The solution curve of the differential equation y(dx/dy) = x( x - y + 1), x > 0, y > 0 passing through the point (e, 1) is

Given DE: (dx/dy) = (x/y) (((x/y)) + 1) Let (x/y) = t x = ty. Differentiating w.r.t y: Integrate both sides. Let t = p (1/t) dt = dp. Given curve passes through (e, 1): Class 12 Maths: Differential Equations

The solution curve of the differential equation y(dx/dy) = x( x - y + 1), x > 0, y > 0 passing through the point (e, 1) is

Given DE: (dx/dy) = (x/y) (((x/y)) + 1) Let (x/y) = t x = ty. Differentiating w.r.t y: Integrate both sides. Let t = p (1/t) dt = dp. Given curve passes through (e, 1): Class 12 Maths: Differential Equations

The solution curve of the differential equation y(dx/dy) = x( x - y + 1), x > 0, y > 0 passing through the point (e, 1) is

Given DE: (dx/dy) = (x/y) (((x/y)) + 1) Let (x/y) = t x = ty. Differentiating w.r.t y: Integrate both sides. Let t = p (1/t) dt = dp. Given curve passes through (e, 1): Class 12 Maths: Differential Equations

The solution curve of the differential equation y(dx/dy) = x( x - y + 1), x > 0, y > 0 passing through the point (e, 1) is

Given DE: (dx/dy) = (x/y) (((x/y)) + 1) Let (x/y) = t x = ty. Differentiating w.r.t y: Integrate both sides. Let t = p (1/t) dt = dp. Given curve passes through (e, 1): Class 12 Maths: Differential Equations

The solution curve of the differential equation y(dx/dy) = x( x - y + 1), x > 0, y

Year	2026	2025	2024	Total
Questions	13	17	16	46

Year

2026

2025

2024

Total

Questions

Solution & Explanation

Core Logic

Given DE: $(dx)/(dy) = (x)/(y) (ln((x)/(y)) + 1)$ Let $(x)/(y) = t x = ty$ $\frac{x}{y} = t \implies x = ty$. Differentiating w.r.t $y$ $y$:

(dx)/(dy) = t + y(dt)/(dy)

Step 1: Substitution and Integration

t + y(dt)/(dy) = t(ln(t) + 1) = tln t + t

y(dt)/(dy) = tln t (dt)/(tln t) = (dy)/(y)

Integrate both sides. Let $ln t = p (1)/(t) dt = dp$ .

\int (dp)/(p) = \int (dy)/(y)

ln|p| = ln y + C ln|ln t| = ln y + C

ln|ln((x)/(y))| = ln y + C

Step 2: Applying Boundary Conditions

Given curve passes through $(e, 1)$ $(e, 1)$:

ln|ln((e)/(1))| = ln(1) + C C = 0

ln|ln((x)/(y))| = ln y

|ln((x)/(y))| = eln y = y

Chapter Mix

Class 12 Maths: Differential Equations

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

The solution curve of the differential equation $y(dx)/(dy) = x( ₑ x - ₑ y + 1), x > 0, y > 0$ passing through the point $(e, 1)$ $(e, 1)$ is

Solution & Explanation

Core Logic

Step 1: Substitution and Integration

Step 2: Applying Boundary Conditions

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_31_jan_morning

Select Your Target Exam

Full Mock Test Hub

Topic-wise Practice Hub

Exam Analysis Tools

Chapter Weightage & Trends

Topic-wise Practice Hub

JEE Main Exam Papers

Chapter Weightage Board

Formula Recognition

Exam Questions

Select an Exam

The solution curve of the differential equation y(dx)/(dy) = x( ₑ x - ₑ y + 1), x > 0, y > 0$y\frac{dx}{dy} = x(\log_e x - \log_e y + 1), x > 0, y > 0$ passing through the point (e, 1)$(e, 1)$ is

Solution & Explanation

Core Logic

Step 1: Substitution and Integration

Step 2: Applying Boundary Conditions

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_31_jan_morning

Select Your Target Exam

Full Mock Test Hub

Topic-wise Practice Hub

Exam Analysis Tools

Chapter Weightage & Trends

Topic-wise Practice Hub

JEE Main Exam Papers

Chapter Weightage Board

Formula Recognition

Exam Questions

Select an Exam

Welcome back

The solution curve of the differential equation $y(dx)/(dy) = x( ₑ x - ₑ y + 1), x > 0, y > 0$ passing through the point $(e, 1)$ $(e, 1)$ is