Let y = y(x) be the solution of the differential equation x , dy + 2(1 - x) x + x(2

Year	2026	2025	2024	Total
Questions	13	17	16	46

Year

2026

2025

2024

Total

Questions

Solution & Explanation

Related Formula

d(u \cdot v) = u dv + v du

Core Logic

Given the differential equation:

x dy + 2(1 - x) x + (2x - x²) dx = 0

Divide entirely by $x$ $\sec x$ (which is $1/ x$ $1/\cos x$):

dy + 2(1 - x) x + (2x - x²) x dx = 0

Rearranging it as an exact differential form:

(dy)/(dx) = 2(x - 1) x + (x² - 2x) x

Step 1: Integration using By-Parts

Now integrate both sides:

y(x) = \int 2(x - 1) x dx + \int (x² - 2x) x dx

Apply integration by parts on the second integral, where $u = x² - 2x$ $u = x^2 - 2x$ and $dv = x dx$ $dv = \cos x \, dx$:

\int (x² - 2x) x dx = (x² - 2x)( x) - \int (2x - 2) x dx

Substituting this back into $y(x)$ $y(x)$:

y(x) = \int 2(x - 1) x dx + [ (x² - 2x) x - \int 2(x - 1) x dx ] + λ

The integrals cancel out perfectly:

y(x) = (x² - 2x) x + λ

Step 2: Finding Constant and Final Value

Use the initial condition $y(0) = 2$ $y(0) = 2$:

y(0) = 0 + λ \Rightarrow λ = 2

Thus, the solution is:

y(x) = (x² - 2x) x + 2

To find $y(2)$ $y(2)$:

y(2) = (2² - 2(2)) 2 + 2 = (4 - 4) 2 + 2 = 2

Pattern Recognition

When an integrand is a mix of polynomial and trigonometric terms, grouping them to form an exact differential or observing that integration by parts on one term perfectly cancels the other term is a classic structural trick.

Chapter Mix

Class 12 Maths: Differential Equations Class 12 Maths: Integrals

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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

Let $y = y(x)$ $y = y(x)$ be the solution of the differential equation $x dy + 2(1 - x) x + x(2 - x) dx = 0$ such that $y(0) = 2$ $y(0) = 2$. Then $y(2)$ $y(2)$ is equal to:

Solution & Explanation

Related Formula

Core Logic

Step 1: Integration using By-Parts

Step 2: Finding Constant and Final Value

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_jan_morning

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Let y = y(x)$y = y(x)$ be the solution of the differential equation x dy + 2(1 - x) x + x(2 - x) dx = 0$\sec x \, dy + \{2(1 - x) \tan x + x(2 - x)\} \, dx = 0$ such that y(0) = 2$y(0) = 2$. Then y(2)$y(2)$ is equal to:

Solution & Explanation

Related Formula

Core Logic

Step 1: Integration using By-Parts

Step 2: Finding Constant and Final Value

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_jan_morning

Select Your Target Exam

Full Mock Test Hub

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Chapter Weightage & Trends

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Formula Recognition

Exam Questions

Select an Exam

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Let $y = y(x)$ $y = y(x)$ be the solution of the differential equation $x dy + 2(1 - x) x + x(2 - x) dx = 0$ such that $y(0) = 2$ $y(0) = 2$. Then $y(2)$ $y(2)$ is equal to: