The slope of the tangent at any point (x, y) on a curve y = y(x) is x^2 + y^22xy, x 0. If y(2) = 0, then a value of y(8) is:

Given: - dydx = x^2 + y^22xy - y(2) = 0 Find: A value of y(8). Use the substitution y = vx, so dydx = v + x dvdx Substituting into the differential equation, v + x dvdx = x^2 + v^2x^22vx^2 which simplifies to v + x dvdx = v^2 + 12v Rearranging, x dvdx = v^2 + 1 - 2v^22v = 1 - v^22v Hence, 2v dv1 - v^2 = dxx Integrating both sides, 2v dv1 - v^2 = dxx So, - |1 - v^2| = |x| + C which gives |x(1 - v^2)| = C Now substitute back v = yx: ( x^2 - y^2x) = C Therefore, x^2 - y^2 = cx Using x = 2 and y = 0, 2^2 - 0^2 = c(2) c = 2 So the curve satisfies x^2 - y^2 = 2x At x = 8, 8^2 - y^2 = 2(8) 64 - y^2 = 16 y^2 = 48 y = 48 = 4 3 Therefore, a value of y(8) is 4 3. The correct option is A.

The slope of the tangent at any point (x, y) on a curve y = y(x)… — Solution | JEE Mathematics

MCQMediumJEE 2023Homogeneous Differential Equations

The slope of the tangent at any point ( $x$ , $y$ ) on a curve $y = y(x)$ is $\frac{x^2 + y^2}{2xy}$ , $x > 0$ . If $y(2) = 0$ , then a value of $y(8)$ is:

A
$4\sqrt{3}$
B
$-4\sqrt{2}$
C
$-2\sqrt{3}$
D
$2\sqrt{3}$

Answer

Correct answer:A

Step-by-step solution

Standard Method

Given:

$\frac{dy}{dx} = \frac{x^2 + y^2}{2xy}$
$y(2) = 0$

Find: A value of $y(8)$ .

Use the substitution $y = vx$ , so

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

Substituting into the differential equation,

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

which simplifies to

v + x\frac{dv}{dx} = \frac{v^2 + 1}{2v}

Separation and Evaluation

Rearranging,

x\frac{dv}{dx} = \frac{v^2 + 1 - 2v^2}{2v} = \frac{1 - v^2}{2v}

Hence,

\frac{2v\,dv}{1 - v^2} = \frac{dx}{x}

Integrating both sides,

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

So,

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

which gives

\ln|x(1 - v^2)| = C

Back-substitution and Final Value

Now substitute back $v = \frac{y}{x}$ :

\ln\left(\frac{x^2 - y^2}{x}\right) = C

Therefore,

x^2 - y^2 = cx

Using $x = 2$ and $y = 0$ ,

2^2 - 0^2 = c(2) \implies c = 2

So the curve satisfies

x^2 - y^2 = 2x

At $x = 8$ ,

8^2 - y^2 = 2(8)

64 - y^2 = 16 \implies y^2 = 48 \implies y = \sqrt{48} = 4\sqrt{3}

Therefore, a value of $y(8)$ is $4\sqrt{3}$ . The correct option is A.

Common mistakes

Taking the substitution as $x = vy$ instead of $y = vx$ . This changes the derivative relation and leads to a different separable equation. Use the homogeneous-form substitution consistent with the ratio $\frac{y}{x}$ .
Forgetting that $\frac{dy}{dx} = v + x\frac{dv}{dx}$ when $y = vx$ . Writing only $\frac{dy}{dx} = v$ ignores the product rule and makes the differential equation incorrect.
Making an error while integrating $\int \frac{2v\,dv}{1 - v^2}$ . The derivative of $1 - v^2$ is $-2v$ , so the integral becomes $-\ln|1 - v^2|$ , not $\ln|1 - v^2|$ .

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Answer

Step-by-step solution

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