JEE Main 2024 · Mathematics · Differential Equations · hard

The curve y=y(x) satisfies (x^2+y^2 dx=2xy dy for x>0 and passes through (2,1). The value of [y(4)]^2 is

Numerical answer

The curve

y=y(x)

satisfies

(x^2+y^2)\,dx=2xy\,dy

for

x>0

and passes through

(2,1)

. The value of

\left[y(4)\right]^2

Show answer and solution

Answer: 10

The equation is homogeneous. Put

y=vx

, so

\dfrac{dy}{dx}=v+x\dfrac{dv}{dx}

and

1+v^2=2v\left(v+x\dfrac{dv}{dx}\right)

, which gives

2vx\dfrac{dv}{dx}=1-v^2

. Separating,

\int\dfrac{2v}{1-v^2}dv=\int\dfrac{dx}{x}

, so

-\ln|1-v^2|=\ln x+c

and

x(1-v^2)=C

. Replacing

v=y/x

gives

x^2-y^2=Cx

. At

(2,1)

4-1=2C

, so

C=\dfrac32

. At

x=4

16-y^2=6

, so

y(4)^2=10

. Differentiating

x^2-y^2=\tfrac32x

gives

2x-2yy'=\tfrac32

, and substituting back verifies the original equation.

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