SATHEE: Differential Equations 2 Question 5

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 $y \frac{π}{3} = 0$ , then $y \frac{π}{6}$ is equal to

(2019 Main, 9 April II)

(a) $\frac{π^{2}}{2 \sqrt{3}}$

(b) $- \frac{π^{2}}{2 \sqrt{3}}$

(d) $- \frac{π^{2}}{2}$

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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

$\begin{aligned} \cos x \frac{d y}{d x} - (\sin x) y & = 6 x \\ \Rightarrow & \frac{d y}{d x} - (\tan x) y & = \frac{6 x}{\cos x}, which is the linear \end{aligned}$

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, $I F = e^{- \int \tan x d x} = e^{- \log (\sec x)} = \cos x$

$∴$ 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,

$y \frac{π}{3} = 0$

So, $0 = 3 {\frac{π}{3}}^{2} + C \Rightarrow C = - \frac{π^{2}}{3}$

$∴ y (\cos x) = 3 x^{2} - \frac{π^{2}}{3}$

Now, at $x = \frac{π}{6}$

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

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 $y \frac{π}{3} = 0$ , then $y \frac{π}{6}$ is equal to

Answer:

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Differential Equations 2 Question 5

5. If cos⁡xdydx−ysin⁡x=6x,0<x<x2 and yπ3=0, then yπ6 is equal to

Answer:

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5. If $\cos x \frac{d y}{d x} - y \sin x = 6 x, 0 < x < \frac{x}{2}$ and $y \frac{π}{3} = 0$ , then $y \frac{π}{6}$ is equal to