Differential Equations 2 Question 4
4. If $y=y(x)$ is the solution of the differential equation $\frac{d y}{d x}=(\tan x-y) \sec ^{2} x, x \in-\frac{\pi}{2}, \frac{\pi}{2}$, such that $y(0)=0$, then $y-\frac{\pi}{4}$ is equal to
(2019 Main, 10 April I)
(a) $\frac{1}{e}-2$
(b) $\frac{1}{2}-e$
(c) $2+\frac{1}{e}$
(d) $e-2$
Show Answer
Answer:
Correct Answer: 4. (d)
Solution:
- Given differential equation
$$ \begin{array}{r} \frac{d y}{d x}=(\tan x-y) \sec ^{2} x \\ \Rightarrow \quad \frac{d y}{d x}+\left(\sec ^{2} x\right) y=\sec ^{2} x \tan x \end{array} $$
which is linear differential equation of the form
$$ \frac{d y}{d x}+P y=Q $$
where $P=\sec ^{2} x$ and $Q=\sec ^{2} x \tan x$
$$ IF=e^{\int \sec ^{2} x d x}=e^{\tan x} $$
So, solution of given differential equation is
$$ \begin{array}{rlrl} y \times IF & =\int(Q \times IF) d x+C \\ \text { Let } \quad y\left(e^{\tan x}\right) & =\int e^{\tan x} \cdot \sec ^{2} x \tan x d x+C \\ \tan x & =t \Rightarrow \sec ^{2} x d x=d t \\ y e^{\tan x} & =\int e^{t} \cdot t d t+C=t e^{t}-\int e^{t} d t+C \\ & & {[\text { using integration by parts method }]} \\ & =e^{t}(t-1)+C \\ \Rightarrow \quad y \cdot e^{\tan x} & =e^{\tan x}(\tan x-1)+C \quad[\because t=\tan x] \\ y(0) & =0 \\ \Rightarrow \quad 0 & =1(0-1)+C \Rightarrow C=1 \\ \therefore \quad y \cdot e^{\tan x} & =e^{\tan x}(\tan x-1)+1 \\ \text { Now, at } x=-\frac{\pi}{4} \quad & \\ \Rightarrow \quad y e^{-1} & =e^{-1}(-1-1)+1 \\ y e^{-1} & =-2 e^{-1}+1 \Rightarrow y=e-2 \end{array} $$
