Differential Equations 2 Question 12
12. If $\frac{d y}{d x}+\frac{3}{\cos ^{2} x} y=\frac{1}{\cos ^{2} x}, x \in \frac{-\pi}{3}, \frac{\pi}{3}$ and $y \frac{\pi}{4}=\frac{4}{3}$, then $y-\frac{\pi}{4}$ equals
(2019 Main, 10 Jan I)
(a) $\frac{1}{3}+e^{6}$
(b) $-\frac{4}{3}$
(c) $\frac{1}{3}+e^{3}$
(d) $\frac{1}{3}$
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Answer:
Correct Answer: 12. (a)
Solution:
- Given, differential equation is $\frac{d y}{d x}+\frac{3}{\cos ^{2} x} y=\frac{1}{\cos ^{2} x}$, which is a linear differential equation of the form $\frac{d y}{d x}+P y=Q$, where $P=\frac{3}{\cos ^{2} x}$ and $Q=\frac{1}{\cos ^{2} x}$.
Now, Integrating factor
$IF=e^{\int \frac{3}{\cos ^{2} x} d x}=e^{\beta \sec ^{2} x d x}=e^{3 \tan x}$ and the solution of differential equation is given by
$$ \begin{aligned} & y(IF)=\int(Q .(IF)) d x \\ & \Rightarrow \quad e^{3 \tan x} \cdot y=\int e^{3 \tan x} \sec ^{2} x d x \\ & \text { Let } \quad I=\int e^{3 \tan x} \sec ^{2} x d x \\ & \text { Put } \quad 3 \tan x=t \\ & \Rightarrow \quad 3 \sec ^{2} x d x=d t \\ & \therefore \quad I=\int \frac{e^{t}}{3} d t=\frac{e^{t}}{3}+C=\frac{e^{3 \tan x}}{3}+C \end{aligned} $$
From Eq. (i)
$$ e^{3 \tan x} \cdot y=\frac{e^{3 \tan x}}{3}+C $$
It is given that when,
$$ \begin{array}{rlrl} x & =\frac{\pi}{4}, y \text { is } \frac{4}{3} \\ \Rightarrow & e^{3} \frac{4}{3} & =\frac{e^{3}}{3}+C \\ \Rightarrow \quad & C & =e^{3} \end{array} $$
Thus, $e^{3 \tan x} y=\frac{e^{3 \tan x}}{3}+e^{3}$
Now, when $x=-\frac{\pi}{4}, e^{-3} y=\frac{e^{-3}}{3}+e^{3}$
$$ \Rightarrow \quad y=e^{6}+\frac{1}{3} \quad \because \tan -\frac{\pi}{4}=-1 $$
