JEE Main 2025 (January)

LEVELJEE Main

Animated Solution for Mathematics - Differential Equations: If for the solution curve $y = f (x)$ of the differential equation $\frac{d y}{d x} + (tan x) y = \frac{2 + s e c x}{( 1 + 2 s e c x ) ^{2}},$ $x \in (\frac{- π}{2}, \frac{π}{2}), f (\frac{π}{3}) = \frac{3}{10} .$ then $f (\frac{π}{4})$ is equal to:

Q: Wait, why is the Integrating Factor just $\sec x$? Shouldn't it be more complicated?

It is a common misconception that the $IF$ must be complex! Remember, the definition is $e^{\int P(x) dx}$. Here, $P(x) = \tan x$. The integral of $\tan x$ is $\ln |\sec x|$. When you raise $e$ to the power of $\ln |\sec x|$, the exponential and logarithmic functions are inverses, so they cancel out perfectly. Trust the process—sometimes the most elegant solutions come from the simplest identities.

Select Answer:

Progressive Hints

Hint 1

The Spark: Look at the structure of the differential equation. It is a linear first-order differential equation of the form

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

. The challenge is not the form, but the complexity of

Q (x)

. Visualize the geometry of the curve; it is defined by this relationship, and your goal is to find the integrating factor to linearize it.

Hint 2

The Arsenal: Your primary tool is the Integrating Factor,

I F = e^{\int P (x) d x}

. In this case,

P (x) = tan x

. Recall that

\int tan x d x = ln ∣ sec x ∣

. Once you have the

I F

, multiply both sides of the equation by it. The right side will transform into an integral that looks daunting but simplifies significantly if you convert everything to

sin x

and

cos x

Hint 3

The Execution: After multiplying by

sec x

, you are left with the integral

\int \frac{2 c o s x + 1}{( c o s x + 2 ) ^{2}} d x

. Do not try to solve this with standard substitution immediately. Instead, observe the derivative of

\frac{s i n x}{c o s x + 2}

using the quotient rule. You will find that it perfectly matches your integrand, allowing you to solve the integral in one step.

Visualized Solution

Identifying the Linear Form

The given equation is $\frac{d y}{d x} + (tan x) y = \frac{2 + s e c x}{( 1 + 2 s e c x ) ^{2}}$
This is a Linear Differential Equation of the form $\frac{d y}{d x} + P (x) y = Q (x)$
Here, $P (x) = tan x$ and $Q (x) = \frac{2 + s e c x}{( 1 + 2 s e c x ) ^{2}}$

Calculating the Integrating Factor $(I F)$

Integrating Factor $I F = e^{\int P (x) d x}$
$I F = e^{\int t a n x d x} = e^{l n ∣ s e c x ∣}$
Therefore, $I F = sec x$

Setting up the General Solution

The general solution is given by: $y \cdot (I F) = \int Q (x) \cdot (I F) d x + C$
Substituting the values: $y sec x = \int \frac{s e c x ( 2 + s e c x )}{( 1 + 2 s e c x ) ^{2}} d x$

Trigonometric Simplification

Convert $sec x$ to $\frac{1}{c o s x}$ in the integral:
$\int \frac{\frac{1}{c o s x} ( 2 + \frac{1}{c o s x} )}{( 1 + \frac{2}{c o s x} ) ^{2}} d x = \int \frac{\frac{2 c o s x + 1}{c o s ^{2} x}}{( \frac{c o s x + 2}{c o s x} ) ^{2}} d x$
Simplifying the fraction: $\int \frac{2 c o s x + 1}{( c o s x + 2 ) ^{2}} d x$

The Quotient Rule Observation

Notice that $\frac{d}{d x} (\frac{s i n x}{c o s x + 2}) = \frac{c o s x ( c o s x + 2 ) - s i n x ( - s i n x )}{( c o s x + 2 ) ^{2}}$
Expanding the numerator: $\frac{c o s ^{2} x + 2 c o s x + s i n ^{2} x}{( c o s x + 2 ) ^{2}} = \frac{2 c o s x + 1}{( c o s x + 2 ) ^{2}}$
So, $\int \frac{2 c o s x + 1}{( c o s x + 2 ) ^{2}} d x = \frac{s i n x}{c o s x + 2} + C$

Writing the General Solution

$y sec x = \frac{s i n x}{c o s x + 2} + C$
Multiplying by $cos x$ : $y = f (x) = \frac{s i n x c o s x}{c o s x + 2} + C cos x$

Finding the Constant $C$

Given $f (\frac{π}{3}) = \frac{3}{10}$
Substitute $x = \frac{π}{3}$ into $f (x)$ :
$\frac{3}{10} = \frac{s i n ( \frac{π}{3} ) c o s ( \frac{π}{3} )}{c o s ( \frac{π}{3} ) + 2} + C cos (\frac{π}{3})$

Solving for $C$

$\frac{3}{10} = \frac{( \frac{3}{2} ) ( \frac{1}{2} )}{\frac{1}{2} + 2} + \frac{C}{2}$
$\frac{3}{10} = \frac{\frac{3}{4}}{\frac{5}{2}} + \frac{C}{2} = \frac{3}{10} + \frac{C}{2}$
This implies $\frac{C}{2} = 0 ⟹ C = 0$

The Final Function

Since $C = 0$ , the solution curve is: $f (x) = \frac{s i n x c o s x}{c o s x + 2}$

Evaluating at $x = \frac{π}{4}$

$f (\frac{π}{4}) = \frac{s i n ( \frac{π}{4} ) c o s ( \frac{π}{4} )}{c o s ( \frac{π}{4} ) + 2}$
$f (\frac{π}{4}) = \frac{\frac{1}{2} \cdot \frac{1}{2}}{\frac{1}{2} + 2} = \frac{\frac{1}{2}}{\frac{1 + 2 2}{2}}$
$f (\frac{π}{4}) = \frac{2}{2 ( 1 + 2 2 )} = \frac{1}{2 ( 1 + 2 2 )} = \frac{1}{4 + 2}$

Final Rationalization

To match the options, rationalize the denominator:
$f (\frac{π}{4}) = \frac{1}{4 + 2} \cdot \frac{4 - 2}{4 - 2}$
$f (\frac{π}{4}) = \frac{4 - 2}{16 - 2} = \frac{4 - 2}{14}$
The correct option is (3).

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The Sigma Insight: Linear Differential Equations

Solution Diagram

Analyzing the Setup

The given differential equation is:

\frac{d y}{d x} + (tan x) y = \frac{2 + sec x}{( 1 + 2 sec x ) ^{2}}

This is a classic Linear Differential Equation of the form

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

. By inspection, we identify the components:

P (x) = tan x

Q (x) = \frac{2 + sec x}{( 1 + 2 sec x ) ^{2}}

Identifying this structure is the first major victory, as it dictates the standard algorithmic path for the solution.

The Magic of the Integrating Factor

To solve this, we calculate the Integrating Factor (

I F

) using the formula

I F = e^{\int P (x) d x}

. Substituting our

P (x)

, we obtain:

I F = e^{\int t a n x d x} = e^{l n ∣ s e c x ∣}

Since the exponential and natural logarithm functions are inverses, they cancel out. This yields the elegant result:

I F = sec x

The Integral Challenge

The general solution is given by

y \cdot (I F) = \int Q (x) \cdot (I F) d x + C

. Substituting our values, we get:

y sec x = \int \frac{sec x ( 2 + sec x )}{( 1 + 2 sec x ) ^{2}} d x

To simplify, we convert the trigonometric functions to sines and cosines. Replacing

sec x

with

\frac{1}{c o s x}

, the expression becomes:

\int \frac{\frac{1}{c o s x} ( 2 + \frac{1}{c o s x} )}{( 1 + \frac{2}{c o s x} ) ^{2}} d x = \int \frac{2 cos x + 1}{( cos x + 2 ) ^{2}} d x

The "Aha!" Moment

Quotient Rule in Reverse

We observe the integrand

\frac{2 c o s x + 1}{( c o s x + 2 ) ^{2}}

. Testing the derivative of

\frac{s i n x}{c o s x + 2}

using the quotient rule:

\frac{d}{d x} (\frac{sin x}{cos x + 2}) = \frac{cos x ( cos x + 2 ) - sin x ( - sin x )}{( cos x + 2 ) ^{2}}

Expanding the numerator, we get

\frac{c o s ^{2} x + 2 c o s x + s i n ^{2} x}{( c o s x + 2 ) ^{2}}

. Since

cos^{2} x + sin^{2} x = 1

, this simplifies perfectly to:

\frac{2 cos x + 1}{( cos x + 2 ) ^{2}}

Thus, the integral is solved in one stroke:

y sec x = \frac{s i n x}{c o s x + 2} + C

Final Calculation

Multiplying by

cos x

, we isolate

y

f (x) = \frac{sin x cos x}{cos x + 2} + C cos x

Given

f (\frac{π}{3}) = \frac{3}{10}

, we substitute

x = \frac{π}{3}

to find

C = 0

. Consequently, the function simplifies to:

f (x) = \frac{sin x cos x}{cos x + 2}

Evaluating at

x = \frac{π}{4}

f (\frac{π}{4}) = \frac{sin ( \frac{π}{4} ) cos ( \frac{π}{4} )}{cos ( \frac{π}{4} ) + 2} = \frac{\frac{1}{2}}{\frac{1}{2} + 2} = \frac{1}{4 + 2}

Rationalizing the denominator, we arrive at the final answer:

\frac{4 - 2}{14}

Common Traps & FAQs

Wait, why is the Integrating Factor just

sec x

? Shouldn't it be more complicated?

It is a common misconception that the

I F

must be complex! Remember, the definition is

e^{\int P (x) d x}

. Here,

P (x) = tan x

. The integral of

tan x

ln ∣ sec x ∣

. When you raise

e

to the power of

ln ∣ sec x ∣

, the exponential and logarithmic functions are inverses, so they cancel out perfectly. Trust the process—sometimes the most elegant solutions come from the simplest identities.

How am I supposed to 'see' that the quotient rule is the key to the integral?

Pattern recognition is the heart of JEE Advanced physics and math. When you see a denominator like

(cos x + 2)^{2}

, it is a massive hint that you are looking at the result of a quotient rule differentiation. Always check if the numerator is the derivative of the numerator of the fraction inside the square. In this case, testing

\frac{d}{d x} (\frac{s i n x}{c o s x + 2})

is a standard, high-level strategy.

Why do we assume

C = 0

after finding the general solution?

We don't assume it; we calculate it! The problem provides the condition

f (\frac{π}{3}) = \frac{3}{10}

. By substituting

x = \frac{π}{3}

into your general solution, you create an algebraic equation where

C

is the only unknown. Solving that equation forces

C

to be

0

. It is not an assumption, but a consequence of the specific boundary condition provided.

Select Answer:

Visualized Solution

Identifying the Linear Form

Calculating the Integrating Factor $(I F)$

Setting up the General Solution

Trigonometric Simplification

The Quotient Rule Observation

Writing the General Solution

Finding the Constant $C$

Solving for $C$

The Final Function

Evaluating at $x = \frac{π}{4}$

Final Rationalization

The Sigma Insight: Linear Differential Equations

Analyzing the Setup

The Magic of the Integrating Factor

The Integral Challenge

The "Aha!" Moment

Final Calculation

Similar Questions

Let $y = y (x)$ be the solution of the differential equation $\frac{d y}{d x} + 3 (tan^{2} x) y + 3 y = sec^{2} x$ , $y (0) = \frac{1}{3} + e^{3}$ . Then $y (\frac{π}{4})$ is equal to

Let $y = y (x)$ be the solution of the differential equation $\frac{d y}{d x} = \frac{( t a n x ) + y}{s i n x ( s e c x - s i n x t an x )}, x \in (0, \frac{π}{2})$ satisfying the condition $y (\frac{π}{4}) = 2$ . Then, $y (\frac{π}{3})$ is

Let $y = y (x)$ be a solution curve of the differential equation $(y + 1) tan^{2} x d x + tan x d y + y d x = 0, x \in (0, \frac{π}{2})$ . If $lim_{x \to 0^{+}} x y (x) = 1$ , then the value of $y (\frac{π}{4})$ is :

If $x = f (y)$ is the solution of the differential equation $(1 + y^{2}) + (x - 2 e^{t a n^{- 1} y}) \frac{d y}{d x} = 0,$ $y \in (- \frac{π}{2}, \frac{π}{2})$ with $f (0) = 1$ , then $f (\frac{1}{3})$ is equal to:

If $y (x)$ satisfies the differential equation $y^{'} - y tan x = 2 x sec x$ and $y (0) = 0$ , then

If $y = y (x)$ is the solution of the differential equation $\frac{d y}{d x} + (tan x) y = sin x, 0 \leq x \leq \frac{π}{3}$ , with $y (0) = 0$ , then $y (\frac{π}{4})$ equal to :

Let the solution curve $y = f (x)$ of the differential equation $\frac{d y}{d x} + \frac{x y}{x ^{2} - 1} = \frac{x ^{4} + 2 x}{1 - x ^{2}}, x \in (- 1, 1)$ pass through the origin. Then $\int_{- \frac{3}{2}}^{\frac{3}{2}} f (x) d x$ is equal to

The function $y = f (x)$ is the solution of the differential equation $\frac{d y}{d x} + \frac{x y}{x ^{2} - 1} = \frac{x ^{4} + 2 x}{1 - x ^{2}}$ in $(- 1, 1)$ satisfying $f (0) = 0$ . Then $\int_{- 3 /2}^{3 /2} f (x) d x$ is

If $y = y (x)$ is the solution of the differential equation $\frac{d y}{d x} = (tan x - y) sec^{2} x, x \in (- π /2, π /2)$ , such that $y (0) = 0$ , then $y (- π /4)$ is equal to :

If $y = y (x)$ is the solution curve of the differential equation $\frac{d y}{d x} + y tan x = x sec x, 0 \leq x \leq \frac{π}{3}, y (0) = 1$ , then $y (\frac{π}{6})$ is equal to

Select Answer:

Visualized Solution

Identifying the Linear Form

Calculating the Integrating Factor .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } (IF)

Setting up the General Solution

Trigonometric Simplification

The Quotient Rule Observation

Writing the General Solution

Finding the Constant .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } C

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } C

The Final Function

Evaluating at .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x=4π​

Final Rationalization

The Sigma Insight: Linear Differential Equations

Analyzing the Setup

The Magic of the Integrating Factor

The Integral Challenge

The "Aha!" Moment

Final Calculation

Similar Questions

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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let y=y(x) be the solution of the differential equation dxdy​=sinx(secx−sinx tanx)(tanx)+y​,x∈(0,2π​) satisfying the condition y(4π​)=2. Then, y(3π​) is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let y=y(x) be a solution curve of the differential equation (y+1)tan2xdx+tanxdy+ydx=0,x∈(0,2π​). If limx→0+​xy(x)=1, then the value of y(4π​) is :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If x=f(y) is the solution of the differential equation (1+y2)+(x−2etan−1y)dxdy​=0, y∈(−2π​,2π​) with f(0)=1, then f(3​1​) is equal to:

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If y(x) satisfies the differential equation y′−ytanx=2xsecx and y(0)=0, then

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If y=y(x) is the solution of the differential equation dxdy​+(tanx)y=sinx,0≤x≤3π​, with y(0)=0, then y(4π​) equal to :

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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If y=y(x) is the solution of the differential equation dxdy​=(tanx−y)sec2x,x∈(−π/2,π/2), such that y(0)=0, then y(−π/4) is equal to :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If y=y(x) is the solution curve of the differential equation dxdy​+ytanx=xsecx,0≤x≤3π​,y(0)=1, then y(6π​) is equal to

Calculating the Integrating Factor $(I F)$

Finding the Constant $C$

Solving for $C$

Evaluating at $x = \frac{π}{4}$

Let $y = y (x)$ be the solution of the differential equation $\frac{d y}{d x} + 3 (tan^{2} x) y + 3 y = sec^{2} x$ , $y (0) = \frac{1}{3} + e^{3}$ . Then $y (\frac{π}{4})$ is equal to

Let $y = y (x)$ be the solution of the differential equation $\frac{d y}{d x} = \frac{( t a n x ) + y}{s i n x ( s e c x - s i n x t an x )}, x \in (0, \frac{π}{2})$ satisfying the condition $y (\frac{π}{4}) = 2$ . Then, $y (\frac{π}{3})$ is

Let $y = y (x)$ be a solution curve of the differential equation $(y + 1) tan^{2} x d x + tan x d y + y d x = 0, x \in (0, \frac{π}{2})$ . If $lim_{x \to 0^{+}} x y (x) = 1$ , then the value of $y (\frac{π}{4})$ is :

If $x = f (y)$ is the solution of the differential equation $(1 + y^{2}) + (x - 2 e^{t a n^{- 1} y}) \frac{d y}{d x} = 0,$ $y \in (- \frac{π}{2}, \frac{π}{2})$ with $f (0) = 1$ , then $f (\frac{1}{3})$ is equal to:

If $y (x)$ satisfies the differential equation $y^{'} - y tan x = 2 x sec x$ and $y (0) = 0$ , then

If $y = y (x)$ is the solution of the differential equation $\frac{d y}{d x} + (tan x) y = sin x, 0 \leq x \leq \frac{π}{3}$ , with $y (0) = 0$ , then $y (\frac{π}{4})$ equal to :

The function $y = f (x)$ is the solution of the differential equation $\frac{d y}{d x} + \frac{x y}{x ^{2} - 1} = \frac{x ^{4} + 2 x}{1 - x ^{2}}$ in $(- 1, 1)$ satisfying $f (0) = 0$ . Then $\int_{- 3 /2}^{3 /2} f (x) d x$ is

If $y = y (x)$ is the solution of the differential equation $\frac{d y}{d x} = (tan x - y) sec^{2} x, x \in (- π /2, π /2)$ , such that $y (0) = 0$ , then $y (- π /4)$ is equal to :

If $y = y (x)$ is the solution curve of the differential equation $\frac{d y}{d x} + y tan x = x sec x, 0 \leq x \leq \frac{π}{3}, y (0) = 1$ , then $y (\frac{π}{6})$ is equal to