Let x=x(t) and y=y(t) be solutions of the differential equations (dx/dt)+ax=0 and (dy/dt)+by=0 respectively, a, b R…

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Year

2026

2025

2024

Total

Questions

Solution & Explanation

Related Formula

\int (1)/(x) dx = ln|x| + C

Core Logic

Solving the first differential equation:

(dx)/(dt) + ax = 0 \Rightarrow (dx)/(x) = -a dt

Integrating both sides:

ln|x| = -at + c₁

Given $x(0) = 2$ $x(0) = 2$, we find $c₁ = ln 2$ $c_1 = \ln 2$. Thus:

ln(x) = -at + ln 2 \Rightarrow x(t) = 2e-at

Step 1: Solving for y(t)

Solving the second differential equation:

(dy)/(dt) + by = 0 \Rightarrow (dy)/(y) = -b dt

Integrating both sides:

ln|y| = -bt + c₂

Given $y(0) = 1$ $y(0) = 1$, we find $c₂ = 0$ $c_2 = 0$. Thus: $y(t) = e-bt$ $y(t) = e^{-bt}$

Step 2: Applying the condition

We are given $3y(1) = 2x(1)$ $3y(1) = 2x(1)$. Substituting our solutions at $t=1$ $t=1$:

3(e-b) = 2(2e-a) \Rightarrow 3e-b = 4e-a

Rearranging to group exponential terms:

e-be-a = (4)/(3) \Rightarrow ea-b = (4)/(3)

Step 3: Finding t for x(t) = y(t)

Set the two trajectory solutions equal:

x(t) = y(t) \Rightarrow 2e-at = e-bt

Rearranging gives:

2 = e-bte-at \Rightarrow 2 = e(a-b)t

Substitute $ea-b = (4)/(3)$ $e^{a-b} = \frac{4}{3}$ from Step 2:

2 = ((4)/(3))^t

Taking $(4)/(3)$ $\log_{\frac{4}{3}}$ on both sides yields:

t = (4)/(3)2

Pattern Recognition

For $(dz)/(dt) + kz = 0$ $\frac{dz}{dt} + kz = 0$, the solution is always an exponential decay $z = z₀ e-kt$ $z = z_0 e^{-kt}$. Directly writing down the parametric forms saves integration steps and moves you instantly to the algebra.

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

Solution & Explanation

Related Formula

Core Logic

Step 1: Solving for y(t)

Step 2: Applying the condition

Step 3: Finding t for x(t) = y(t)

Pattern Recognition

Chapter Mix

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: Solving for y(t)

Step 2: Applying the condition

Step 3: Finding t for x(t) = y(t)

Pattern Recognition

Chapter Mix

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

Select Your Target Exam

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