Differential Equations jee mains questions | JEE Main Maths Differential Equations Previous Year Question

Q 71 :

Let $y = y (x)$ be the solution of the differential equation $(x^{2} - 3 y^{2}) d x + 3 x y d y = 0, y (1) = 1 .$ Then $6 y^{2} (e)$ is equal to [2023]

$e^{2}$
$\frac{3}{2} e^{2}$
$3 e^{2}$
$2 e^{2}$

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(4)

$(x^{2} - 3 y^{2}) d x + 3 x y d y = 0$

$\Rightarrow x^{2} d x - 3 y^{2} d x + 3 x y d y = 0 \Rightarrow x^{2} - 3 y^{2} + 3 x y \frac{d y}{d x} = 0$

Put $t = y^{2} \Rightarrow 2 y \frac{d y}{d x} = \frac{d t}{d x}$ $\Rightarrow x^{2} - 3 t + \frac{3 x}{2} \frac{d t}{d x} = 0$

$\Rightarrow \frac{2 x}{3} - \frac{2 t}{x} + \frac{d t}{d x} = 0 \Rightarrow \frac{d t}{d x} - \frac{2 t}{x} = - \frac{2 x}{3},$

which is a linear differential equation.

$I.F. = e^{\int - \frac{2}{x} d x} = e^{- 2 \ln | x |} = e^{\ln \frac{1}{x^{2}}} = \frac{1}{x^{2}}$

So, $t \cdot \frac{1}{x^{2}} = \int \frac{1}{x^{2}} (\frac{- 2 x}{3}) d x$ $\Rightarrow \frac{y^{2}}{x^{2}} = - \frac{2}{3} \ln | x | + C$

When, $x = 1, y = 1$

$1 = \frac{- 2}{3} \ln (1) + C \Rightarrow C = 1$ $∴$ $\frac{y^{2}}{x^{2}} = \frac{- 2}{3} \ln | x | + 1$

At $x = e,$ $\frac{y^{2} (e)}{e^{2}} = \frac{- 2}{3} + 1 \Rightarrow y^{2} (e) = \frac{e^{2}}{3} \Rightarrow 6 y^{2} (e) = 2 e^{2}$

Q 72 :

Let $y = y (x)$ be the solution curve of the differential equation $\frac{d y}{d x} = \frac{y}{x} (1 + x y^{2} (1 + \log_{e} x)), x > 0, y (1) = 3 .$ Then, $\frac{y^{2} (x)}{9}$ is equal to [2023]

$\frac{x^{2}}{5 - 2 x^{3} (2 + \log_{e} x^{3})}$
$\frac{x^{2}}{3 x^{3} (1 + \log_{e} x^{2}) - 2}$
$\frac{x^{2}}{7 - 3 x^{3} (2 + \log_{e} x^{2})}$
$\frac{x^{2}}{2 x^{3} (2 + \log_{e} x^{3}) - 3}$

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(1)

$Let y = v x; \frac{d y}{d x} = v + x \frac{d v}{d x}$

$v + x \frac{d v}{d x} = v + v^{3} x^{3} (1 + \log_{e} x)$

$\int v^{- 3} d v = \int (x^{2} + x^{2} \log_{e} x) d x$

$\frac{v^{- 2}}{- 2} = \frac{x^{3}}{3} + \log x \frac{x^{3}}{3} - \int \frac{x^{2}}{3} d x = \frac{x^{3}}{3} + \frac{x^{3}}{3} \log x - \frac{x^{3}}{9} + C$

$- \frac{x^{2}}{2 y^{2}} = \frac{2}{9} x^{3} + \frac{x^{3}}{3} \log x + c; y (1) = 3$

$\frac{- 1}{2 \times 9} = \frac{2}{9} + C \Rightarrow C = \frac{- 5}{18}$

$\frac{- x^{2}}{2 y^{2}} = \frac{2}{9} x^{3} + \frac{x^{3}}{3} \log x - \frac{5}{18} \Rightarrow \frac{9}{y^{2}} = \frac{4 x^{3} + 6 x^{3} \log x - 5}{- x^{2}}$

$\Rightarrow \frac{y^{2} (x)}{9} = \frac{x^{2}}{5 - 2 x^{3} (2 + \log x^{3})}$

Q 73 :

Let $y = y (t)$ be a solution of the differential equation $\frac{d y}{d t} + α y = γ e^{- β t}$ where, $α > 0, β > 0$ and $γ > 0$ . Then $\lim_{t \to \infty} y (t)$ [2023]

is -1
is 0
is 1
does not exist

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$\frac{d y}{d t} + α y = γ e^{- β t}$

$I.F. = e^{\int α d t} = e^{α t}$

Solution is given by

$y \cdot e^{α t} = \int e^{α t} \cdot γ e^{- β t} d t = γ \int e^{(α - β) t} d t$

$\Rightarrow y \cdot e^{α t} = γ \frac{e^{(α - β) t}}{α - β} + c$ $\Rightarrow y = \frac{γ}{α - β} e^{- β t} + c e^{- α t}$

$So, \lim_{t \to \infty} y (t) = \frac{γ}{α - β} \lim_{t \to \infty} e^{- β t} + c \lim_{t \to \infty} e^{- α t} = 0$

Q 74 :

Let $y = f (x)$ be the solution of the differential equation $y (x + 1) d x - x^{2} d y = 0, y (1) = e .$ Then $\lim_{x \to 0^{+}} f (x)$ is equal to [2023]

$\frac{1}{e^{2}}$
$\frac{1}{e}$
$0$
$e^{2}$

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(3)

$Given differential equation,$ $y (x + 1) d x - x^{2} d y = 0$

$\Rightarrow \frac{d y}{d x} = \frac{y (x + 1)}{x^{2}}$

$On integrating,$ $\int \frac{1}{y} d y = \int \frac{x + 1}{x^{2}} d x \Rightarrow \ln y = \ln x - \frac{1}{x} + C$

$Given y (1) = e, that is at x = 1, y = e$

So, $\ln e = \ln 1 - 1 + C \Rightarrow C = 2$

$So, \ln y = \ln x - \frac{1}{x} + 2$

$\Rightarrow y = x e^{- \frac{1}{x} + 2}$ $\Rightarrow \lim_{x \to 0^{+}} x e^{- \frac{1}{x} + 2} = 0$

Q 75 :

Let $y = y (x)$ be the solution of the differential equation $x \log_{e} x \frac{d y}{d x} + y = x^{2} \log_{e} x, (x > 1) .$ If $y (2) = 2$ , then $y (e)$ is equal to [2023]

$\frac{1 + e^{2}}{2}$
$\frac{4 + e^{2}}{4}$
$\frac{1 + e^{2}}{4}$
$\frac{2 + e^{2}}{2}$

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(2)

$Given: x \log_{e} x \frac{d y}{d x} + y = x^{2} \log_{e} x (x > 1)$

$\Rightarrow \frac{d y}{d x} + \frac{y}{x \ln x} = x$ which is a linear differential equation.

whose Integrating factor is given by

$I.F. = e^{\int \frac{1}{x \ln x} d x} =$ $\ln$ $| x |$

$y \ln$ $| x |$ $= \frac{x^{2}}{2} \ln x - \int \frac{1}{x} \cdot \frac{x^{2}}{2} d x$ ...(i)

$\Rightarrow y \ln$ $| x |$ $= \ln$ $| x |$ $\frac{x^{2}}{2} - \frac{x^{2}}{4} + C$

$Also, y (2) = 2 \Rightarrow C = 1$

On substituting the value of $C$ in (i), we get

$y (x) = \frac{x^{2}}{2} - \frac{x^{2}}{4 | \ln (x) |} + \frac{1}{| \ln (x) |}$

$\Rightarrow y (e) = \frac{e^{2}}{2} - \frac{e^{2}}{4} + 1 = 1 + \frac{e^{2}}{4}$ or $\frac{4 + e^{2}}{4}$

Q 76 :

Let the solution curve $y = y (x)$ of the differential equation $\frac{d y}{d x} - \frac{3 x^{5} \tan^{- 1} (x^{3})}{{(1 + x^{6})}^{3 / 2}} y = 2 x \exp {\frac{x^{3} - \tan^{- 1} x^{3}}{\sqrt{(1 + x^{6})}}}$ pass through the origin. Then $y (1)$ is equal to [2023]

$\exp {\frac{4 + π}{4 \sqrt{2}}}$
$\exp {\frac{4 - π}{4 \sqrt{2}}}$
$\exp {\frac{1 - π}{4 \sqrt{2}}}$
$\exp {\frac{π - 4}{4 \sqrt{2}}}$

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(2)

Given differentiate equation is

$\frac{d y}{d x} - \frac{3 x^{5} \tan^{- 1} (x^{3})}{{(1 + x^{6})}^{3 / 2}} y = 2 x \exp {\frac{x^{3} - \tan^{- 1} x^{3}}{\sqrt{(1 + x^{6})}}}$ ,

Which is in the form $\frac{d y}{d x} + P y = Q$

Here, $P = - \frac{3 x^{5} \tan^{- 1} (x^{3})}{{(1 + x^{6})}^{3 / 2}}, Q = 2 x \exp {\frac{x^{3} - \tan^{- 1} x^{3}}{\sqrt{1 + x^{6}}}}$

Let $I = - \int \frac{3 x^{5} \tan^{- 1} x^{3}}{{(1 + x^{6})}^{3 / 2}} d x$

Put $x^{3} = \tan θ \Rightarrow 3 x^{2} d x = \sec^{2} θ d θ$

$I = - \int \frac{\tan θ \cdot θ}{\sec^{3} θ} \times \sec^{2} θ d θ = - \int θ \cdot \sin θ d θ$

$= - [θ \int \sin θ d θ - \int (\int \sin θ d θ) d θ]$

$= - [- θ \cdot \cos θ + \sin θ] = - [\frac{x^{3}}{\sqrt{1 + x^{6}}} - \tan^{- 1} (x^{3}) \cdot \frac{1}{\sqrt{1 + x^{6}}}]$

$∴$ $I.F. = e^{\int p d x} = e^{I}$

$Solution of the D.E. is$

$y \cdot e^{I} = \int e^{I} \cdot 2 x \cdot \exp {\frac{x^{3} - \tan^{- 1} (x^{3})}{\sqrt{1 + x^{6}}}} d x$

$= \int 2 x \cdot e^{\frac{- x^{3} + \tan^{- 1} (x^{3})}{\sqrt{1 + x^{6}}}} \cdot e^{\frac{x^{3} - \tan^{- 1} (x^{3})}{\sqrt{1 + x^{6}}}} d x$

$= \int 2 x \cdot d x = x^{2} + c$ $∴$ $y e^{I} = x^{2} + c$

$∴$ $y = x^{2} \exp {\frac{- \tan^{- 1} (x^{3}) + x^{3}}{\sqrt{1 + x^{6}}}} + c \exp {\frac{x^{3} - \tan^{- 1} (x^{3})}{\sqrt{1 + x^{6}}}}$

This curve passes through the origin.

So, $0 = 0 + c \Rightarrow c = 0$

The required solution is

$y (x) = x^{2} \exp {\frac{x^{3} - \tan^{- 1} (x^{3})}{\sqrt{1 + x^{6}}}}$

At $x = 1$ $y (1) = \exp (\frac{1 - \frac{π}{4}}{\sqrt{2}}) = \exp (\frac{4 - π}{4 \sqrt{2}})$

Q 77 :

The solution of the differential equation $\frac{d y}{d x} = - (\frac{x^{2} + 3 y^{2}}{3 x^{2} + y^{2}}), y (1) = 0$ is [2023]

$\log_{e}$ $| x + y |$ $- \frac{x y}{{(x + y)}^{2}} = 0$
$\log_{e}$ $| x + y |$ $- \frac{2 x y}{{(x + y)}^{2}} = 0$
$\log_{e}$ $| x + y |$ $+ \frac{x y}{{(x + y)}^{2}} = 0$
$\log_{e}$ $| x + y |$ $+ \frac{2 x y}{{(x + y)}^{2}} = 0$

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Given, $\frac{d y}{d x} = - [\frac{x^{2} + 3 y^{2}}{3 x^{2} + y^{2}}], y (1) = 0$

$Substitute y = v x \Rightarrow \frac{d y}{d x} = v + x \frac{d v}{d x}$

$v + x \frac{d v}{d x} = - [\frac{x^{2} + 3 v^{2} x^{2}}{3 x^{2} + v^{2} x^{2}}] = - [\frac{1 + 3 v^{2}}{3 + v^{2}}]$

$\Rightarrow x \frac{d v}{d x} = - [\frac{1 + 3 v^{2}}{3 + v^{2}}] - v$ $= - [\frac{1 + 3 v^{2} + 3 v + v^{3}}{3 + v^{2}}]$

$\Rightarrow x \frac{d v}{d x} = - \frac{{(1 + v)}^{3}}{3 + v^{2}} \Rightarrow \int \frac{3 + v^{2}}{{(1 + v)}^{3}} d v = - \int \frac{d x}{x}$

$\Rightarrow \int \frac{1 + v^{2} + 2 v - 2 v + 2}{{(1 + v)}^{3}} d v = - \int \frac{d x}{x}$

$\Rightarrow \int \frac{1}{(1 + v)} d v + 2 \int \frac{1 - v}{{(1 + v)}^{3}} d v = - \int \frac{d x}{x}$

$Substitute 1 + v = t \Rightarrow v = t - 1 \Rightarrow d v = d t$

$∴$ $\int \frac{d t}{t} + 2 \int \frac{1 - t + 1}{t^{3}} d t = - \ln x + C$

$\Rightarrow \log_{e} t + 2 \int (\frac{2 - t}{t^{3}}) d t = - \log_{e} x + C$

$\Rightarrow \log_{e} t + 2 \int (2 t^{- 3} - t^{- 2}) d t = - \log_{e} x + C$

$\Rightarrow \log_{e}$ $| \frac{y}{x} + 1 |$ $+ 2 [- t^{- 2} + t^{- 1}] = - \log_{e} x + C$

$\Rightarrow \log_{e}$ $| \frac{y + x}{x} |$ $+ 2$ $[\frac{- 1}{{(1 + \frac{y}{x})}^{2}} + \frac{1}{\frac{y}{x} + 1}] = - \log_{e} x + C$

$\Rightarrow \log_{e}$ $| y + x |$ $+ 2$ $[\frac{- x^{2}}{{(y + x)}^{2}} + \frac{x}{y + x}]$ $= C$

$\Rightarrow \log_{e}$ $| y + x |$ $+ \frac{2 x y}{{(y + x)}^{2}} = C$

$We have y (1) = 0 \Rightarrow C = 0 .$

$\Rightarrow \log_{e}$ $| x + y |$ $+ \frac{2 x y}{{(x + y)}^{2}} = 0$

Q 78 :

Let a differentiable function $f$ satisfy $f (x) + \int_{3}^{x} \frac{f (t)}{t} d t = \sqrt{x + 1}, x \geq 3 .$ Then $12 f (8)$ is equal to: [2023]

19
17
1
34

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(2)

$We have, f (x) + \int_{3}^{x} \frac{f (t)}{t} d t = \sqrt{x + 1}, x \geq 3$

On differentiating, we get

$f' (x) + \frac{f (x)}{x} = \frac{1}{2 \sqrt{x + 1}}$ $\Rightarrow \frac{d y}{d x} + \frac{1}{x} y = \frac{1}{2 \sqrt{x + 1}}$

$I . F . = e^{\int \frac{1}{x} d x} = e^{\ln x} = x$

Solution; $x y = \int \frac{x}{2 \sqrt{x + 1}} d x$

Put $x + 1 = t^{2};$ $d x = 2 t d t$

$x y = \int \frac{(t^{2} - 1) 2 t d t}{2 t} = \int (t^{2} - 1) d t = \frac{t^{3}}{3} - t + C$

$= \frac{{(\sqrt{x + 1})}^{3}}{3} - \sqrt{x + 1} + C$

At $x = 3, y = 2$

$∴$ $3 \times 2 = \frac{{(\sqrt{3 + 1})}^{3}}{3} - \sqrt{3 + 1} + C \Rightarrow$ $6 = \frac{8}{3} - 2 + C \Rightarrow C = \frac{16}{3}$

$∴$ $f (x) = \frac{{(\sqrt{x + 1})}^{3}}{3 x} - \frac{\sqrt{x + 1}}{x} + \frac{16}{3 x}$

Now, $12 f (8) = 12 [\frac{{(\sqrt{8 + 1})}^{3}}{3 \times 8} - \frac{\sqrt{8 + 1}}{8} + \frac{16}{3 \times 8}] = 17$

Q 79 :

Let $y = y (x)$ be the solution of the differential equation $(3 y^{2} - 5 x^{2}) y d x + 2 x (x^{2} - y^{2}) d y = 0$ such that $y (1) = 1$ . Then $| (y {(2))}^{3} - 12 y (2) |$ is equal to [2023]

$64$
$32 \sqrt{2}$
$32$
$16 \sqrt{2}$

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(2)

$Given, (3 y^{2} - 5 x^{2}) y d x + 2 x (x^{2} - y^{2}) d y = 0$

$\Rightarrow \frac{d y}{d x} = \frac{y (5 x^{2} - 3 y^{2})}{2 x (x^{2} - y^{2})}$

It is a homogeneous differential equation.

Put $y = m x \Rightarrow$ $\frac{d y}{d x} = m + x \frac{d m}{d x}$

$\Rightarrow m + x \frac{d m}{d x} = \frac{m (5 - 3 m^{2})}{2 (1 - m^{2})}$ $\Rightarrow \frac{x d m}{d x} = \frac{m (5 - 3 m^{2})}{2 (1 - m^{2})} - m$

$\Rightarrow x \frac{d m}{d x} = \frac{(5 - 3 m^{2}) m - 2 m (1 - m^{2})}{2 (1 - m^{2})}$

$\Rightarrow \frac{d x}{x} = \frac{2 (m^{2} - 1)}{m (m^{2} - 3)} d m$ $\Rightarrow \frac{d x}{x} = [\frac{2}{m} - \frac{\frac{4}{3}}{m} + \frac{\frac{4 m}{3}}{m^{2} - 3}] d m$

Integrating both sides $\int \frac{d x}{x} = \int \frac{(\frac{2}{3})}{m} d m + \int \frac{2}{3} (\frac{2 m}{m^{2} - 3}) d m$

$\Rightarrow \ln$ $| x |$ $= \frac{2}{3} \ln$ $| m |$ $+ \frac{2}{3} \ln$ $| m^{2} - 3 |$ $+ C$

$\Rightarrow \ln$ $| x |$ $= \frac{2}{3} \ln$ $| \frac{y}{x} |$ $+ \frac{2}{3} \ln$ $| {(\frac{y}{x})}^{2} - 3 |$ $+ C$

Put $(x = 1, y = 1)$ , $\ln (1) = \frac{2}{3} \ln (1) + \frac{2}{3} \ln$ $| (1 - 3) |$ $+ C$

$C = - \frac{2}{3} \ln (2) \Rightarrow \ln$ $| x |$ $= \frac{2}{3} \ln$ $| \frac{y}{x} |$ $+ \frac{2}{3} \ln$ $| {(\frac{y}{x})}^{2} - 3 |$ $- \frac{2}{3} \ln (2)$

$\Rightarrow (\frac{y}{x}) [{(\frac{y}{x})}^{2} - 3] = 2 (x^{3 / 2})$

Put $x = 2$ to get $y (2)$ , $\Rightarrow y (y^{2} - 12) = 4 \times 2 \times 2 \times 2 \sqrt{2}$

$\Rightarrow y^{3} - 12 y = 32 \sqrt{2} \Rightarrow$ $| y^{3} (2) - 12 y (2) |$ $= 32 \sqrt{2}$

Q 80 :

Let $y = y (x)$ be a solution of the differential equation $(x \cos x) d y + (x y \sin x + y \cos x - 1) d x = 0, 0 < x < \frac{π}{2} .$ If $\frac{π}{3} y (\frac{π}{3}) = \sqrt{3},$ then $| \frac{π}{6} y'' (\frac{π}{6}) + 2 y' (\frac{π}{6}) |$ is equal to ______ . [2023]

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(2)

$(x \cos x) d y + (x y \sin x + y \cos x - 1) d x = 0,$

$0 < x < \frac{π}{2}$ $\Rightarrow \frac{d y}{d x} + \frac{(x \sin x + \cos x)}{x \cos x} y = \frac{1}{x \cos x}$

$I.F. = e^{\int (\frac{x \sin x + \cos x}{x \cos x}) d x} = e^{\int (\tan x + \frac{1}{x}) d x} = e^{\ln | \sec x | + \ln | x |} = e^{\ln | x \sec x |}$

$I.F. = x \sec x$

$y \times x \sec x = \int \frac{x \sec x}{x \cos x} d x \Rightarrow y \times x$ $s e c x = \tan x + C$

Given, $y (\frac{π}{3}) = \frac{3 \sqrt{3}}{π}$

$So, \frac{3 \sqrt{3}}{π} \times \frac{π}{3} \cdot 2 = \sqrt{3} + C$ $\Rightarrow 2 \sqrt{3} = \sqrt{3} + C \Rightarrow C = \sqrt{3}$

$∴$ $y x \sec x = \tan x + \sqrt{3} \Rightarrow x y = \sin x + \sqrt{3} \cos x$

$\Rightarrow x y' + y = \cos x - \sqrt{3} \sin x$

$\Rightarrow x y'' + 2 y' = - \sin x - \sqrt{3} \cos x$

$∴$ $| \frac{π}{6} \cdot y'' (\frac{π}{6}) + 2 y' (\frac{π}{6}) |$ $=$ $| - \frac{1}{2} - \frac{3}{2} |$ $= 2$

Topic Question Set

Q 71 :

Let $y = y (x)$ be the solution of the differential equation $(x^{2} - 3 y^{2}) d x + 3 x y d y = 0, y (1) = 1 .$ Then $6 y^{2} (e)$ is equal to [2023]

Q 72 :

Let $y = y (x)$ be the solution curve of the differential equation $\frac{d y}{d x} = \frac{y}{x} (1 + x y^{2} (1 + \log_{e} x)), x > 0, y (1) = 3 .$ Then, $\frac{y^{2} (x)}{9}$ is equal to [2023]

Q 73 :

Let $y = y (t)$ be a solution of the differential equation $\frac{d y}{d t} + α y = γ e^{- β t}$ where, $α > 0, β > 0$ and $γ > 0$ . Then $\lim_{t \to \infty} y (t)$ [2023]

Q 74 :

Let $y = f (x)$ be the solution of the differential equation $y (x + 1) d x - x^{2} d y = 0, y (1) = e .$ Then $\lim_{x \to 0^{+}} f (x)$ is equal to [2023]

Q 75 :

Let $y = y (x)$ be the solution of the differential equation $x \log_{e} x \frac{d y}{d x} + y = x^{2} \log_{e} x, (x > 1) .$ If $y (2) = 2$ , then $y (e)$ is equal to [2023]

Q 76 :

Let the solution curve $y = y (x)$ of the differential equation $\frac{d y}{d x} - \frac{3 x^{5} \tan^{- 1} (x^{3})}{{(1 + x^{6})}^{3 / 2}} y = 2 x \exp {\frac{x^{3} - \tan^{- 1} x^{3}}{\sqrt{(1 + x^{6})}}}$ pass through the origin. Then $y (1)$ is equal to [2023]

Q 77 :

The solution of the differential equation $\frac{d y}{d x} = - (\frac{x^{2} + 3 y^{2}}{3 x^{2} + y^{2}}), y (1) = 0$ is [2023]

Q 78 :

Let a differentiable function $f$ satisfy $f (x) + \int_{3}^{x} \frac{f (t)}{t} d t = \sqrt{x + 1}, x \geq 3 .$ Then $12 f (8)$ is equal to: [2023]

Q 79 :

Let $y = y (x)$ be the solution of the differential equation $(3 y^{2} - 5 x^{2}) y d x + 2 x (x^{2} - y^{2}) d y = 0$ such that $y (1) = 1$ . Then $| (y {(2))}^{3} - 12 y (2) |$ is equal to [2023]

Q 80 :

Let $y = y (x)$ be a solution of the differential equation $(x \cos x) d y + (x y \sin x + y \cos x - 1) d x = 0, 0 < x < \frac{π}{2} .$ If $\frac{π}{3} y (\frac{π}{3}) = \sqrt{3},$ then $| \frac{π}{6} y'' (\frac{π}{6}) + 2 y' (\frac{π}{6}) |$ is equal to ______ . [2023]

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