JEE Main 2024 (08 April Shift 1)

LEVELJEE Main

Animated Solution for Mathematics - Differential Equations: Let $y = y (x)$ be the solution of the differential equation $(1 + y^{2}) e^{t a n x} d x + cos^{2} x (1 + e^{2 t a n x}) d y = 0, y (0) = 1$ . Then $y (\frac{π}{4})$ is equal to

Select Answer:

Visualized Solution

The Differential Equation

Given: $(1 + y^{2}) e^{t a n x} d x + cos^{2} x (1 + e^{2 t a n x}) d y = 0$
Initial Condition: $y (0) = 1$
Goal: Find $y (\frac{π}{4})$

Variable Separable Form

The equation can be rearranged to separate $x$ and $y$ terms.
Divide the entire equation by $(1 + y^{2}) cos^{2} x (1 + e^{2 t a n x})$ .

Rearranging the Equation

$\frac{e ^{t a n x}}{c o s ^{2} x ( 1 + e ^{2 t a n x} )} d x + \frac{1}{1 + y ^{2}} d y = 0$

Simplifying the $x$ -term

Use the trigonometric identity: $\frac{1}{c o s ^{2} x} = sec^{2} x$
$\frac{s e c ^{2} x e ^{t a n x}}{1 + ( e ^{t a n x} ) ^{2}} d x + \frac{1}{1 + y ^{2}} d y = 0$

Setting up the Integrals

Integrate both sides of the equation:
$\int \frac{s e c ^{2} x e ^{t a n x}}{1 + ( e ^{t a n x} ) ^{2}} d x + \int \frac{1}{1 + y ^{2}} d y = C$

Substitution Method

Let $u = e^{t a n x}$
Differentiating both sides: $d u = e^{t a n x} \cdot sec^{2} x d x$
The first integral transforms to: $\int \frac{d u}{1 + u ^{2}}$

The General Solution

Standard integral: $\int \frac{1}{1 + u ^{2}} d u = tan^{- 1} u$
$\int \frac{1}{1 + y ^{2}} d y = tan^{- 1} y$
General Solution: $tan^{- 1} (e^{t a n x}) + tan^{- 1} y = C$

Applying Initial Condition

Given: $y (0) = 1$
Substitute $x = 0$ and $y = 1$ into the general solution:
$tan^{- 1} (e^{t a n 0}) + tan^{- 1} (1) = C$

Calculating Constant $C$

Since $tan 0 = 0$ , $e^{0} = 1$
$tan^{- 1} (1) + tan^{- 1} (1) = C$
$\frac{π}{4} + \frac{π}{4} = C ⟹ C = \frac{π}{2}$
Particular Solution: $tan^{- 1} (e^{t a n x}) + tan^{- 1} y = \frac{π}{2}$

Finding $y (\frac{π}{4})$

We need to find $y$ when $x = \frac{π}{4}$ .
Substitute $x = \frac{π}{4}$ into the particular solution:
$tan^{- 1} (e^{t a n \frac{π}{4}}) + tan^{- 1} y = \frac{π}{2}$

Solving for $y$

Since $tan \frac{π}{4} = 1$ , the equation becomes:
$tan^{- 1} (e^{1}) + tan^{- 1} y = \frac{π}{2}$
$tan^{- 1} y = \frac{π}{2} - tan^{- 1} e$

Final Answer

Using the identity: $tan^{- 1} A + cot^{- 1} A = \frac{π}{2}$
$\frac{π}{2} - tan^{- 1} e = cot^{- 1} e$
$tan^{- 1} y = cot^{- 1} e$
Since $cot^{- 1} e = tan^{- 1} (\frac{1}{e})$ , we get $y = \frac{1}{e}$

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The Sigma Insight: Variable Separable Method

Analyzing the Setup

The given differential equation is:

(1 + y^{2}) e^{t a n x} d x + cos^{2} x (1 + e^{2 t a n x}) d y = 0

with the initial condition

y (0) = 1

To solve this, we employ the strategy of variable separability. We divide the entire equation by

(1 + y^{2}) cos^{2} x (1 + e^{2 t a n x})

to isolate the variables:

\frac{e ^{t a n x}}{cos ^{2} x ( 1 + e ^{2 t a n x} )} d x + \frac{1}{1 + y ^{2}} d y = 0

The Master Equation

We simplify the

x

-term by noting that

\frac{1}{c o s ^{2} x} = sec^{2} x

and

e^{2 t a n x} = (e^{t a n x})^{2}

. The equation becomes:

\int \frac{sec ^{2} x e ^{t a n x}}{1 + ( e ^{t a n x} ) ^{2}} d x + \int \frac{1}{1 + y ^{2}} d y = C

To solve the integral involving

x

, we use the substitution

u = e^{t a n x}

. Consequently, the differential is

d u = e^{t a n x} sec^{2} x d x

The integral transforms into the standard form:

\int \frac{d u}{1 + u ^{2}} + \int \frac{d y}{1 + y ^{2}} = C

Integrating both sides yields:

tan^{- 1} (e^{t a n x}) + tan^{- 1} y = C

Applying Initial Conditions

We use the initial condition

y (0) = 1

to determine the constant

C

. Substituting

x = 0

and

y = 1

tan^{- 1} (e^{t a n 0}) + tan^{- 1} (1) = C

tan^{- 1} (1) + tan^{- 1} (1) = C

\frac{π}{4} + \frac{π}{4} = C \Rightarrow C = \frac{π}{2}

Thus, the particular solution is:

tan^{- 1} (e^{t a n x}) + tan^{- 1} y = \frac{π}{2}

Final Calculation

We are tasked to find the value of

y

when

x = \frac{π}{4}

. Substituting this into our particular solution:

tan^{- 1} (e^{t a n (π /4)}) + tan^{- 1} y = \frac{π}{2}

tan^{- 1} (e) + tan^{- 1} y = \frac{π}{2}

Using the trigonometric identity

tan^{- 1} A + cot^{- 1} A = \frac{π}{2}

, we identify that

tan^{- 1} y = cot^{- 1} e

. Since

cot^{- 1} e = tan^{- 1} (\frac{1}{e})

, we conclude:

$y = \frac{1}{e}$

Common Traps & FAQs

Wait, why do we divide by

(1 + y^{2}) cos^{2} x (1 + e^{2 t a n x})

? Can't we just rearrange terms?

Great question! In differential equations, 'rearranging' is just a casual way of saying 'dividing to separate variables.' We divide by that specific product because it is the only way to clear the

y

terms from the

d x

side and the

x

terms from the

d y

side. It is the surgical tool that isolates our variables.

I got stuck at

\int \frac{s e c ^{2} x e ^{t a n x}}{1 + ( e ^{t a n x} ) ^{2}} d x

. How do I know when to use substitution?

Pattern recognition is the heart of JEE Advanced. Whenever you see a function and its derivative multiplied together—like

e^{t a n x}

and

sec^{2} x

—that is a neon sign screaming 'Substitution!' Always look for the derivative of the inner function.

Why did we use

tan^{- 1} y = cot^{- 1} e

at the end? Couldn't we just leave it as

tan^{- 1} y = \frac{π}{2} - tan^{- 1} e

You absolutely could! However, in competitive exams, options are often simplified. Knowing the identity

tan^{- 1} A + cot^{- 1} A = \frac{π}{2}

is a powerful shortcut that transforms a messy expression into a clean, elegant result. It shows the examiner you understand the geometry of inverse functions.

Animated Solution for Mathematics - Differential Equations: Let $y = y (x)$ be the solution of the differential equation $(1 + y^{2}) e^{t a n x} d x + cos^{2} x (1 + e^{2 t a n x}) d y = 0, y (0) = 1$ . Then $y (\frac{π}{4})$ is equal to

Select Answer:

Visualized Solution

The Differential Equation

Variable Separable Form

Rearranging the Equation

Simplifying the $x$ -term

Setting up the Integrals

Substitution Method

The General Solution

Applying Initial Condition

Calculating Constant $C$

Finding $y (\frac{π}{4})$

Solving for $y$

Final Answer

The Sigma Insight: Variable Separable Method

Analyzing the Setup

The Master Equation

Applying Initial Conditions

Final Calculation

Similar Questions

If $y = y (x)$ is the solution of the differential equation $(1 + e^{2 x}) \frac{d y}{d x} + 2 (1 + y^{2}) e^{x} = 0$ and $y (0) = 0$ , then $6 [y^{'} (0) + (y (lo g_{e} 3))^{2}]$ is equal to:

If $y = y (x)$ is the solution of the differential equation, $e^{y} (\frac{d y}{d x} - 1) = e^{x}$ such that $y (0) = 0$ , then $y (1)$ is equal to

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The solution of the equation $\frac{d ^{2} y}{d x ^{2}} = e^{- 2 x}$

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Select Answer:

Visualized Solution

The Differential Equation

Variable Separable Form

Rearranging the Equation

Simplifying the .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x-term

Setting up the Integrals

Substitution Method

The General Solution

Applying Initial Condition

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

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

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

Final Answer

The Sigma Insight: Variable Separable Method

Analyzing the Setup

The Master Equation

Applying Initial Conditions

Final Calculation

Similar Questions

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Finding $y (\frac{π}{4})$

Solving for $y$

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