JEE Main 2021 (16 March Shift 1)

LEVELJEE Main

Animated Solution for Mathematics - Differential Equations: If $y = y (x)$ is the solution of the differential equation, $\frac{d y}{d x} + 2 y tan x = sin x, y (\frac{π}{3}) = 0$ , then the maximum value of the function $y (x)$ over $R$ is equal to :

Q: Why can't I just separate the variables like I do in simpler problems?

Separation of variables works only when you can isolate $x$ and $y$ completely on opposite sides of the equation. Here, the term $2y \tan x$ couples them inextricably. The Linear Differential Equation method is our surgical tool designed specifically for this kind of coupling, allowing us to 'linearize' the behavior.

Q: Why did we substitute $t = \cos x$? Is that always necessary?

It is not strictly necessary, but it is a master's technique. By mapping the trigonometric function to a simple algebraic parabola, we turn a complex calculus problem into a basic vertex-finding exercise. It reduces the cognitive load and significantly lowers the chance of making a silly arithmetic error.

Q: How do I know if the maximum is at the vertex or the boundaries of the domain?

That is a brilliant question! Since $t = \cos x$ ranges from $-1$ to $1$, we must always check if the vertex of our parabola falls within this range. Since our vertex was at $t = 1/4$, which is well within $[-1, 1]$, we are guaranteed that the vertex is the global maximum.

Select Answer:

Visualized Solution

Identify the Differential Equation

Given equation: $\frac{d y}{d x} + 2 y tan x = sin x$
This is a Linear Differential Equation of the form: $\frac{d y}{d x} + P y = Q$
Comparing the two, we get:
$P = 2 tan x$
$Q = sin x$

Calculate Integrating Factor (I.F.)

Integrating Factor (I.F.) $= e^{\int P d x}$
Substitute $P = 2 tan x$ :
I.F. $= e^{\int 2 t a n x d x}$
I.F. $= e^{2 l n ∣ s e c x ∣}$
Using property $n ln a = ln a^{n}$ :
I.F. $= e^{l n (s e c^{2} x)} = sec^{2} x$

General Solution Setup

The general solution is: $y \cdot (I.F.) = \int Q \cdot (I.F.) d x + C$
Substitute I.F. $= sec^{2} x$ and $Q = sin x$ :
$y \cdot sec^{2} x = \int sin x \cdot sec^{2} x d x + C$

Solve the Integral

Simplify the integrand:
$\int sin x sec^{2} x d x = \int \frac{s i n x}{c o s ^{2} x} d x = \int sec x tan x d x$
Integrating gives:
$y sec^{2} x = sec x + C$

Apply Initial Condition

Given: $y (\frac{π}{3}) = 0$
Substitute $x = \frac{π}{3}$ and $y = 0$ into $y sec^{2} x = sec x + C$ :
$0 \cdot sec^{2} (\frac{π}{3}) = sec (\frac{π}{3}) + C$
$0 = 2 + C ⟹ C = - 2$

Express y as a Function of x

Substitute $C = - 2$ back:
$y sec^{2} x = sec x - 2$
Divide by $sec^{2} x$ to isolate $y$ :
$y = \frac{s e c x}{s e c ^{2} x} - \frac{2}{s e c ^{2} x}$
$y = cos x - 2 cos^{2} x$

Transform to Quadratic Form

Let $t = cos x$ , where $t \in [- 1, 1]$
The function becomes: $y (t) = t - 2 t^{2}$
This is a downward-opening parabola because the coefficient of $t^{2}$ is negative.

Find the Critical Point

Differentiate $y$ with respect to $t$ :
$\frac{d y}{d t} = 1 - 4 t$
Set $\frac{d y}{d t} = 0$ for maximum:
$1 - 4 t = 0 ⟹ t = \frac{1}{4}$
Check second derivative: $\frac{d ^{2} y}{d t ^{2}} = - 4 < 0$ (Confirmed Maximum)

Calculate Maximum Value

Substitute $t = \frac{1}{4}$ into $y = t - 2 t^{2}$ :
$y_{ma x} = \frac{1}{4} - 2 (\frac{1}{4})^{2}$
$y_{ma x} = \frac{1}{4} - 2 (\frac{1}{16})$
$y_{ma x} = \frac{1}{4} - \frac{1}{8}$
$y_{ma x} = \frac{2 - 1}{8} = \frac{1}{8}$

Conclusion and Key Takeaway

Key Takeaway:
1. Identify the LDE form and use the Integrating Factor method.
2. Use initial conditions to find the constant $C$ .
3. Convert trigonometric expressions into algebraic ones (like $t = cos x$ ) to simplify optimization.
Final Answer: The maximum value is $\frac{1}{8}$ .

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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} + 2 y tan x = sin x

This is a classic Linear Differential Equation of the form

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

. By comparing terms, we identify:

P = 2 tan x, Q = sin x

The Magic of the Integrating Factor

To solve this, we calculate the Integrating Factor (I.F.), defined as

e^{\int P d x}

I . F . = e^{\int 2 t a n x d x} = e^{2 l n ∣ s e c x ∣}

Using the logarithmic property

n ln a = ln a^{n}

, the expression simplifies elegantly:

I . F . = e^{l n (s e c^{2} x)} = sec^{2} x

The Integration and the Constant

The general solution is given by

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

. Substituting our values:

y sec^{2} x = \int sin x \cdot sec^{2} x d x + C

We simplify the integral by rewriting the integrand:

\int \frac{sin x}{cos ^{2} x} d x = \int tan x sec x d x = sec x

Thus, the general solution is:

y sec^{2} x = sec x + C

We apply the initial condition

y (\frac{π}{3}) = 0

to determine the constant

C

0 = sec (\frac{π}{3}) + C \Rightarrow 0 = 2 + C \Rightarrow C = - 2

The Final Optimization

Substituting

C = - 2

into our equation, we obtain the specific function:

y sec^{2} x = sec x - 2 \Rightarrow y = cos x - 2 cos^{2} x

To find the maximum value, let

t = cos x

. The function becomes a downward-opening parabola:

y (t) = t - 2 t^{2}

The maximum occurs at the vertex, where the derivative with respect to

t

is zero:

\frac{d y}{d t} = 1 - 4 t = 0 \Rightarrow t = \frac{1}{4}

Substituting

t = \frac{1}{4}

back into the expression for

y

y_{ma x} = \frac{1}{4} - 2 (\frac{1}{4})^{2} = \frac{1}{4} - \frac{2}{16} = \frac{1}{4} - \frac{1}{8} = \frac{1}{8}

The maximum value of the function is $\frac{1}{8}$ .

Common Traps & FAQs

Why can't I just separate the variables like I do in simpler problems?

Separation of variables works only when you can isolate

x

and

y

completely on opposite sides of the equation. Here, the term

2 y tan x

couples them inextricably. The Linear Differential Equation method is our surgical tool designed specifically for this kind of coupling, allowing us to 'linearize' the behavior.

Why did we substitute

t = cos x

? Is that always necessary?

It is not strictly necessary, but it is a master's technique. By mapping the trigonometric function to a simple algebraic parabola, we turn a complex calculus problem into a basic vertex-finding exercise. It reduces the cognitive load and significantly lowers the chance of making a silly arithmetic error.

How do I know if the maximum is at the vertex or the boundaries of the domain?

That is a brilliant question! Since

t = cos x

ranges from

- 1

1

, we must always check if the vertex of our parabola falls within this range. Since our vertex was at

t = 1/4

, which is well within

[- 1, 1]

, we are guaranteed that the vertex is the global maximum.

Animated Solution for Mathematics - Differential Equations: If $y = y (x)$ is the solution of the differential equation, $\frac{d y}{d x} + 2 y tan x = sin x, y (\frac{π}{3}) = 0$ , then the maximum value of the function $y (x)$ over $R$ is equal to :

Select Answer:

Visualized Solution

Identify the Differential Equation

Calculate Integrating Factor (I.F.)

General Solution Setup

Solve the Integral

Apply Initial Condition

Express y as a Function of x

Transform to Quadratic Form

Find the Critical Point

Calculate Maximum Value

Conclusion and Key Takeaway

The Sigma Insight: Linear Differential Equations

Analyzing the Setup

The Magic of the Integrating Factor

The Integration and the Constant

The Final Optimization

Similar Questions

If $\frac{d y}{d x} + 2 y tan x = sin x, 0 < x < \frac{π}{2}$ and $y (\frac{π}{3}) = 0$ , then the maximum value of $y (x)$ is

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 $y = y (x)$ 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

For $x \in R$ , let $y (x)$ be a solution of the differential equation $(x^{2} - 5) \frac{d y}{d x} - 2 x y = - 2 x (x^{2} - 5)^{2}$ such that $y (2) = 7$ . Then the maximum value of the function $y (x)$ is

If the solution curve $y = f (x)$ of the differential equation $(x^{2} - 4) y^{'} - 2 x y + 2 x (4 - x^{2})^{2} = 0, x > 2$ , passes through the point $(3, 15)$ , then the local maximum value of $f$ 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 :

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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 $y = y (x)$ satisfies the equation $\frac{d y}{d x} - ∣ A ∣ = 0$ , for all $x > 0$ , where $A = y 02 sin x - 1 0 11 \frac{1}{x}$ . If $y (π) = π + 2$ , then the value of $y (\frac{π}{2})$ is :

Select Answer:

Visualized Solution

Identify the Differential Equation

Calculate Integrating Factor (I.F.)

General Solution Setup

Solve the Integral

Apply Initial Condition

Express y as a Function of x

Transform to Quadratic Form

Find the Critical Point

Calculate Maximum Value

Conclusion and Key Takeaway

The Sigma Insight: Linear Differential Equations

Analyzing the Setup

The Magic of the Integrating Factor

The Integration and the Constant

The Final Optimization

Similar Questions

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