JEE Main 2024 (30 Jan Shift 2)

LEVELJEE Advanced

Animated Solution for Mathematics - Differential Equations: Let $Y = Y (X)$ be a curve lying in the first quadrant such that the area enclosed by the line $Y - y = Y^{'} (x) (X - x)$ and the co-ordinate axes, where $(x, y)$ is any point on the curve, is always $\frac{- y ^{2}}{2 Y ^{'} ( x )} + 1, Y^{'} (x) \neq = 0$ . If $Y (1) = 1$ , then $12 Y (2)$ equals ________.

Q: How do we know to divide by $y'$ in the differential equation step?

The problem statement explicitly gives the condition $Y'(x) \ne 0$. In differential equations, when we have a term like $f(x)y' = 0$, and we know $y' \ne 0$, we can safely divide by $y'$ without losing any valid solutions. It is a standard technique to simplify the equation to a linear form.

Q: Is there a trick to the integration? It looks like a standard linear DE.

You are absolutely right! Once you reach the form $\frac{dy}{dx} - \frac{2}{x}y = -\frac{2}{x^2}$, it is a classic linear differential equation. The key is to correctly identify $P(x) = -\frac{2}{x}$ and $Q(x) = -\frac{2}{x^2}$. The integrating factor $IF = e^{\int P(x) dx} = e^{-2 \ln x} = x^{-2}$ makes the integration straightforward. Just remember to add the constant of integration $C$!

Enter Numerical Value:

Progressive Hints

Hint 1

Hint 1 (The Spark): Visualize the tangent line at any point

(x, y)

on the curve. This line intersects the coordinate axes at two points, forming a right-angled triangle. The area of this triangle is given by

\frac{1}{2} ∣ X_{in t} \cdot Y_{in t} ∣

. Express the intercepts

X_{in t}

and

Y_{in t}

in terms of

x, y,

and

y^{'}

Hint 2

Hint 2 (The Arsenal): The problem provides an algebraic expression for the area. Equate your geometric expression for the area to the given expression:

\frac{1}{2} ∣ (x - \frac{y}{y ^{'}}) (y - x y^{'}) ∣ = \frac{- y ^{2}}{2 y ^{'}} + 1

. Simplify this equation carefully, keeping in mind that

y^{'} \neq = 0

Hint 3

Hint 3 (The Execution): After simplifying the equation, you will arrive at a first-order linear differential equation. Rearrange it into the standard form

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

and solve it using the integrating factor method, where

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

Visualized Solution

Visualizing the Tangent and Triangle

Let the point on the curve be $(x, y)$ .
The equation of the tangent at $(x, y)$ is given as:
$Y - y = Y^{'} (x) (X - x)$

Finding the Intercepts

To find the X-intercept, set $Y = 0$ in the tangent equation:
$X = x - \frac{y}{y ^{'}}$
To find the Y-intercept, set $X = 0$ in the tangent equation:
$Y = y - x y^{'}$

Calculating the Area of the Triangle

Area of the triangle $A = \frac{1}{2} ∣ X_{in t} \cdot Y_{in t} ∣$
Substitute the intercepts:
$A = \frac{1}{2} ∣ (x - \frac{y}{y ^{'}}) (y - x y^{'}) ∣$

Simplifying the Area Expression

Take $y^{'}$ as common denominator in the first bracket:
$A = \frac{1}{2} ∣ \frac{x y ^{'} - y}{y ^{'}} (y - x y^{'}) ∣$
Simplify to form a perfect square:
$A = \frac{( y - x y ^{'} ) ^{2}}{- 2 y ^{'}}$

Equating to the Given Area

Given Area: $A = \frac{- y ^{2}}{2 y ^{'}} + 1$
Equating both expressions:
$\frac{( y - x y ^{'} ) ^{2}}{- 2 y ^{'}} = \frac{- y ^{2}}{2 y ^{'}} + 1$

Expanding the Equation

Multiply by $- 2 y^{'}$ :
$(y - x y^{'})^{2} = y^{2} - 2 y^{'}$
Expand the left side:
$y^{2} + x^{2} (y^{'})^{2} - 2 x y y^{'} = y^{2} - 2 y^{'}$

Forming the Differential Equation

Cancel $y^{2}$ and rearrange:
$x^{2} (y^{'})^{2} - 2 x y y^{'} + 2 y^{'} = 0$
Divide by $y^{'}$ (since $y^{'} \neq = 0$ ):
$x^{2} y^{'} - 2 x y + 2 = 0$

Standard Form of Linear DE

Rearrange to standard form:
$\frac{d y}{d x} - \frac{2}{x} y = - \frac{2}{x ^{2}}$
Compare with $\frac{d y}{d x} + P (x) y = Q (x)$
Here, $P (x) = - \frac{2}{x}$ and $Q (x) = - \frac{2}{x ^{2}}$

Calculating the Integrating Factor

Integrating Factor $I F = e^{\int P (x) d x} = e^{\int - \frac{2}{x} d x}$
$I F = e^{- 2 l n x} = e^{l n (x^{- 2})} = \frac{1}{x ^{2}}$

Solving the Integral

Solution: $y \cdot I F = \int Q (x) \cdot I F d x$
$y \cdot \frac{1}{x ^{2}} = \int - \frac{2}{x ^{2}} \cdot \frac{1}{x ^{2}} d x = \int - 2 x^{- 4} d x$
$\frac{y}{x ^{2}} = \frac{- 2 x ^{- 3}}{- 3} + C = \frac{2}{3 x ^{3}} + C$

Applying the Boundary Condition

Given $Y (1) = 1$ , substitute $x = 1, y = 1$ :
$\frac{1}{1 ^{2}} = \frac{2}{3 ( 1 ) ^{3}} + C$
$1 = \frac{2}{3} + C ⟹ C = \frac{1}{3}$

Final Equation of the Curve

Substitute $C = \frac{1}{3}$ back:
$\frac{y}{x ^{2}} = \frac{2}{3 x ^{3}} + \frac{1}{3}$
Multiply by $x^{2}$ to get $Y (x)$ :
$Y (x) = \frac{2}{3 x} + \frac{x ^{2}}{3}$

Calculating $12 Y (2)$

Substitute $x = 2$ :
$Y (2) = \frac{2}{3 ( 2 )} + \frac{2 ^{2}}{3} = \frac{1}{3} + \frac{4}{3} = \frac{5}{3}$
Calculate $12 Y (2)$ :
$12 Y (2) = 12 \times \frac{5}{3} = 20$
Final Answer: 20

00:00 / 00:00

The Sigma Insight: Linear Differential Equations

Solution Diagram

Analyzing the Setup

Imagine standing on a curve at an arbitrary point

(x, y)

. You draw a tangent line, which serves as the local linear approximation of the curve. The equation of this tangent is given by:

Y - y = y^{'} (X - x)

This line intersects the coordinate axes. To find the X-intercept, we set

Y = 0

, yielding

X = x - \frac{y}{y ^{'}}

. Similarly, setting

X = 0

provides the Y-intercept

Y = y - x y^{'}

These two intercepts form the legs of a right-angled triangle. The area

A

of this triangle is defined as:

A = \frac{1}{2} (x - \frac{y}{y ^{'}}) (y - x y^{'})

The Algebraic Bridge

To simplify the expression for the area, we take

y^{'}

as a common denominator in the first term:

A = \frac{1}{2} \frac{x y ^{'} - y}{y ^{'}} (y - x y^{'})

Notice the symmetry: the term

(x y^{'} - y)

is the negative of

(y - x y^{'})

. Multiplying these terms results in

- (y - x y^{'})^{2}

. Thus, the area simplifies to:

A = \frac{( y - x y ^{'} ) ^{2}}{- 2 y ^{'}}

The problem states that this area is equal to

\frac{- y ^{2}}{2 y ^{'}} + 1

. Equating these expressions provides our bridge:

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

The Calculus Journey

We clear the denominators by multiplying by

- 2 y^{'}

, resulting in:

(y - x y^{'})^{2} = y^{2} - 2 y^{'}

Expanding the left side, we obtain

y^{2} + x^{2} (y^{'})^{2} - 2 x y y^{'} = y^{2} - 2 y^{'}

. The

y^{2}

terms cancel out, leaving:

x^{2} (y^{'})^{2} - 2 x y y^{'} + 2 y^{'} = 0

Assuming $y' eq 0$, we divide by

y^{'}

to arrive at the first-order linear differential equation:

x^{2} y^{'} - 2 x y + 2 = 0

Rearranging into standard form, we get:

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

The integrating factor is

I F = e^{\int - \frac{2}{x} d x} = e^{- 2 l n x} = \frac{1}{x ^{2}}

. Multiplying the equation by

\frac{1}{x ^{2}}

yields:

\frac{d}{d x} (\frac{y}{x ^{2}}) = - \frac{2}{x ^{4}}

Integrating both sides, we find:

\frac{y}{x ^{2}} = \frac{2}{3 x ^{3}} + C

Final Calculation

We are given the condition

Y (1) = 1

. Plugging in

x = 1

and

y = 1

, we find

1 = \frac{2}{3} + C

, which implies

C = \frac{1}{3}

. The equation of our curve is:

Y (x) = \frac{2}{3 x} + \frac{x ^{2}}{3}

To find

12 Y (2)

, we substitute

x = 2

Y (2) = \frac{2}{3 ( 2 )} + \frac{4}{3} = \frac{1}{3} + \frac{4}{3} = \frac{5}{3}

Therefore, the final result is:

12 Y (2) = 12 \times \frac{5}{3} = 20

Common Traps & FAQs

Wait, why do we assume the area is

\frac{1}{2} ∣ X_{in t} \cdot Y_{in t} ∣

? Isn't the area formula different?

That is a great question! The tangent line and the coordinate axes form a right-angled triangle in the first quadrant. The base of this triangle is the distance from the origin to the X-intercept, and the height is the distance from the origin to the Y-intercept. Since it is a right triangle, the area is simply

\frac{1}{2} \times base \times height

. The absolute value is used because the intercepts could be negative depending on the slope, but area must always be positive.

How do we know to divide by

y^{'}

in the differential equation step?

The problem statement explicitly gives the condition

Y^{'} (x) \neq = 0

. In differential equations, when we have a term like

f (x) y^{'} = 0

, and we know

y^{'} \neq = 0

, we can safely divide by

y^{'}

without losing any valid solutions. It is a standard technique to simplify the equation to a linear form.

Is there a trick to the integration? It looks like a standard linear DE.

You are absolutely right! Once you reach the form

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

, it is a classic linear differential equation. The key is to correctly identify

P (x) = - \frac{2}{x}

and

Q (x) = - \frac{2}{x ^{2}}

. The integrating factor

I F = e^{\int P (x) d x} = e^{- 2 l n x} = x^{- 2}

makes the integration straightforward. Just remember to add the constant of integration

C

Enter Numerical Value:

Visualized Solution

Visualizing the Tangent and Triangle

Finding the Intercepts

Calculating the Area of the Triangle

Simplifying the Area Expression

Equating to the Given Area

Expanding the Equation

Forming the Differential Equation

Standard Form of Linear DE

Calculating the Integrating Factor

Solving the Integral

Applying the Boundary Condition

Final Equation of the Curve

Calculating $12 Y (2)$

The Sigma Insight: Linear Differential Equations

Analyzing the Setup

The Algebraic Bridge

The Calculus Journey

Final Calculation

Similar Questions

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__________

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 curve of the differential equation, $(y^{2} - x) \frac{d y}{d x} = 1$ , satisfying $y (0) = 1$ . This curve intersects the $x$ -axis at a point whose abscissa is:

Consider the differential equation, $y^{2} d x + (x - \frac{1}{y}) d y = 0$ . If value of y is 1 when x = 1, then the value of x for which y = 2, is :

Let $y = y (x)$ be the solution curve of the differential equation, $(y^{2} - x) \frac{d y}{d x} = 1$ , satisfying $y (0) = 1$ . This curve intersects the $x$ -axis at a point whose abscissa is :

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 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 $f : [0, \infty) \to R$ 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 of the differential equation $\frac{d y}{d x} = (y + 1) ((y + 1) e^{x^{2} /2} - x), 0 < x < 2.1$ , with $y (2) = 0$ . Then the value of $\frac{d y}{d x}$ at $x = 1$ 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

Visualizing the Tangent and Triangle

Finding the Intercepts

Calculating the Area of the Triangle

Simplifying the Area Expression

Equating to the Given Area

Expanding the Equation

Forming the Differential Equation

Standard Form of Linear DE

Calculating the Integrating Factor

Solving the Integral

Applying the Boundary Condition

Final Equation of the Curve

Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 12Y(2)

The Sigma Insight: Linear Differential Equations

Analyzing the Setup

The Algebraic Bridge

The Calculus Journey

Final Calculation

Similar Questions

.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 curve of the differential equation, (y2−x)dxdy​=1, satisfying y(0)=1. This curve intersects the x-axis at a point whose abscissa is:

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Consider the differential equation, y2dx+(x−y1​)dy=0. If value of y is 1 when x = 1, then the value of x for which y = 2, 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 curve of the differential equation, (y2−x)dxdy​=1, satisfying y(0)=1. This curve intersects the x-axis at a point whose abscissa 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 (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 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 dxdy​=(y+1)((y+1)ex2/2−x),0<x<2.1, with y(2)=0. Then the value of dxdy​ at x=1 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 :

Calculating $12 Y (2)$

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 curve of the differential equation, $(y^{2} - x) \frac{d y}{d x} = 1$ , satisfying $y (0) = 1$ . This curve intersects the $x$ -axis at a point whose abscissa is:

Consider the differential equation, $y^{2} d x + (x - \frac{1}{y}) d y = 0$ . If value of y is 1 when x = 1, then the value of x for which y = 2, is :

Let $y = y (x)$ be the solution curve of the differential equation, $(y^{2} - x) \frac{d y}{d x} = 1$ , satisfying $y (0) = 1$ . This curve intersects the $x$ -axis at a point whose abscissa is :

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 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 $\frac{d y}{d x} = (y + 1) ((y + 1) e^{x^{2} /2} - x), 0 < x < 2.1$ , with $y (2) = 0$ . Then the value of $\frac{d y}{d x}$ at $x = 1$ 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 :