Question

If $\dfrac{dy}{dx}+\dfrac{3}{{{\cos }^{2}}x}y=\dfrac{1}{{{\cos }^{2}}x}$, $x\in \left( \dfrac{-\pi }{3},\dfrac{\pi }{3} \right)$ and $y\left( \dfrac{\pi }{4} \right)=\dfrac{4}{3}$, then $y\left( \dfrac{-\pi }{4} \right)$ equals to?
(a) $\dfrac{1}{3}+{{e}^{6}}$
(b) $\dfrac{1}{3}$
(c) $\dfrac{-4}{3}$
(d) $\dfrac{1}{3}+{{e}^{3}}$

Answer 1

First, before proceeding for this, we must know the following trigonometric conversion as $\sec x=\dfrac{1}{\cos x}$. Then, to get the solution of the above differential equation in the form $\dfrac{dy}{dx}+Py=Q$, we need a integrating factor(IF) given by the formula as $IF={{e}^{\int{Pdx}}}$. Then, to get the solution of the above differential equation in the form $\dfrac{dy}{dx}+Py=Q$, we have the form of solution as $y\times IF=\int{Q\times IF}dx+c$. Then, by using the condition given in the question as $y\left( \dfrac{\pi }{4} \right)=\dfrac{4}{3}$which means at $x=\dfrac{\pi }{4}$, y is $\dfrac{4}{3}$, we get the value of c and then we get the desired value.

Complete step by step answer:
In this question, we are supposed to find the value of $y\left( \dfrac{-\pi }{4} \right)$ when $\dfrac{dy}{dx}+\dfrac{3}{{{\cos }^{2}}x}y=\dfrac{1}{{{\cos }^{2}}x}$and $y\left( \dfrac{\pi }{4} \right)=\dfrac{4}{3}$.
So, before proceeding for this, we must know the following trigonometric conversion as:
$\sec x=\dfrac{1}{\cos x}$
So, by using it in the given question, we get the differential equation as:
$\dfrac{dy}{dx}+3{{\sec }^{2}}xy={{\sec }^{2}}x$
Now, to get the solution of the above differential equation in the form $\dfrac{dy}{dx}+Py=Q$, we need a integrating factor(IF) given by the formula as:
$IF={{e}^{\int{Pdx}}}$
So, the value of P from the above differential equation is $3{{\sec }^{2}}x$ to get the value of IF as:
$\begin{aligned} & IF={{e}^{\int{3{{\sec }^{2}}xdx}}} \\\ & \Rightarrow IF={{e}^{3\tan x}} \\\ \end{aligned}$
Now, to get the solution of the above differential equation in the form $\dfrac{dy}{dx}+Py=Q$, we have the form of solution as:
$y\times IF=\int{Q\times IF}dx+c$
Then, by substituting the value of IF and Q, we get:
$y\times {{e}^{3\tan x}}=\int{{{\sec }^{2}}x\times {{e}^{3\tan x}}}dx+c$
Now, by using the substitution as let tan x=u, we get the differentiation as:
${{\sec }^{2}}xdx=du$
Then, by substituting the value in the above expression, we get:
$y\times {{e}^{3\tan x}}=\int{{{e}^{3u}}du}+c$
Then, b y solving the integral, we get:
$y\times {{e}^{3\tan x}}=\dfrac{{{e}^{3u}}}{3}+c$
Now, by substituting the value of assumed u as tan x, we get:
$\begin{aligned} & y\times {{e}^{3\tan x}}=\dfrac{{{e}^{3\tan x}}}{3}+c \\\ & \Rightarrow y=\dfrac{1}{3}+\dfrac{c}{{{e}^{3\tan x}}} \\\ \end{aligned}$
Now, by using the condition given in the question as $y\left( \dfrac{\pi }{4} \right)=\dfrac{4}{3}$which means at $x=\dfrac{\pi }{4}$, y is $\dfrac{4}{3}$ as:
$\begin{aligned} & \dfrac{4}{3}=\dfrac{1}{3}+\dfrac{c}{{{e}^{3\tan \dfrac{\pi }{4}}}} \\\ & \Rightarrow \dfrac{4}{3}-\dfrac{1}{3}=\dfrac{c}{{{e}^{3}}} \\\ & \Rightarrow \dfrac{3}{3}=\dfrac{c}{{{e}^{3}}} \\\ & \Rightarrow c={{e}^{3}} \\\ \end{aligned}$
Then, by substituting the value of c as ${{e}^{3}}$, we get:
$y=\dfrac{1}{3}+\dfrac{{{e}^{3}}}{{{e}^{3\tan x}}}$
Now, we are asked to find the value of $y\left( \dfrac{-\pi }{4} \right)$which means the value of y at $x=\dfrac{-\pi }{4}$, we get:
$\begin{aligned} & y=\dfrac{1}{3}+\dfrac{{{e}^{3}}}{{{e}^{3\tan \left( \dfrac{-\pi }{4} \right)}}} \\\ & \Rightarrow y=\dfrac{1}{3}+\dfrac{{{e}^{3}}}{{{e}^{-3}}} \\\ & \Rightarrow y=\dfrac{1}{3}+{{e}^{6}} \\\ \end{aligned}$
So, we get the value of $y\left( \dfrac{-\pi }{4} \right)$ as $\dfrac{1}{3}+{{e}^{6}}$.

So, the correct answer is “Option A”.

Note: Now, to solve these types of questions we need to know some of the basic integration and differentiation formulas beforehand to solve accurately. So, the required formula is as:
$\begin{aligned} & \int{{{\sec }^{2}}xdx=\tan x} \\\ & \int{{{e}^{x}}dx={{e}^{x}}} \\\ \end{aligned}$
Similarly, the required formula for differentiation is:
$\dfrac{d}{dx}\left( \tan x \right)={{\sec }^{2}}x$