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 $x=\frac{\pi}{6}$

$$ y \frac{\sqrt{3}}{2}=3 \frac{\pi^{2}}{36}-\frac{\pi^{2}}{3}=-\frac{\pi^{2}}{4} \Rightarrow y=-\frac{\pi^{2}}{2 \sqrt{3}} $$



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