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

Year	2026	2025	2024	Total
Questions	13	17	16	46

Year

2026

2025

2024

Total

Questions

Solution & Explanation

Related Formula

Equation of tangent at (x, y): Y - y = (dy)/(dx)(X - x)

Area of Right Triangle: (1)/(2) \times base \times height

Core Logic

The line $Y - y = y'(X - x)$ $Y - y = y'(X - x)$ intersects the axes to form a triangle. Let's find the intercepts: X-intercept (set $Y=0$ $Y=0$): $-y = y'(X - x) \Rightarrow X = x - (y)/(y')$ Y-intercept (set $X=0$ $X=0$): $Y - y = y'(-x) \Rightarrow Y = y - xy'$

The area of the triangle bounded by the coordinate axes is:

Area = (1)/(2) ( x - (y)/(y') ) (y - xy') = (-y²)/(2y') + 1

Linear Differential Equation diagram for Q21 - JEE Main 2024 Evening

Step 1: Setting up the Differential Equation

Expand the Area equation:

(1)/(2) [ xy - x²y' - (y²)/(y') + xy ] = (-y²)/(2y') + 1

Multiply by 2:

2xy - x²y' - (y²)/(y') = -(y²)/(y') + 2

Cancel $-(y²)/(y')$ $-\frac{y^2}{y'}$ from both sides: $2xy - x²y' = 2$ $2xy - x^2y' = 2$

Step 2: Solving the Linear Differential Equation

Rearrange into standard LDE form:

x² (dy)/(dx) - 2xy = -2

(dy)/(dx) - (2)/(x)y = -(2)/(x²)

Find the Integrating Factor (IF):

I.F. = e\int -(2)/(x) dx = e-2ln x = x⁻² = (1)/(x²)

Multiply the LDE by the IF:

y \cdot (1)/(x²) = \int ( -(2)/(x²) ) (1)/(x²) dx + C

y x⁻² = -2 \int x⁻⁴ dx + C

y x⁻² = -2 ( x⁻³-3 ) + C

(y)/(x²) = (2)/(3x³) + C

y = (2)/(3x) + Cx²

Step 3: Applying Initial Conditions

Given $Y(1) = 1$ $Y(1) = 1$ (i.e., when $x=1, y=1$ $x=1, y=1$):

1 = (2)/(3(1)) + C(1)² \Rightarrow 1 = (2)/(3) + C \Rightarrow C = (1)/(3)

Thus, the curve is:

y = (2)/(3x) + (x²)/(3)

We need to find $12Y(2)$ $12Y(2)$:

Y(2) = (2)/(3(2)) + (2²)/(3) = (1)/(3) + (4)/(3) = (5)/(3)

12Y(2) = 12 ( (5)/(3) ) = 4 \times 5 = 20

Pattern Recognition

Geometrical word problems forming an Area differential equation almost always reduce to a first-order Linear Differential Equation (LDE) or an Exact differential form once you cleanly substitute the axis intercepts.

Chapter Mix

Class 12 Maths: Differential Equations Class 12 Maths: Application of Derivatives

More Differential Equations Previous-Year Questions — Page 10

Q11 jee_main_2024_31_jan_morning Linear Differential Equations

Let

y = y(x)

be the solution of the differential equation

(dy)/(dx) = (( x) + y)/( x( x - x x)), x in (0, (π)/(2))

satisfying the condition

y((π)/(4)) = 2

. Then,

y((π)/(3))

A. $\sqrt(3)(2 + ₑ\sqrt(3))$ $$\sqrt{3}(2 + \log_e\sqrt{3})$$
B. $\sqrt(3)2(2 + ₑ 3)$ $$\frac{\sqrt{3}}{2}(2 + \log_e 3)$$
C. $\sqrt(3)(1 + 2 ₑ 3)$ $$\sqrt{3}(1 + 2\log_e 3)$$
D. $\sqrt(3)(2 + ₑ 3)$ $$\sqrt{3}(2 + \log_e 3)$$

Solution

Core Logic

(dy)/(dx) = (( x)/( x) + y)/( x ((1)/( x) - ( ² x)/( x))) = ( x + y x)/( x (1 - ² x))

(dy)/(dx) = ( x + y x)/( x ² x) = ² x + (2y)/( 2x)

(dy)/(dx) - 2 (2x)y = ² x

Step 1: Integrating Factor

This is an LDE of form $(dy)/(dx) + Py = Q$ $\frac{dy}{dx} + Py = Q$.

I.F. = e\int -2 (2x) dx

Let $2x = t 2dx = dt$ $2x = t \implies 2dx = dt$.

I.F. = e-\int t dt = e-ln| (t/2)| = e-ln| x| = (1)/(| x|)

Step 2: Solution of LDE

y(I.F.) = \int Q(I.F.) dx + C

y(1)/( x) = \int ² x (1)/( x) dx + C

Let $x = t ² x dx = dt$ $\tan x = t \implies \sec^2 x dx = dt$.

y(1)/( x) = \int (dt)/(t) + C = ln| x| + C

y = x(ln| x| + C)

Step 3: Boundary Value

Given $y(π/4) = 2$ $y(\pi/4) = 2$:

2 = 1(ln 1 + C) C = 2

Thus, $y = x (ln| x| + 2)$ $y = \tan x (\ln|\tan x| + 2)$. At $x = π/3$ $x = \pi/3$:

y(π/3) = \sqrt(3)(ln\sqrt(3) + 2)

Chapter Mix

Class 12 Maths: Differential Equations

Solution & Explanation

Related Formula

Core Logic

Step 1: Setting up the Differential Equation

Step 2: Solving the Linear Differential Equation

Step 3: Applying Initial Conditions

Pattern Recognition

Chapter Mix

Reference Study Guides

More Differential Equations Previous-Year Questions — Page 10

Solution

Core Logic

Step 1: Integrating Factor

Step 2: Solution of LDE

Step 3: Boundary Value

Chapter Mix

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Solution & Explanation

Related Formula

Core Logic

Step 1: Setting up the Differential Equation

Step 2: Solving the Linear Differential Equation

Step 3: Applying Initial Conditions

Pattern Recognition

Chapter Mix

Reference Study Guides

More Differential Equations Previous-Year Questions — Page 10

Solution

Core Logic

Step 1: Integrating Factor

Step 2: Solution of LDE

Step 3: Boundary Value

Chapter Mix

More Differential Equations Questions — jee_main_2024_30_january_evening

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