JEE Main 2019 (10 April Shift 1)

LEVELJEE Main

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

Q: Why do we need the initial condition $y(0) = 0$ at the end?

Great question. When we integrate, we introduce a constant $C$. This $C$ represents an infinite family of curves—every possible solution to the differential equation. The initial condition $y(0) = 0$ acts like a 'GPS coordinate' for our specific curve. It allows us to lock onto the one unique path that passes through the origin, turning a general family of solutions into the one specific solution we need.

Select Answer:

Visualized Solution

Analyze the Differential Equation

Given equation: $\frac{d y}{d x} = (tan x - y) sec^{2} x$
Interval: $x \in (- \frac{π}{2}, \frac{π}{2})$
Initial Condition: $y (0) = 0$
Objective: Find $y (- \frac{π}{4})$

Rearranging to Standard Form

Distribute $sec^{2} x$ : $\frac{d y}{d x} = tan x sec^{2} x - y sec^{2} x$
Rearrange terms: $\frac{d y}{d x} + y sec^{2} x = tan x sec^{2} x$
This is a Linear Differential Equation of the form: $\frac{d y}{d x} + P (x) y = Q (x)$

Identify $P (x)$ and $Q (x)$

Comparing with standard form:
$P (x) = sec^{2} x$
$Q (x) = tan x sec^{2} x$

Calculate Integrating Factor (I.F.)

Formula: $I . F . = e^{\int P (x) d x}$
Substitute $P (x)$ : $I . F . = e^{\int s e c^{2} x d x}$
Integrating: $I . F . = e^{t a n x}$

Set up the General Solution

General Solution: $y \cdot (I . F .) = \int Q (x) \cdot (I . F .) d x + C$
Substitute values: $y \cdot e^{t a n x} = \int (tan x sec^{2} x) e^{t a n x} d x + C$

Integration by Substitution

Let $t = tan x$
Then $d t = sec^{2} x d x$
The integral becomes: $\int t e^{t} d t$

Evaluate the Integral

Using Integration by Parts: $\int uv d t = u \int v d t - \int (u^{'} \int v d t) d t$
$\int t e^{t} d t = t e^{t} - \int (1 \cdot e^{t}) d t = t e^{t} - e^{t} = e^{t} (t - 1)$
Back-substituting $t = tan x$ :
$\int (tan x sec^{2} x) e^{t a n x} d x = e^{t a n x} (tan x - 1)$

The General Solution Equation

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

Apply Initial Condition $y (0) = 0$

Substitute $x = 0$ and $y = 0$ :
$0 \cdot e^{t a n 0} = e^{t a n 0} (tan 0 - 1) + C$
$0 = 1 (0 - 1) + C$

Solve for Constant $C$

Calculation: $0 = - 1 + C \Rightarrow C = 1$

The Particular Solution

Particular Solution:
$y e^{t a n x} = e^{t a n x} (tan x - 1) + 1$
Divide by $e^{t a n x}$ :
$y (x) = tan x - 1 + e^{- t a n x}$

Evaluate at $x = - \frac{π}{4}$

Substitute $x = - \frac{π}{4}$ :
$y (- \frac{π}{4}) = tan (- \frac{π}{4}) - 1 + e^{- t a n (- \frac{π}{4})}$
Recall: $tan (- \frac{π}{4}) = - 1$

Final Calculation

Calculation:
$y (- \frac{π}{4}) = - 1 - 1 + e^{- (- 1)}$
$y (- \frac{π}{4}) = - 2 + e^{1}$
$y (- \frac{π}{4}) = e - 2$

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The Sigma Insight: Linear Differential Equations

Solution Diagram

Analyzing the Setup

We are given the differential equation:

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

with the initial constraint

y (0) = 0

. At first glance, this equation appears complex, but it follows a standard structure waiting to be revealed.

The Art of Rearrangement

First, we distribute the

sec^{2} x

term:

\frac{d y}{d x} = tan x sec^{2} x - y sec^{2} x

Next, we move the

y

term to the left side to align with the standard form of a Linear Differential Equation:

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

This matches the form

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

, where

P (x) = sec^{2} x

and

Q (x) = tan x sec^{2} x

The Magic of the Integrating Factor

To solve this, we calculate the Integrating Factor (

I . F .

I . F . = e^{\int P (x) d x} = e^{\int s e c^{2} x d x} = e^{t a n x}

Multiplying the entire differential equation by

e^{t a n x}

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

\frac{d}{d x} (y \cdot e^{t a n x}) = (tan x sec^{2} x) e^{t a n x}

The Integration Challenge

Integrating both sides with respect to

x

, we obtain:

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

To solve the integral on the right, we use the substitution

t = tan x

, which implies

d t = sec^{2} x d x

\int t e^{t} d t = t e^{t} - \int e^{t} d t = e^{t} (t - 1)

Substituting back

t = tan x

, the general solution is:

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

Finding the Specific Path

We apply the initial condition

y (0) = 0

to determine the constant

C

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

Since

tan 0 = 0

and

e^{0} = 1

, we find

0 = 1 (0 - 1) + C

, which yields $C = 1$ . The particular solution is:

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

Final Calculation

Finally, we evaluate the function at

x = - \frac{π}{4}

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

Given

tan (- \frac{π}{4}) = - 1

, we substitute to get:

y (- \frac{π}{4}) = - 1 - 1 + e^{- (- 1)} = - 2 + e

The final answer is $e - 2$ .

Common Traps & FAQs

Why can't we just separate the variables directly? It looks like a simple equation.

That is a fantastic instinct to check for! However, notice the term

(tan x - y)

. Because

y

is subtracted inside the parenthesis and then multiplied by

sec^{2} x

, the variables

x

and

y

are 'entangled.' You cannot isolate all

y

terms on one side and all

x

terms on the other without the

y

being stuck with an

x

function. This entanglement is exactly why we must use the Linear Differential Equation method.

How do I know when to use Integration by Parts versus simple substitution?

Think of it as a hierarchy of tools. Always look for substitution first—it is the 'surgical' tool. In this problem, we used substitution (

t = tan x

) to simplify the integral into

\int t e^{t} d t

. Once we reached that form, we realized we had a product of an algebraic function (

t

) and an exponential function (

e^{t}

). That is the classic signature for Integration by Parts. Always simplify first, then apply the heavy machinery if needed!

Why do we need the initial condition

y (0) = 0

at the end?

Great question. When we integrate, we introduce a constant

C

. This

C

represents an infinite family of curves—every possible solution to the differential equation. The initial condition

y (0) = 0

acts like a 'GPS coordinate' for our specific curve. It allows us to lock onto the one unique path that passes through the origin, turning a general family of solutions into the one specific solution we need.

Animated Solution for Mathematics - Differential Equations: 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 :

Select Answer:

Visualized Solution

Analyze the Differential Equation

Rearranging to Standard Form

Identify $P (x)$ and $Q (x)$

Calculate Integrating Factor (I.F.)

Set up the General Solution

Integration by Substitution

Evaluate the Integral

The General Solution Equation

Apply Initial Condition $y (0) = 0$

Solve for Constant $C$

The Particular Solution

Evaluate at $x = - \frac{π}{4}$

Final Calculation

The Sigma Insight: Linear Differential Equations

Analyzing the Setup

The Art of Rearrangement

The Magic of the Integrating Factor

The Integration Challenge

Finding the Specific Path

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

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

Let $y = y (x)$ be the solution of the differential equation $(1 + x^{2}) \frac{d y}{d x} + y = e^{t a n^{- 1} x}, y (1) = 0$ . Then $y (0)$ 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 :

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 :

Let $y = y (x)$ be the solution of the differential equation $\frac{d y}{d x} + 2 y sec^{2} x = 2 sec^{2} x + 3 tan x \cdot sec^{2} x$ such that $y (0) = \frac{5}{4}$ . Then $12 (y (\frac{π}{4}) - e^{- 2})$ is equal to _______.

Let $y = y (x)$ be the solution curve of the differential equation $(1 + x^{2}) d y + (y - tan^{- 1} x) d x = 0, y (0) = 1$ . Then the value of $y (1)$ is :

Let $y = y (x)$ be the solution of the differential equation $x lo g_{e} x \frac{d y}{d x} + y = x^{2} lo g_{e} x, (x > 1)$ . If $y (2) = 2$ , then $y (e)$ is equal to

Select Answer:

Visualized Solution

Analyze the Differential Equation

Rearranging to Standard Form

Identify .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)

Calculate Integrating Factor (I.F.)

Set up the General Solution

Integration by Substitution

Evaluate the Integral

The General Solution Equation

Apply Initial Condition .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } y(0)=0

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

The Particular Solution

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

Final Calculation

The Sigma Insight: Linear Differential Equations

Analyzing the Setup

The Art of Rearrangement

The Magic of the Integrating Factor

The Integration Challenge

Finding the Specific Path

Final Calculation

Similar Questions

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Identify $P (x)$ and $Q (x)$

Apply Initial Condition $y (0) = 0$

Solve for Constant $C$

Evaluate at $x = - \frac{π}{4}$

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

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

Let $y = y (x)$ be the solution of the differential equation $(1 + x^{2}) \frac{d y}{d x} + y = e^{t a n^{- 1} x}, y (1) = 0$ . Then $y (0)$ 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 :

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 :

Let $y = y (x)$ be the solution of the differential equation $\frac{d y}{d x} + 2 y sec^{2} x = 2 sec^{2} x + 3 tan x \cdot sec^{2} x$ such that $y (0) = \frac{5}{4}$ . Then $12 (y (\frac{π}{4}) - e^{- 2})$ is equal to _______.

Let $y = y (x)$ be the solution curve of the differential equation $(1 + x^{2}) d y + (y - tan^{- 1} x) d x = 0, y (0) = 1$ . Then the value of $y (1)$ is :

Let $y = y (x)$ be the solution of the differential equation $x lo g_{e} x \frac{d y}{d x} + y = x^{2} lo g_{e} x, (x > 1)$ . If $y (2) = 2$ , then $y (e)$ is equal to