Let y = y(x) be the solution of he differential equation , y(0) = –2. Let the maximum and minimum values of the function y = y(x) in be and , respectively. If , then equals __________. [2024]

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Solution

Q.
Let y = y(x) be the solution of the differential equation ${(x + y + 2)}^{2} d x = d y$ , y(0) = –2. Let the maximum and minimum values of the function y = y(x) in $[0, \frac{π}{3}]$ be $α$ and $β$ , respectively. If ${(3 α + π)}^{2} + β^{2} = γ + δ \sqrt{3}, γ, δ \in Z$ , then $γ + δ$ equals __________. [2024]

Ans.
(31)

We have, $\frac{d y}{d x} = {(x + y + 2)}^{2}$ ... (i)

Put x + y + 2 = t

$\Rightarrow 1 + \frac{d y}{d x} = \frac{d t}{d x}$

$∴$ From (i) $\frac{d t}{d x} - 1 = t^{2} \Rightarrow \frac{d t}{1 + t^{2}} = d x \Rightarrow \tan^{- 1} t = x + C$

$\Rightarrow x + y + 2 = \tan (x + C) \Rightarrow y = \tan (x + C) - x - 2$

$∵ y (0) = - 2 \Rightarrow - 2 = \tan C - 0 - 2$

$\Rightarrow \tan C = 0 \Rightarrow C = 0 \Rightarrow y = \tan x - x - 2$

$\Rightarrow \frac{d y}{d x} = \sec^{2} x - 1 \geq 0$

$\Rightarrow y (x) is increasing if x \in (0, \frac{π}{3}) \Rightarrow α = y (\frac{π}{3}), β = y (0)$

$\Rightarrow α = \tan \frac{π}{3} - \frac{π}{3} - 2 = - \frac{π}{3} - 2 + \sqrt{3}$

and $β$ = tan 0 – 0 – 2 = –2

Hence, ${(3 α + π)}^{2} + β^{2} = {(- π - 6 + 3 \sqrt{3} + π)}^{2} + (- 2)^{2}$

$= {(- 6 + 3 \sqrt{3})}^{2} + 4 = 36 + 27 - 36 \sqrt{3} + 4$

$= 67 - 36 \sqrt{3} = y + δ \sqrt{3}$ [Given]

On comparing, we get

$γ$ = 67 and $δ$ = –36

Hence, $γ + δ = 67 - 36 = 31$ .

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