Logo
Jeeify
MCQMediumJEE 2025Linear Differential Equations

JEE Mathematics 2025 Question with Solution

Let y=y(x)y = y(x) be the solution of the differential equation (xy−5x21+x2)dx+(1+x2)dy=0,y(0)=0.\left( xy - 5x^2 \sqrt{1 + x^2} \right) dx + (1 + x^2) dy = 0, \quad y(0) = 0. Then y(3)y(\sqrt{3}) is equal to:

  • A

    532\frac{5\sqrt{3}}{2}

  • B

    143\sqrt{\frac{14}{3}}

  • C

    222\sqrt{2}

  • D

    152\sqrt{\frac{15}{2}}

Answer

Correct answer:A

Step-by-step solution

Standard Method

Given:

(xy−5x21+x2)dx+(1+x2)dy=0,\left( xy - 5x^2 \sqrt{1 + x^2} \right) dx + (1 + x^2) dy = 0,

with y(0)=0y(0) = 0.

Find: y(3)y(\sqrt{3}).

Rearrange the differential equation in linear form:

(1+x2)dydx+xy−5x21+x2=0(1+x^2)\frac{dy}{dx} + xy - 5x^2\sqrt{1+x^2} = 0

so

dydx+x1+x2y=5x21+x2.\frac{dy}{dx} + \frac{x}{1+x^2}y = \frac{5x^2}{\sqrt{1+x^2}}.

This is a first-order linear differential equation with

P(x)=x1+x2,Q(x)=5x21+x2.P(x)=\frac{x}{1+x^2}, \qquad Q(x)=\frac{5x^2}{\sqrt{1+x^2}}.

Its integrating factor is

I.F.=e∫x1+x2 dx=e12ln⁡(1+x2)=1+x2.\text{I.F.} = e^{\int \frac{x}{1+x^2} \, dx} = e^{\frac{1}{2}\ln(1+x^2)} = \sqrt{1+x^2}.

Multiplying the equation by 1+x2\sqrt{1+x^2} gives

1+x2 dydx+x1+x2y=5x2.\sqrt{1+x^2}\,\frac{dy}{dx} + \frac{x}{\sqrt{1+x^2}}y = 5x^2.

The left-hand side is

ddx(y1+x2).\frac{d}{dx}\left(y\sqrt{1+x^2}\right).

Hence,

ddx(y1+x2)=5x2.\frac{d}{dx}\left(y\sqrt{1+x^2}\right)=5x^2.

Integrating,

y1+x2=∫5x2 dx=5x33+C. y\sqrt{1+x^2} = \int 5x^2 \, dx = \frac{5x^3}{3} + C.

Using y(0)=0y(0)=0, we get

0=0+C⇒C=0.0 = 0 + C \Rightarrow C=0.

Therefore,

y=5x331+x2.y = \frac{5x^3}{3\sqrt{1+x^2}}.

Now substitute x=3x=\sqrt{3}:

y(3)=5(3)331+3=5(33)3⋅2=532.y(\sqrt{3}) = \frac{5(\sqrt{3})^3}{3\sqrt{1+3}} = \frac{5(3\sqrt{3})}{3\cdot 2} = \frac{5\sqrt{3}}{2}.

Therefore, the correct option is A.

Using the derivative-product form

Starting from

dydx+x1+x2y=5x21+x2,\frac{dy}{dx} + \frac{x}{1+x^2}y = \frac{5x^2}{\sqrt{1+x^2}},

notice that multiplying by 1+x2\sqrt{1+x^2} is natural because

ddx(1+x2)=x1+x2.\frac{d}{dx}\left(\sqrt{1+x^2}\right)=\frac{x}{\sqrt{1+x^2}}.

So,

1+x2 dydx+yx1+x2=5x2.\sqrt{1+x^2}\,\frac{dy}{dx} + y\frac{x}{\sqrt{1+x^2}} = 5x^2.

By the product rule,

ddx(y1+x2)=5x2.\frac{d}{dx}\left(y\sqrt{1+x^2}\right)=5x^2.

Integrating both sides,

y1+x2=5x33+C.y\sqrt{1+x^2}=\frac{5x^3}{3}+C.

Using the initial condition y(0)=0y(0)=0, we get C=0C=0, and hence

y=5x331+x2.y=\frac{5x^3}{3\sqrt{1+x^2}}.

Finally,

y(3)=532.y(\sqrt{3})=\frac{5\sqrt{3}}{2}.

This matches option A.

The exactness-based working shown in one provided approach is inconsistent; the linear differential equation method above gives the correct result.

Common mistakes

  • Treating the equation as exact is incorrect because for M=xy−5x21+x2M=xy-5x^2\sqrt{1+x^2} and N=1+x2N=1+x^2, we have ∂M∂y=x\frac{\partial M}{\partial y}=x and ∂N∂x=2x\frac{\partial N}{\partial x}=2x, which are not equal. Instead, first rewrite it as a linear differential equation in yy.

  • Missing the correct integrating factor. For dydx+x1+x2y=Q(x)\frac{dy}{dx}+\frac{x}{1+x^2}y=Q(x), the integrating factor is e∫x1+x2 dx=1+x2e^{\int \frac{x}{1+x^2} \, dx}=\sqrt{1+x^2}, not xx. Using a wrong integrating factor leads to an invalid product derivative.

  • While evaluating y(3)y(\sqrt{3}), students may simplify 1+3\sqrt{1+3} incorrectly or forget that (3)3=33(\sqrt{3})^3=3\sqrt{3}. Compute the substitution carefully after obtaining the explicit formula for y(x)y(x).

Practice more Linear Differential Equations questions

Get unlimited AI-adaptive practice, mastery tracking, and an AI tutor that explains every step — free to start.

Start free trialMore on Linear Differential Equations

Related questions