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 :
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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