Differential Equations 2 Question 22
23. If $y(x)$ satisfies the differential equation $y^{\prime}-y \tan x=2 x \sec x$ and $y(0)$, then
(2012)
(a) $y \frac{\pi}{4}=\frac{\pi^{2}}{8 \sqrt{2}}$
(b) $y^{\prime} \quad \frac{\pi}{4}=\frac{\pi^{2}}{18}$
(c) $y \frac{\pi}{3}=\frac{\pi^{2}}{9}$
(d) $y^{\prime} \frac{\pi}{3}=\frac{4 \pi}{3}+\frac{2 \pi^{2}}{3 \sqrt{3}}$
(2016 Adv.)
Analytical & Descriptive Question
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Answer:
Correct Answer: 23. (a, d)
Solution:
- PLAN Linear differential equation under one variable.
$$ \begin{array}{cc} & \frac{d y}{d x}+P y=Q ; \quad I F=e^{\int P d x} \\ & \therefore \quad \text { Solution is, } y(IF)=\int Q \cdot(IF) d x+C \\ & y^{\prime}-y \tan x=2 x \sec x \text { and } y(0)=0 \\ \therefore \quad & \frac{d y}{d x}-y \tan x=2 x \sec x \\ \therefore \quad & IF=\int e^{-\tan x} d x=e^{\log |\cos x|}=\cos x \end{array} $$
Solution is $y \cdot \cos x=\int 2 x \sec x \cdot \cos x d x+C$
$$ \begin{aligned} & \Rightarrow \quad y \cdot \cos x=x^{2}+C \\ & \text { As } \quad y(0)=0 \Rightarrow \quad C=0 \\ & \therefore \quad y=x^{2} \sec x \\ & \text { Now, } \quad y \frac{\pi}{4}=\frac{\pi^{2}}{8 \sqrt{2}} \\ & \Rightarrow \quad y^{\prime} \quad \frac{\pi}{4}=\frac{\pi}{\sqrt{2}}+\frac{\pi^{2}}{8 \sqrt{2}} \\ & y \frac{\pi}{3}=\frac{2 \pi^{2}}{9} \Rightarrow y^{\prime} \frac{\pi}{3}=\frac{4 \pi}{3}+\frac{2 \pi^{2}}{3 \sqrt{3}} \end{aligned} $$
