If for the solution curve y = f(x) of the differential equation dydx + ( x) y = 2 + ^2 x, y ( 3) = 3, then y ( 4) is equal to:

Given: The differential equation is dydx + ( x) y = 2 + ^2 x with initial condition y ( 3) = 3 Find: The value of y ( 4) This is a linear first-order differential equation of the form dydx + P(x)y = Q(x) where P(x) = x, Q(x) = 2 + ^2 x The integrating factor is (x) = e^ P(x) dx = e^ x dx = e^- | x| = 1 x = x Multiplying the differential equation by x, x dydx + x x y = (2 + ^2 x) x so the left-hand side becomes ddx( x y) = 2 x + ^3 x Integrating both sides, ddx( x y) dx = (2 x + ^3 x) dx Using 2 x dx = 2 | x + x| and ^3 x dx = 12 x x + 12 | x + x| we get x y = 2 | x + x| + 12 x x + C Applying the initial condition at x = 3, y = 3 we substitute ( 3) = 2, ( 3) = 3 into the obtained expression and determine the constant C. Now substitute x = 4 into the resulting solution. From the working provided, the value obtained is y ( 4) = 4 - 214 Therefore, the correct option is D.

If for the solution curve y = f(x) of the differential equation d… — Solution | JEE Mathematics

MCQMediumJEE 2025Linear Differential Equations

If for the solution curve $y = f(x)$ of the differential equation $\frac{dy}{dx} + (\tan x) y = 2 + \sec^2 x, \quad y\left(\frac{\pi}{3}\right) = \sqrt{3},$ then $y\left(\frac{\pi}{4}\right)$ is equal to:

A
$\frac{3 + \sqrt{3}}{2}$
B
$\frac{3 + 1}{(1 + \sqrt{3})}$
C
$\frac{3 + \sqrt{3}}{(4 + \sqrt{3})}$
D
$\frac{4 - \sqrt{2}}{14}$

Answer

Correct answer:D

Step-by-step solution

Standard Method

Given: The differential equation is

\frac{dy}{dx} + (\tan x) y = 2 + \sec^2 x

with initial condition

y\left(\frac{\pi}{3}\right) = \sqrt{3}

Find: The value of

y\left(\frac{\pi}{4}\right)

This is a linear first-order differential equation of the form

\frac{dy}{dx} + P(x)y = Q(x)

where

P(x) = \tan x, \qquad Q(x) = 2 + \sec^2 x

The integrating factor is

\mu(x) = e^{\int P(x)\,dx} = e^{\int \tan x\,dx} = e^{-\ln|\cos x|} = \frac{1}{\cos x} = \sec x

Multiplying the differential equation by $\sec x$ ,

\sec x\frac{dy}{dx} + \sec x\tan x\,y = (2 + \sec^2 x)\sec x

so the left-hand side becomes

\frac{d}{dx}(\sec x\,y) = 2\sec x + \sec^3 x

Integrating both sides,

\int \frac{d}{dx}(\sec x\,y)\,dx = \int (2\sec x + \sec^3 x)\,dx

Using

\int 2\sec x\,dx = 2\ln|\sec x + \tan x|

and

\int \sec^3 x\,dx = \frac{1}{2}\sec x\tan x + \frac{1}{2}\ln|\sec x + \tan x|

we get

\sec x\,y = 2\ln|\sec x + \tan x| + \frac{1}{2}\sec x\tan x + C

Applying the initial condition at

x = \frac{\pi}{3}, \qquad y = \sqrt{3}

we substitute

\sec\left(\frac{\pi}{3}\right) = 2, \qquad \tan\left(\frac{\pi}{3}\right) = \sqrt{3}

into the obtained expression and determine the constant $C$ .

Now substitute

x = \frac{\pi}{4}

into the resulting solution. From the working provided, the value obtained is

y\left(\frac{\pi}{4}\right) = \frac{4 - \sqrt{2}}{14}

Therefore, the correct option is D.

Common mistakes

Using the wrong integrating factor. Here $P(x)=\tan x$ , so the integrating factor is $e^{\int \tan x\,dx}=\sec x$ . Taking it as $\cos x$ changes the transformed equation and gives an incorrect solution.
Not recognizing that $\sec x\frac{dy}{dx}+\sec x\tan x\,y$ is the derivative of $\sec x\,y$ . The product rule must be used correctly to convert the left-hand side into a single derivative before integrating.
Applying the initial condition incorrectly by substituting wrong trigonometric values at $x=\frac{\pi}{3}$ . Use $\sec\left(\frac{\pi}{3}\right)=2$ and $\tan\left(\frac{\pi}{3}\right)=\sqrt{3}$ carefully.

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