 is

Single correct option

Let

y=y(x)

be the solution of

2xy\,\dfrac{dy}{dx}=x^{2}+y^{2}

for

x>0

with

y(1)=2

and

y>0

. The slope of the tangent to the curve at the point where

x=3

Show answer and solution

Answer: (D) $\dfrac{3\sqrt2}{4}$

Put

v=y^2

. Then

\frac{dv}{dx}=2y\frac{dy}{dx}

, so

x\frac{dv}{dx}=x^2+v

, i.e.

\frac{dv}{dx}-\frac vx=x

. The integrating factor is

\frac1x

, giving

\frac{d}{dx}\!\left(\frac vx\right)=1

, so

v=x^2+cx

.
Using

y(1)=2

4=1+c

, so

c=3

and

y^2=x^2+3x

.
At

x=3

y^2=18

y=3\sqrt2

.
Differentiating,

2y\,y'=2x+3

, so

y'(3)=\dfrac{9}{2\cdot3\sqrt2}=\dfrac{3}{2\sqrt2}=\dfrac{3\sqrt2}{4}

.
The value

\frac32

comes from dropping the

\sqrt2

, and

\frac{3\sqrt2}{2}

from a slip of a factor of

2

in the denominator. The substitution

v=y^2

is the key step because it converts the homogeneous equation into a linear first-order one, which is solved by the integrating factor

1/x

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