Let y = y(x) be the solution of the differential equation (dy/dx) + 3(2 x)y + 3y = 2 x, y(0) = (1/3) + e3. Then…

Q: Let y = y(x) be the solution of the differential equation (dy/dx) + 3(2 x)y + 3y = 2 x, y(0) = (1/3) + e3. Then y((pi/4)) is equal to

### Related Formula For a first-order linear differential equation (dy/dx) + P(x)y = Q(x): - Integrating Factor (I.F.) = e P(x) , dx - Solution is y cdot I.F. = Q(x) cdot I.F. , dx + C ### Core Logic Let's simplify the coefficient of y: Thus, the equation is: ### Step 1: Finding Integrating Factor and General Solution I.F. = e 32 x , dx = e3 x The general solution is: Substitute u = 3 x du = 32 x , dx: ### Step 2: Solving for boundary conditions Given y(0) = (1/3) + e3: Thus, the explicit functi

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2026

2025

2024

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Questions

Solution & Explanation

Related Formula

For a first-order linear differential equation $(dy)/(dx) + P(x)y = Q(x)$ $\frac{dy}{dx} + P(x)y = Q(x)$:

Integrating Factor (I.F.) = $e\int P(x) dx$ $e^{\int P(x) \, dx}$
Solution is $y \cdot I.F. = \int Q(x) \cdot I.F. dx + C$

Core Logic

Let's simplify the coefficient of $y$ $y$:

3 ² x + 3 = 3( ² x + 1) = 3 ² x

Thus, the equation is:

(dy)/(dx) + 3( ² x)y = ² x

Step 1: Finding Integrating Factor and General Solution

I.F. = $e\int 3 ² x dx = e3 x$

The general solution is:

y \cdot e3 x = \int ² x \cdot e3 x dx + C

Substitute $u = 3 x du = 3 ² x dx$ :

y \cdot e3 x = (1)/(3) \int e^u du + C = (1)/(3) e3 x + C

Step 2: Solving for boundary conditions

Given $y(0) = (1)/(3) + e³$ $y(0) = \frac{1}{3} + e^3$:

((1)/(3) + e³) \cdot e⁰ = (1)/(3) e⁰ + C C = e³

Thus, the explicit function is:

y = (1)/(3) + e3 - 3 x

Evaluating at $x = (π)/(4)$ $x = \frac{\pi}{4}$:

y((π)/(4)) = (1)/(3) + e3 - 3 (π/4) = (1)/(3) + e³⁻³ = (1)/(3) + 1 = (4)/(3)

Pattern Recognition

Recognizing that $3 ² x + 3 = 3 ² x$ $3\tan^2 x + 3 = 3\sec^2 x$ converts the system immediately into a classic linear differential equation where the coefficient of $y$ $y$ is exactly the derivative of the exponent of I.F. This makes integration virtually instantaneous.

Chapter Mix

Class 12 Mathematics: Differential Equations Class 11 Mathematics: Trigonometric Functions

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

Let $y = y(x)$ $y = y(x)$ be the solution of the differential equation $(dy)/(dx) + 3( ² x)y + 3y = ² x, y(0) = (1)/(3) + e³$ . Then $y((π)/(4))$ $y\left(\frac{\pi}{4}\right)$ is equal to

Solution & Explanation

Related Formula

Core Logic

Step 1: Finding Integrating Factor and General Solution

Step 2: Solving for boundary conditions

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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Let y = y(x)$y = y(x)$ be the solution of the differential equation (dy)/(dx) + 3( ² x)y + 3y = ² x, y(0) = (1)/(3) + e³$\frac{dy}{dx} + 3(\tan^2 x)y + 3y = \sec^2 x, y(0) = \frac{1}{3} + e^3$. Then y((π)/(4))$y\left(\frac{\pi}{4}\right)$ is equal to

Solution & Explanation

Related Formula

Core Logic

Step 1: Finding Integrating Factor and General Solution

Step 2: Solving for boundary conditions

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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Let $y = y(x)$ $y = y(x)$ be the solution of the differential equation $(dy)/(dx) + 3( ² x)y + 3y = ² x, y(0) = (1)/(3) + e³$ . Then $y((π)/(4))$ $y\left(\frac{\pi}{4}\right)$ is equal to