Medium MCQ +4 / -1 PYQ · JEE Mains 2025

Let $y=y(x)$ be the solution of the differential equation

$\frac{d y}{d x}+3\left(\tan ^2 x\right) y+3 y=\sec ^2 x, y(0)=\frac{1}{3}+e^3$. Then $y\left(\frac{\pi}{4}\right)$ is equal to :

  1. A $\frac{4}{3}$ Correct answer
  2. B $\frac{2}{3}+e^3$
  3. C $\frac{4}{3}+e^3$
  4. D $\frac{2}{3}$

Solution

<p>$$\begin{aligned} & \frac{d y}{d x}+3\left(\tan ^2 x\right) y+3 y=\sec ^2 x \\ & \Rightarrow \frac{d y}{d x}+3 \sec ^2 x y=\sec ^2 x \\ & \text { I.F }=e^{\int 3 \sec ^2 x d x} \\ & \quad=e^{3 \tan x} \end{aligned}$$</p> <p>$$\begin{aligned} & y \cdot e^{\tan x}=\int e^{3 \tan x} \cdot \sec ^2 x d x+c \\ & y \cdot e^{3 \tan x}=\frac{e^{3 \tan x}}{3}+c \\ & \text { Also } f(0)=\frac{1}{3}+e^3 \\ & \Rightarrow\left(\frac{1}{3}+e^3\right)=\frac{1}{3}+c \\ & \Rightarrow c=e^3 \\ & \therefore y \cdot e^{3 \tan x}=\frac{e^{3 \tan x}}{3}+e^3 \\ & \text { Put } x=\frac{\pi}{4} \end{aligned}$$</p> <p>$y e^3=\frac{e^3}{3}+e^3 \Rightarrow y=\frac{4}{3}$</p>

About this question

Subject: Mathematics · Chapter: Differential Equations · Topic: Order and Degree

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