JEE Advanced 2012

LEVELJEE Main

Animated Solution for Mathematics - Differential Equations: If $y (x)$ satisfies the differential equation $y^{'} - y tan x = 2 x sec x$ and $y (0) = 0$ , then

Q: Do I really need to find $C$? Can't I just leave it as a general solution?

In this specific problem, yes, you absolutely need to find $C$! The problem provides an initial condition: $y(0) = 0$. This is a 'particular solution' problem. If you leave $C$ as an arbitrary constant, you haven't fully solved the problem. The initial condition acts like a key that locks the general solution into one specific curve that passes through the origin. Without it, you cannot evaluate the specific values requested in the options.

Select Answer:

* Multiple Correct

Visualized Solution

Identify the Differential Equation

The given equation is $y^{'} - y tan x = 2 x sec x$
This is a Linear Differential Equation of the form $\frac{d y}{d x} + P (x) y = Q (x)$

Extract $P (x)$ and $Q (x)$

Comparing with standard form:
$P (x) = - tan x$
$Q (x) = 2 x sec x$

Setup Integrating Factor

$I . F . = e^{\int P (x) d x}$
$I . F . = e^{\int - t a n x d x}$

Compute Integrating Factor

Using $\int - tan x d x = ln ∣ cos x ∣$
$I . F . = e^{l n ∣ c o s x ∣} = cos x$

General Solution Setup

The solution is $y \cdot (I . F .) = \int Q (x) \cdot (I . F .) d x + C$
Substituting values: $y cos x = \int (2 x sec x) cos x d x + C$

Simplify the Integral

Since $sec x \cdot cos x = 1$
The equation simplifies to: $y cos x = \int 2 x d x + C$

Execute Integration

Integrating the right side:
$y cos x = x^{2} + C$

Apply Initial Condition

Given $y (0) = 0$
Substitute $x = 0, y = 0$ : $0 \cdot cos (0) = 0^{2} + C ⟹ C = 0$

Specific Solution

The specific solution is $y cos x = x^{2}$
Which can be written as $y = x^{2} sec x$

Evaluate $y (\frac{π}{4})$

At $x = \frac{π}{4}$ : $y (\frac{π}{4}) = (\frac{π}{4})^{2} sec (\frac{π}{4})$
$y (\frac{π}{4}) = \frac{π ^{2}}{16} \cdot 2 = \frac{π ^{2}}{8 2}$
Option A is correct.

Find the Derivative $y^{'} (x)$

Differentiating $y = x^{2} sec x$ using the Product Rule:
$y^{'} = \frac{d}{d x} (x^{2}) \cdot sec x + x^{2} \cdot \frac{d}{d x} (sec x)$
$y^{'} = 2 x sec x + x^{2} sec x tan x$

Evaluate $y^{'} (\frac{π}{3})$

At $x = \frac{π}{3}$ :
$y^{'} (\frac{π}{3}) = 2 (\frac{π}{3}) sec (\frac{π}{3}) + (\frac{π}{3})^{2} sec (\frac{π}{3}) tan (\frac{π}{3})$

Final Calculation

$y^{'} (\frac{π}{3}) = \frac{2 π}{3} (2) + \frac{π ^{2}}{9} (2) (3)$
$y^{'} (\frac{π}{3}) = \frac{4 π}{3} + \frac{2 π ^{2}}{3 3}$
Option D is correct.

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The Sigma Insight: Linear Differential Equations

Solution Diagram

Analyzing the Setup

The given differential equation is

y^{'} - y tan x = 2 x sec x

. This is a classic Linear Differential Equation of the form:

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

Here, our

P (x) = - tan x

and

Q (x) = 2 x sec x

. Note the negative sign attached to

tan x

; failing to account for this sign is a common error that will lead to an incorrect Integrating Factor.

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

P (x) = - tan x

, we evaluate the integral:

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

The exponential and logarithmic functions cancel out, simplifying the I.F. to

cos x

. This reduction is the key to simplifying the entire differential equation.

The Master Equation

We multiply the original differential equation by the I.F. (

cos x

), which allows us to express the left side as the derivative of a product:

\frac{d}{d x} (y \cdot cos x) = (2 x sec x) \cdot cos x

Since

sec x = \frac{1}{c o s x}

, the right side simplifies significantly:

\frac{d}{d x} (y cos x) = 2 x

Now, we integrate both sides with respect to

x

y cos x = \int 2 x d x + C

y cos x = x^{2} + C

Final Calculation

We apply the initial condition

y (0) = 0

to determine the constant

C

0 \cdot cos (0) = 0^{2} + C ⟹ C = 0

Substituting

C = 0

back into our general solution, we obtain the particular solution:

y cos x = x^{2}

Rearranging for

y

, we arrive at the final result:

$y = x^{2} sec x$

Common Traps & FAQs

Why is this considered a linear differential equation? It looks so complicated with the

tan x

and

sec x

terms!

That is a fantastic observation! In the world of differential equations, 'linear' doesn't mean the equation looks simple; it means the dependent variable

y

and its derivative

y^{'}

appear only to the first power and are not multiplied together. Even though we have trigonometric functions like

tan x

and

sec x

, they are functions of

x

only. Since

y

and

y^{'}

are 'clean' (no

y^{2}

, no

sin y

, no

y \cdot y^{'}

), it fits the linear form

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

perfectly. You are doing great—keep looking for that structure!

Wait, why did the integral of

sec x \cdot cos x

become just

1

? Is that always true?

It is always true because of the fundamental definition of reciprocal trigonometric functions! Remember that

sec x = \frac{1}{c o s x}

. When you multiply

sec x

cos x

, you are essentially multiplying

\frac{1}{c o s x}

cos x

, which cancels out to

1

. In JEE Advanced problems, examiners often hide these beautiful simplifications inside complex-looking expressions. Always look for reciprocal identities—they are your best friends in simplifying integrals.

Do I really need to find

C

? Can't I just leave it as a general solution?

In this specific problem, yes, you absolutely need to find

C

! The problem provides an initial condition:

y (0) = 0

. This is a 'particular solution' problem. If you leave

C

as an arbitrary constant, you haven't fully solved the problem. The initial condition acts like a key that locks the general solution into one specific curve that passes through the origin. Without it, you cannot evaluate the specific values requested in the options.

Animated Solution for Mathematics - Differential Equations: If $y (x)$ satisfies the differential equation $y^{'} - y tan x = 2 x sec x$ and $y (0) = 0$ , then

Select Answer:

Visualized Solution

Identify the Differential Equation

Extract $P (x)$ and $Q (x)$

Setup Integrating Factor

Compute Integrating Factor

General Solution Setup

Simplify the Integral

Execute Integration

Apply Initial Condition

Specific Solution

Evaluate $y (\frac{π}{4})$

Find the Derivative $y^{'} (x)$

Evaluate $y^{'} (\frac{π}{3})$

Final Calculation

The Sigma Insight: Linear Differential Equations

Analyzing the Setup

The Magic of the Integrating Factor

The Master Equation

Final Calculation

Similar Questions

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 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, $\frac{d y}{d x} + y tan x = 2 x + x^{2} tan x$ , $x \in (- π /2, π /2)$ , such that $y (0) = 1$ . Then :

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 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 :

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:

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

Let $y = y (x)$ be the solution of the differential equation $cos x \frac{d y}{d x} + 2 y sin x = sin 2 x$ , $x \in (0, \frac{π}{2})$ . If $y (π /3) = 0$ , then $y (π /4)$ is equal to :

If $y = y (x)$ is the solution of the differential equation $\frac{d y}{d x} + 2 y = sin (2 x), y (0) = \frac{3}{4}$ , then $y (\frac{π}{8})$ is equal to:

Let $y = y (x)$ be the solution of the differential equation $(x^{2} + 4)^{2} d y + (2 x^{3} y + 8 x y - 2) d x = 0$ . If $y (0) = 0$ , then $y (2)$ is equal to

Animated Solution for Mathematics - Differential Equations: .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

Select Answer:

Visualized Solution

Identify the Differential Equation

Extract .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P(x) and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Q(x)

Setup Integrating Factor

Compute Integrating Factor

General Solution Setup

Simplify the Integral

Execute Integration

Apply Initial Condition

Specific Solution

Evaluate .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } y(4π​)

Find the Derivative .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } y′(x)

Evaluate .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } y′(3π​)

Final Calculation

The Sigma Insight: Linear Differential Equations

Analyzing the Setup

The Magic of the Integrating Factor

The Master Equation

Final Calculation

Similar Questions

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Animated Solution for Mathematics - Differential Equations: If $y (x)$ satisfies the differential equation $y^{'} - y tan x = 2 x sec x$ and $y (0) = 0$ , then

Extract $P (x)$ and $Q (x)$

Evaluate $y (\frac{π}{4})$

Find the Derivative $y^{'} (x)$

Evaluate $y^{'} (\frac{π}{3})$

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 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, $\frac{d y}{d x} + y tan x = 2 x + x^{2} tan x$ , $x \in (- π /2, π /2)$ , such that $y (0) = 1$ . Then :

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 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 :

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:

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

Let $y = y (x)$ be the solution of the differential equation $cos x \frac{d y}{d x} + 2 y sin x = sin 2 x$ , $x \in (0, \frac{π}{2})$ . If $y (π /3) = 0$ , then $y (π /4)$ is equal to :

If $y = y (x)$ is the solution of the differential equation $\frac{d y}{d x} + 2 y = sin (2 x), y (0) = \frac{3}{4}$ , then $y (\frac{π}{8})$ is equal to:

Let $y = y (x)$ be the solution of the differential equation $(x^{2} + 4)^{2} d y + (2 x^{3} y + 8 x y - 2) d x = 0$ . If $y (0) = 0$ , then $y (2)$ is equal to