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) $y^{\prime} (\frac{\pi}{4})=\frac{\pi^{2}}{18}$
(c) $y (\frac{\pi}{3})=\frac{\pi^{2}}{9}$
(d) $y^{\prime} (\frac{\pi}{3})=\frac{4 \pi}{3}+\frac{2 \pi^{2}}{3 \sqrt{3}}$
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Answer:
Correct Answer: 23. (a, d)
Solution:
- PLAN: Linear differential equation under one variable.
$ \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 $
$\therefore \quad IF=\int e^{-\tan x} d x=e^{\log |\cos x|}=\cos x$
Solution is $y \cdot \cos x=\int 2 x \sec x \cdot \cos x d x+C$
$ \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} (\frac{\pi}{4})=\frac{\pi}{\sqrt{2}}+\frac{\pi^{2}}{8 \sqrt{2}} $
$y (\frac{\pi}{3})=\frac{2 \pi^{2}}{9} \Rightarrow y^{\prime} (\frac{\pi}{3})=\frac{4 \pi}{3}+\frac{2 \pi^{2}}{3 \sqrt{3}}$
