JEE Main 2024 (05 April Shift 2)

LEVELJEE Advanced

Animated Solution for Mathematics - Differential Equations: Let $y = y (x)$ be the solution of the differential equation $\frac{d y}{d x} + \frac{2 x}{( 1 + x ^{2} ) ^{2}} y = x e^{- \frac{1}{( 1 + x ^{2} )}}; y (0) = 0$ . Then the area enclosed by the curve $f (x) = y (x) e^{\frac{1}{( 1 + x ^{2} )}}$ and the line $y - x = 4$ is__________

Q: Is there a shortcut for the area calculation?

For a parabola and a line, you can use the formula $\frac{|a|}{6}(\beta - \alpha)^3$, where $\alpha$ and $\beta$ are the roots of the quadratic equation formed by equating the two curves. Here, $a = -1/2$, so the area is $\frac{|-1/2|}{6}(4 - (-2))^3 = \frac{1}{12}(6^3) = \frac{216}{12} = 18$. It is a great way to verify your answer!

Enter Numerical Value:

Progressive Hints

Hint 1

Hint 1: The Spark. Do not let the exponential term

e^{- \frac{1}{1 + x ^{2}}}

intimidate you. This is a classic Linear Differential Equation of the form

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

. Your first task is to identify

P (x)

and

Q (x)

correctly. Once you do, the structure will start to look much friendlier.

Hint 2

Hint 2: The Arsenal. To solve this, you need the Integrating Factor,

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

. When you integrate

P (x) = \frac{2 x}{( 1 + x ^{2} ) ^{2}}

, use the substitution

t = 1 + x^{2}

. This will simplify the integral significantly and reveal the Integrating Factor.

Hint 3

Hint 3: The Execution. After finding the general solution

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

, you will notice that the exponential terms on the right-hand side cancel out perfectly. This leaves you with a simple integral. Once you find

y (x)

and apply the initial condition

y (0) = 0

, you can easily define

f (x)

and proceed to calculate the area.

Visualized Solution

Identify the Linear Differential Equation

The given equation is a Linear Differential Equation (LDE) of the form: $\frac{d y}{d x} + P (x) y = Q (x)$
Here, $P (x) = \frac{2 x}{( 1 + x ^{2} ) ^{2}}$ and $Q (x) = x e^{\frac{1}{1 + x ^{2}}}$

Calculate the Integrating Factor

Integrating Factor $I F = e^{\int P (x) d x} = e^{\int \frac{2 x}{( 1 + x ^{2} ) ^{2}} d x}$
Let $1 + x^{2} = t \Rightarrow 2 x d x = d t$
$\int \frac{d t}{t ^{2}} = - \frac{1}{t} = - \frac{1}{1 + x ^{2}}$
So, $I F = e^{- \frac{1}{1 + x ^{2}}}$

Solve the Differential Equation

The general solution is $y \cdot I F = \int Q (x) \cdot I F d x$
Simplifying the right hand side, the exponential terms cancel out:
$y e^{- \frac{1}{1 + x ^{2}}} = \int x d x$
$y e^{- \frac{1}{1 + x ^{2}}} = \frac{x ^{2}}{2} + C$

Apply Initial Condition $y (0) = 0$

Given the initial condition: $y (0) = 0$
Substitute $x = 0$ and $y = 0$ into the general solution:
$0 \cdot e^{- 1} = \frac{0}{2} + C$
$C = 0$

Define the Function $f (x)$

The target function is given as $f (x) = y (x) e^{\frac{1}{1 + x ^{2}}}$
From our solution, we have: $y e^{- \frac{1}{1 + x ^{2}}} = \frac{x ^{2}}{2}$
Multiplying both sides by $e^{\frac{1}{1 + x ^{2}}}$ gives:
$f (x) = \frac{x ^{2}}{2}$

Find Intersection Points

The given line is $y = x + 4$
To find intersection points, equate the curve and the line: $\frac{x ^{2}}{2} = x + 4$
$x^{2} - 2 x - 8 = 0$
$(x - 4) (x + 2) = 0 \Rightarrow x = 4, x = - 2$

Set up the Area Integral

The area $A$ between the curves is $\int_{a}^{b} (y_{upper} - y_{lower}) d x$
From the graph, the line is above the parabola between $x = - 2$ and $x = 4$ .
$A = \int_{- 2}^{4} ((x + 4) - \frac{x ^{2}}{2}) d x$

Integrate the Expression

Now, integrate each term separately:
$\int x d x = \frac{x ^{2}}{2}$
$\int 4 d x = 4 x$
$\int \frac{x ^{2}}{2} d x = \frac{x ^{3}}{6}$
$A = [\frac{x ^{2}}{2} + 4 x - \frac{x ^{3}}{6}]_{- 2}^{4}$

Evaluate at Upper Limit

Substitute the upper limit $x = 4$ :
$A_{upper} = \frac{4 ^{2}}{2} + 4 (4) - \frac{4 ^{3}}{6}$
$A_{upper} = \frac{16}{2} + 16 - \frac{64}{6}$
$A_{upper} = 8 + 16 - \frac{32}{3} = 24 - \frac{32}{3} = \frac{40}{3}$

Evaluate at Lower Limit

Substitute the lower limit $x = - 2$ :
$A_{lower} = \frac{( - 2 ) ^{2}}{2} + 4 (- 2) - \frac{( - 2 ) ^{3}}{6}$
$A_{lower} = \frac{4}{2} - 8 - \frac{- 8}{6}$
$A_{lower} = 2 - 8 + \frac{4}{3} = - 6 + \frac{4}{3} = - \frac{14}{3}$

Final Calculation

Subtract the lower limit value from the upper limit value:
$A = A_{upper} - A_{lower}$
$A = \frac{40}{3} - (- \frac{14}{3})$
$A = \frac{40 + 14}{3} = \frac{54}{3} = 18$
Final Answer: 18

00:00 / 00:00

The Sigma Insight: Linear Differential Equations

Solution Diagram

Analyzing the Setup

The given differential equation is:

\frac{d y}{d x} + \frac{2 x}{( 1 + x ^{2} ) ^{2}} y = x e^{\frac{1}{1 + x ^{2}}}

This is a classic Linear Differential Equation (LDE) of the form

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

. By inspection, we identify:

P (x) = \frac{2 x}{( 1 + x ^{2} ) ^{2}} and Q (x) = x e^{\frac{1}{1 + x ^{2}}}

The Magic of the Integrating Factor

To solve this, we calculate the Integrating Factor (IF) using the formula

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

. We evaluate the integral:

\int P (x) d x = \int \frac{2 x}{( 1 + x ^{2} ) ^{2}} d x

Using the substitution

t = 1 + x^{2}

, where

d t = 2 x d x

, the integral simplifies to:

\int \frac{d t}{t ^{2}} = - \frac{1}{t} = - \frac{1}{1 + x ^{2}}

Thus, the integrating factor is:

I F = e^{- \frac{1}{1 + x ^{2}}}

The Elegant Cancellation

The general solution is given by

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

. Substituting our values, the right-hand side becomes:

\int (x e^{\frac{1}{1 + x ^{2}}}) \cdot (e^{- \frac{1}{1 + x ^{2}}}) d x = \int x d x

The exponential terms cancel out perfectly, leaving us with:

y e^{- \frac{1}{1 + x ^{2}}} = \frac{x ^{2}}{2} + C

Applying the initial condition

y (0) = 0

, we find

C = 0

. Therefore, the function is

y = \frac{x ^{2}}{2} e^{\frac{1}{1 + x ^{2}}}

The Final Geometry

The problem asks for the area enclosed by

f (x) = y (x) e^{- \frac{1}{1 + x ^{2}}}

and the line

y = x + 4

. Note that

f (x) = \frac{x ^{2}}{2}

To find the intersection points, we set the parabola equal to the line:

\frac{x ^{2}}{2} = x + 4 \Rightarrow x^{2} - 2 x - 8 = 0

Factoring the quadratic gives

(x - 4) (x + 2) = 0

, yielding intersection points at

x = - 2

and

x = 4

The area

A

is the integral of the upper curve (the line) minus the lower curve (the parabola):

A = \int_{- 2}^{4} (x + 4 - \frac{x ^{2}}{2}) d x

Evaluating the definite integral:

A = [\frac{x ^{2}}{2} + 4 x - \frac{x ^{3}}{6}]_{- 2}^{4}

A = (8 + 16 - \frac{64}{6}) - (2 - 8 + \frac{8}{6}) = \frac{40}{3} - (- \frac{14}{3}) = \frac{54}{3} = 18

The final area is 18 square units.

Common Traps & FAQs

Wait, why do the exponential terms cancel out when we multiply

Q (x)

by the Integrating Factor?

That is the beauty of this specific differential equation! The Integrating Factor is designed to be the inverse of the exponential part of

Q (x)

. When you multiply

Q (x) = x e^{\frac{1}{1 + x ^{2}}}

I F = e^{- \frac{1}{1 + x ^{2}}}

, the exponents add up to zero, and

e^{0} = 1

. It is a deliberate, elegant construction meant to simplify your life!

Why do we reject the negative root or handle the intersection points the way we do?

We don't reject any roots here! We found two intersection points:

x = - 2

and

x = 4

. These are the boundaries of our integration. The area is the integral of the upper curve minus the lower curve between these two points. Always sketch the graph to see which function is on top.

Is there a shortcut for the area calculation?

For a parabola and a line, you can use the formula

\frac{∣ a ∣}{6} (β - α)^{3}

, where

α

and

β

are the roots of the quadratic equation formed by equating the two curves. Here,

a = - 1/2

, so the area is

\frac{∣ - 1/2∣}{6} (4 - (- 2))^{3} = \frac{1}{12} (6^{3}) = \frac{216}{12} = 18

. It is a great way to verify your answer!

Enter Numerical Value:

Visualized Solution

Identify the Linear Differential Equation

Calculate the Integrating Factor

Solve the Differential Equation

Apply Initial Condition $y (0) = 0$

Define the Function $f (x)$

Find Intersection Points

Set up the Area Integral

Integrate the Expression

Evaluate at Upper Limit

Evaluate at Lower Limit

Final Calculation

The Sigma Insight: Linear Differential Equations

Analyzing the Setup

The Magic of the Integrating Factor

The Elegant Cancellation

The Final Geometry

Similar Questions

Let the solution curve $y = y (x)$ of the differential equation $(4 + x^{2}) d y - 2 x (x^{2} + 3 y + 4) d x = 0$ pass through the origin. Then $y (2)$ is equal to _______.

Let the solution curve $y = f (x)$ of the differential equation $\frac{d y}{d x} + \frac{x y}{x ^{2} - 1} = \frac{x ^{4} + 2 x}{1 - x ^{2}}, x \in (- 1, 1)$ pass through the origin. Then $\int_{- \frac{3}{2}}^{\frac{3}{2}} f (x) d x$ is equal to

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 $f : [0, \infty) \to R$ be differentiable function such that $f (x) = 1 - 2 x + \int_{0}^{x} e^{x - t} f (t) d t$ for all $x \in [0, \infty)$ . Then the area of the region bounded by $y = f (x)$ and the coordinate axes is

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 $(x^{2} + 1) y^{'} - 2 x y = (x^{4} + 2 x^{2} + 1) cos x, y (0) = 1$ . Then $\int_{- 3}^{3} y (x) d x$ is :

The function $y = f (x)$ is the solution of the differential equation $\frac{d y}{d x} + \frac{x y}{x ^{2} - 1} = \frac{x ^{4} + 2 x}{1 - x ^{2}}$ in $(- 1, 1)$ satisfying $f (0) = 0$ . Then $\int_{- 3 /2}^{3 /2} f (x) d x$ is

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 $y$ be the solution of the differential equation $(1 - x^{2}) d y = [x y + (x^{3} + 2) 1 - x^{2}] d x, - 1 < x < 1$ and $y (0) = 0$ . If $\int_{- 1/2}^{1/2} 1 - x^{2} y (x) d x = k$ , then $k^{- 1}$ is equal to :

Enter Numerical Value:

Visualized Solution

Identify the Linear Differential Equation

Calculate the Integrating Factor

Solve the Differential 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

Define the Function .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(x)

Find Intersection Points

Set up the Area Integral

Integrate the Expression

Evaluate at Upper Limit

Evaluate at Lower Limit

Final Calculation

The Sigma Insight: Linear Differential Equations

Analyzing the Setup

The Magic of the Integrating Factor

The Elegant Cancellation

The Final Geometry

Similar Questions

.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 (4+x2)dy−2x(x2+3y+4)dx=0 pass through the origin. Then y(2) 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​+2ysec2x=2sec2x+3tanx⋅sec2x such that y(0)=45​. Then 12(y(4π​)−e−2) 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 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 y=y(x) be the solution of the differential equation (x2+1)y′−2xy=(x4+2x2+1)cosx,y(0)=1. Then ∫−33​y(x)dx is :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The function y=f(x) is the solution of the differential equation dxdy​+x2−1xy​=1−x2​x4+2x​ in (−1,1) satisfying f(0)=0. Then ∫−3​/23​/2​f(x)dx 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 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 be the solution of the differential equation (1−x2)dy=[xy+(x3+2)1−x2​]dx,−1<x<1 and y(0)=0. If ∫−1/21/2​1−x2​y(x)dx=k, then k−1 is equal to :

Apply Initial Condition $y (0) = 0$

Define the Function $f (x)$

Let the solution curve $y = y (x)$ of the differential equation $(4 + x^{2}) d y - 2 x (x^{2} + 3 y + 4) d x = 0$ pass through the origin. Then $y (2)$ is equal to _______.

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 $(x^{2} + 1) y^{'} - 2 x y = (x^{4} + 2 x^{2} + 1) cos x, y (0) = 1$ . Then $\int_{- 3}^{3} y (x) d x$ is :

The function $y = f (x)$ is the solution of the differential equation $\frac{d y}{d x} + \frac{x y}{x ^{2} - 1} = \frac{x ^{4} + 2 x}{1 - x ^{2}}$ in $(- 1, 1)$ satisfying $f (0) = 0$ . Then $\int_{- 3 /2}^{3 /2} f (x) d x$ is

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 $y$ be the solution of the differential equation $(1 - x^{2}) d y = [x y + (x^{3} + 2) 1 - x^{2}] d x, - 1 < x < 1$ and $y (0) = 0$ . If $\int_{- 1/2}^{1/2} 1 - x^{2} y (x) d x = k$ , then $k^{- 1}$ is equal to :