JEE Main 2021 (26 August Shift 1)

LEVELJEE Main

Animated Solution for Mathematics - Differential Equations: 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 :

Q: Why can't we just separate the variables $x$ and $y$?

That is a very common instinct! However, notice the term $y dx$ and $(y+1) \tan^2 x dx$. The $y$ is trapped inside a sum with $x$-dependent terms. Separation of variables works when you can write the equation as $f(x) dx = g(y) dy$. Here, the variables are 'entangled' linearly, which is the hallmark of a Linear Differential Equation. We must use the integrating factor method to 'disentangle' them.

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Visualized Solution

Introduction to the Equation

Given differential equation: $(y + 1) tan^{2} x d x + tan x d y + y d x = 0$
Rearrange terms to group $d x$ and $d y$ :
$tan x d y + [(y + 1) tan^{2} x + y] d x = 0$

Simplifying the Coefficient of $d x$

Expand the bracket: $y tan^{2} x d x + tan^{2} x d x + tan x d y + y d x = 0$
Factor out $y$ from the $d x$ terms:
$tan x d y + [y (tan^{2} x + 1) + tan^{2} x] d x = 0$
Use the identity $1 + tan^{2} x = sec^{2} x$ :
$tan x d y + [y sec^{2} x + tan^{2} x] d x = 0$

Converting to Linear Form $\frac{d y}{d x} + P (x) y = Q (x)$

Divide by $d x$ and rearrange: $tan x \frac{d y}{d x} + y sec^{2} x = - tan^{2} x$
Divide by $tan x$ to normalize the coefficient of $\frac{d y}{d x}$ :
$\frac{d y}{d x} + y \frac{s e c ^{2} x}{t a n x} = - tan x$
This is a Linear Differential Equation of the form $\frac{d y}{d x} + P (x) y = Q (x)$

Identifying $P (x)$ and $Q (x)$

Identify $P (x)$ and $Q (x)$ :
$P (x) = \frac{s e c ^{2} x}{t a n x}$
$Q (x) = - tan x$
The Integrating Factor (IF) is given by $e^{\int P (x) d x}$

Calculating the Integrating Factor (IF)

Calculate the integral: $\int \frac{s e c ^{2} x}{t a n x} d x$
Let $u = tan x \Rightarrow d u = sec^{2} x d x$
$\int \frac{1}{u} d u = ln ∣ tan x ∣$
$I F = e^{l n (t a n x)} = tan x$

Writing the General Solution

The general solution is $y \cdot (I F) = \int Q (x) \cdot (I F) d x + C$
Substitute the values: $y tan x = \int (- tan x) tan x d x + C$
$y tan x = - \int tan^{2} x d x + C$

Evaluating the Integral $\int tan^{2} x d x$

Use identity $tan^{2} x = sec^{2} x - 1$ :
$y tan x = - \int (sec^{2} x - 1) d x + C$
$y tan x = \int (1 - sec^{2} x) d x + C$
$y tan x = x - tan x + C$

Expressing $y (x)$ Explicitly

Divide by $tan x$ : $y = \frac{x - t a n x + C}{t a n x}$
$y = \frac{x + C}{t a n x} - 1$

Finding the Constant $C$ using Limits

Given: $lim_{x \to 0^{+}} x y (x) = 1$
Substitute $y$ : $lim_{x \to 0^{+}} x (\frac{x + C}{t a n x} - 1) = 1$
$lim_{x \to 0^{+}} (\frac{x ^{2} + C x}{t a n x} - x) = 1$
Since $lim_{x \to 0} \frac{x}{t a n x} = 1$ :
$0 + C (1) - 0 = 1 \Rightarrow C = 1$

The Particular Solution for $y (x)$

Substitute $C = 1$ back into the equation for $y$ :
$y (x) = \frac{x + 1}{t a n x} - 1$

Calculating $y (\frac{π}{4})$

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

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The Sigma Insight: Linear Differential Equations

Analyzing the Setup

We are presented with the differential equation:

(y + 1) tan^{2} x d x + tan x d y + y d x = 0

Our first objective is to group the terms associated with

d x

and isolate the

d y

term to impose order on the expression:

tan x d y + [(y + 1) tan^{2} x + y] d x = 0

Expanding the bracket yields

y tan^{2} x + tan^{2} x + y

. By factoring out

y

, we obtain

y (tan^{2} x + 1) + tan^{2} x

. Applying the trigonometric identity

tan^{2} x + 1 = sec^{2} x

, the equation simplifies to:

tan x d y + [y sec^{2} x + tan^{2} x] d x = 0

The Transformation to Standard Form

To reach the standard form of a Linear Differential Equation,

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

, we divide by

d x

and rearrange:

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

Dividing the entire equation by

tan x

ensures the coefficient of

\frac{d y}{d x}

1

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

Here, we identify our components:

P (x) = \frac{s e c ^{2} x}{t a n x}

and

Q (x) = - tan x

The Integrating Factor

The Integrating Factor (

I F

) is defined as

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

. We calculate the exponent integral:

\int \frac{sec ^{2} x}{tan x} d x

Since the numerator is the derivative of the denominator, we use the substitution

u = tan x

, which yields

\int \frac{1}{u} d u = ln ∣ tan x ∣

. Consequently, the integrating factor is:

I F = e^{l n (t a n x)} = tan x

The General Solution

The general solution is given by

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

. Substituting our known values:

y tan x = \int (- tan x) tan x d x + C

y tan x = - \int tan^{2} x d x + C

Using the identity

tan^{2} x = sec^{2} x - 1

, we integrate:

y tan x = - \int (sec^{2} x - 1) d x + C

y tan x = - (tan x - x) + C = x - tan x + C

Dividing by

tan x

, we arrive at the general solution:

y = \frac{x + C}{tan x} - 1

The Final Boundary

We apply the condition

lim_{x \to 0^{+}} x y (x) = 1

x \to 0^{+} lim x (\frac{x + C}{tan x} - 1) = 1

x \to 0^{+} lim (\frac{x ^{2} + C x}{tan x} - x) = 1

Using the standard limit

lim_{x \to 0} \frac{x}{t a n x} = 1

, the expression simplifies to

C = 1

. The particular solution is therefore:

y (x) = \frac{x + 1}{tan x} - 1

Evaluating at

x = \frac{π}{4}

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

Common Traps & FAQs

Why can't we just separate the variables

x

and

y

That is a very common instinct! However, notice the term

y d x

and

(y + 1) tan^{2} x d x

. The

y

is trapped inside a sum with

x

-dependent terms. Separation of variables works when you can write the equation as

f (x) d x = g (y) d y

. Here, the variables are 'entangled' linearly, which is the hallmark of a Linear Differential Equation. We must use the integrating factor method to 'disentangle' them.

I got stuck on the limit

lim_{x \to 0^{+}} x y (x) = 1

. How do I handle that?

Think of this limit as a boundary condition. It is the key to unlocking the value of

C

. Substitute your general solution

y (x)

into the expression

x y (x)

. As

x \to 0

, use the standard limit

lim_{x \to 0} \frac{x}{t a n x} = 1

. This will make the expression collapse into a simple algebraic equation for

C

. It is a beautiful moment where calculus meets algebra!

Why did we use the identity

tan^{2} x = sec^{2} x - 1

during integration?

Great observation! When we integrated the right side, we were left with

\int tan^{2} x d x

. There is no standard elementary integral for

tan^{2} x

directly. By converting it to

sec^{2} x - 1

, we transform a difficult integral into two very simple ones:

\int sec^{2} x d x

(which is

tan x

) and

\int 1 d x

(which is

x

). Always look for trigonometric identities to simplify your integrand!

Animated Solution for Mathematics - Differential Equations: 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 :

Select Answer:

Visualized Solution

Introduction to the Equation

Simplifying the Coefficient of $d x$

Converting to Linear Form $\frac{d y}{d x} + P (x) y = Q (x)$

Identifying $P (x)$ and $Q (x)$

Calculating the Integrating Factor (IF)

Writing the General Solution

Evaluating the Integral $\int tan^{2} x d x$

Expressing $y (x)$ Explicitly

Finding the Constant $C$ using Limits

The Particular Solution for $y (x)$

Calculating $y (\frac{π}{4})$

The Sigma Insight: Linear Differential Equations

Analyzing the Setup

The Transformation to Standard Form

The Integrating Factor

The General Solution

The Final Boundary

Similar Questions

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 the solution curve $y = y (x)$ of the differential equation $(1 + e^{2 x}) (\frac{d y}{d x} + y) = 1$ pass through the point $(0, \frac{π}{2})$ . Then, $lim_{x \to \infty} e^{x} y (x)$ is equal to :

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 curve of the differential equation $x^{2} d y + (y - \frac{1}{x}) d x = 0; x > 0$ and $y (1) = 1$ , then $y (\frac{1}{2})$ is equal to :

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

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, $\frac{d y}{d x} + y tan x = 2 x + x^{2} tan x$ , $x \in (- π /2, π /2)$ , 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 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:

Let $y = y (x)$ be the solution curve of the differential equation $sin (2 x^{2}) lo g_{e} (tan x^{2}) d y + (4 x y - 42 x sin (x^{2} - \frac{π}{4})) d x = 0, 0 < x < \frac{π}{2}$ , which passes through the point $(\frac{π}{6}, 1)$ . Then $∣ y (\frac{π}{3}) ∣$ is equal to

Select Answer:

Visualized Solution

Introduction to the Equation

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

Converting to Linear Form .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } dxdy​+P(x)y=Q(x)

Identifying .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)

Calculating the Integrating Factor (IF)

Writing the General Solution

Evaluating the Integral .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ∫tan2xdx

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

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

The Particular Solution for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } y(x)

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

The Sigma Insight: Linear Differential Equations

Analyzing the Setup

The Transformation to Standard Form

The Integrating Factor

The General Solution

The Final Boundary

Similar Questions

.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 curve of the differential equation (1+x2)dy+(y−tan−1x)dx=0,y(0)=1. Then the value of y(1) is :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let the solution curve y=y(x) of the differential equation (1+e2x)(dxdy​+y)=1 pass through the point (0,2π​). Then, limx→∞​exy(x) is equal to :

.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​+3(tan2x)y+3y=sec2x, y(0)=31​+e3. Then y(4π​) 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 x2dy+(y−x1​)dx=0;x>0 and y(1)=1, then y(21​) is equal to :

.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 (1+x2)dxdy​+y=etan−1x,y(1)=0. Then y(0) is

.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 :

.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​+ytanx=2x+x2tanx, x∈(−π/2,π/2), such that y(0)=1. 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)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 for the solution curve y=f(x) of the differential equation dxdy​+(tanx)y=(1+2secx)22+secx​, x∈(2−π​,2π​),f(3π​)=103​​. then f(4π​) is equal to:

Simplifying the Coefficient of $d x$

Converting to Linear Form $\frac{d y}{d x} + P (x) y = Q (x)$

Identifying $P (x)$ and $Q (x)$

Evaluating the Integral $\int tan^{2} x d x$

Expressing $y (x)$ Explicitly

Finding the Constant $C$ using Limits

The Particular Solution for $y (x)$

Calculating $y (\frac{π}{4})$

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 the solution curve $y = y (x)$ of the differential equation $(1 + e^{2 x}) (\frac{d y}{d x} + y) = 1$ pass through the point $(0, \frac{π}{2})$ . Then, $lim_{x \to \infty} e^{x} y (x)$ is equal to :

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 curve of the differential equation $x^{2} d y + (y - \frac{1}{x}) d x = 0; x > 0$ and $y (1) = 1$ , then $y (\frac{1}{2})$ is equal to :

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

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, $\frac{d y}{d x} + y tan x = 2 x + x^{2} tan x$ , $x \in (- π /2, π /2)$ , 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 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: