Differential Equations 2 Question 5
5. If $\cos x \frac{d y}{d x}-y \sin x=6 x, 0<x<\frac{x}{2}$ and $y \frac{\pi}{3}=0$, then $y \frac{\pi}{6}$ is equal to
(2019 Main, 9 April II)
(a) $\frac{\pi^{2}}{2 \sqrt{3}}$
(b) $-\frac{\pi^{2}}{2 \sqrt{3}}$
(c) $-\frac{\pi^{2}}{4 \sqrt{3}}$
(d) $-\frac{\pi^{2}}{2}$
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Answer:
Correct Answer: 5. (b)
Solution:
Key Idea (i) First convert the given differential equation into linear differential equation of the form $\frac{d y}{d x}+P y=Q$
(ii) Find IF
(iii) Apply formula, $y(I F)=\int Q(I F) d x+C$
Given differential equation
$$ \begin{aligned} & \cos x \frac{d y}{d x}-(\sin x) y & =6 x \\ \Rightarrow \quad & \quad \frac{d y}{d x}-(\tan x) y & =\frac{6 x}{\cos x}, \text { which is the linear } \end{aligned} $$
differential equation of the form
$$ \frac{d y}{d x}+P x=Q $$
where $P=-\tan x$ and $Q=\frac{6 x}{\cos x}$
So, $\quad IF=e^{-\int \tan x d x}=e^{-\log (\sec x)}=\cos x$
$\therefore$ Required solution of differential equation is $y(\cos x)=\int(6 x) \frac{\cos x}{\cos x} d x+C=6 \frac{x^{2}}{2}+C=3 x^{2}+C$
Given,
$$ y \frac{\pi}{3}=0 $$
So, $\quad 0=3 \frac{\pi}{3}^{2}+C \Rightarrow C=-\frac{\pi^{2}}{3}$
$\therefore \quad y(\cos x)=3 x^{2}-\frac{\pi^{2}}{3}$
Now, at $x=\frac{\pi}{6}$
$$ y \frac{\sqrt{3}}{2}=3 \frac{\pi^{2}}{36}-\frac{\pi^{2}}{3}=-\frac{\pi^{2}}{4} \Rightarrow y=-\frac{\pi^{2}}{2 \sqrt{3}} $$
