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

Q 21 :
The temperature T(t) of a body at time t = 0 is 160º F and it decreases continuously as per the differential equation $\frac{d T}{d t} = - K (T - 80)$ , where K is positive constant. If T(15) = 120º F, then T(45) is equal to [2024]

80º F
90º F
85º F
95º F

1

2

3

4

(2)

Given differential equation is

$\frac{d T}{d t} = - K (T - 80)$

Now, $\frac{d T}{T - 80} = - K d t$

On integrating, we get

$\int \frac{d T}{T - 80} = - K \int d t$

$\Rightarrow \log | T - 80 | = - K t + C$ ... (i)

When t = 0, T = 160º F

$∴ \log | 160 - 80 | = - K \times 0 + C$

$\Rightarrow C = \log (80)$

From (i),

log |T – 80| = – Kt + log (80) ...(ii)

Now, when t = 15, T = 120º F

$∴$ From (ii),

$\log (120 - 80) = - K \times 15 + \log (80)$

$\Rightarrow 15 K = \log (\frac{80}{40}) = \log (2) ∴ k = \frac{\log (2)}{15}$

When t = 45, then from (ii),

$\log (T - 80) = \frac{- \log (2)}{15} \times 45 + \log (80)$

$\Rightarrow \log (T - 80) = - 3 \log (2) + \log (80) = \log (\frac{80}{8}) = \log (10)$

$\Rightarrow \log (T - 80) = 1 \Rightarrow T - 80 = 10^{1}$

$\Rightarrow T = 90 ° F$

Q 22 :
Let the solution y = y(x) of the differential equation $\frac{d y}{d x} - y = 1 + 4 \sin x$ satisfy y( $π$ ) = 1. Then $y (\frac{π}{2}) + 10$ is equal to __________. [2024]

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(7)

Given, $\frac{d y}{d x} - y = 1 + 4 \sin x$ , which is a linear differential equation.

$∴$ Integrating factor = $= e^{- \int d x} = e^{- x}$

Hence, solution is $y e^{- x} = \int (1 + 4 \sin x) e^{- x} d x$

$= - e^{- x} + 2 e^{- x} (- \sin x - \cos x) + C$ ... (i)

y( $π$ ) = 1 i.e., at x = $π$ , y = 1

$∴$ From (i), we get

$(1) e^{- π} = - e^{- π} + 2 e^{- π} (- \sin π - \cos π) + C$

$\Rightarrow e^{- π} = - e^{- π} + 2 e^{- π} + C \Rightarrow C = 0$

Hence, y(x) = –1 – 2(sin x + cos x)

Now, $y (\frac{π}{2}) + 10 = - 1 - 2 (\sin \frac{π}{2} + \cos \frac{π}{2}) + 10$

= –1 – 2(1) + 10 = 7.

Q 23 :
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]

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(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, $γ + δ = 67 - 36 = 31$ .

Q 24 :
Let f be a differentiable function in the interval (0, $\infty$ ) such that f(1) =1 and $\lim_{t \to x} \frac{t^{2} f (x) - x^{2} f (t)}{t - x} = 1$ for each x > 0. Then 2f(2) + 3f(3) is equal to __________. [2024]

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(24)

We Have $\underset{t \to x}{l t} \frac{t^{2} f (x) - x^{2} f (t)}{t - x} = 1 (\frac{0}{0} form)$

$\Rightarrow \underset{t \to x}{l t} \frac{2 t f (x) - x^{2} f' (t)}{1} = 1 [L' Hospital Rule]$

$\Rightarrow 2 x f (x) - x^{2} f' (x) = 1 \Rightarrow f' (x) - \frac{2}{x} f (x) = \frac{- 1}{x^{2}}$

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

$∴ f (x) \frac{1}{x^{2}} = \int \frac{- 1}{x^{2}} \cdot \frac{1}{x^{2}} d x + C \Rightarrow f (x) \frac{1}{x^{2}} = \frac{1}{3 x^{3}} + C$

Now, f(1) = 1

$\Rightarrow \frac{2}{3} = C \Rightarrow f (x) = \frac{1}{3 x} + \frac{2 x^{2}}{3}$

Now, $2 f (2) + 3 f (3) = 2 [\frac{1}{6} + \frac{8}{3}] + 3 [\frac{1}{9} + \frac{18}{3}]$

$= \frac{1}{3} + \frac{16}{3} + \frac{1}{3} + 18 = 6 + 18 = 24$ .

Q 25 :
Let y = y(x) be the solution of the differential equation $\frac{d y}{d x} + \frac{2 x}{{(1 + x^{2})}^{2}} y = x e^{\frac{1}{(1 + x^{2})}}$ ; y(0) = 0. Then the area enclosed by the curve $f (x) = y (x) e^{- \frac{1}{(1 + x^{2})}}$ and the line y – x = 4 is __________. [2024]

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(18)

We have, $\frac{d y}{d x} + \frac{2 x}{{(1 + x^{2})}^{2}} y = x e^{\frac{1}{(1 + x^{2})}}$

$IF = e^{\int \frac{2 x}{{(1 + x^{2})}^{2}} d x} = e^{\frac{- 1}{1 + x^{2}}}$

$∴$ General solution of given differential equation is,

$y \cdot e^{\frac{- 1}{1 + x^{2}}} = \int x e^{\frac{1}{1 + x^{2}}} \cdot e^{\frac{- 1}{1 + x^{2}}} d x$

$\Rightarrow y e^{\frac{- 1}{1 + x^{2}}} = \frac{x^{2}}{2} + c$

Since y(0) = 0

$\Rightarrow 0 = 0 + c$

$\Rightarrow c = 0$

$∴ y (x) = \frac{x^{2}}{2} e^{\frac{1}{1 + x^{2}}}$

$\Rightarrow f (x) = \frac{x^{2}}{2}$

So, intersection of curve $y = f (x) = \frac{x^{2}}{2}$ and y = x + 4 is given by $\frac{x^{2}}{2} = x + 4$

$\Rightarrow x^{2} - 2 x - 8 = 0 \Rightarrow (x - 4) (x + 2) = 0$

$\Rightarrow x = 4, - 2 \Rightarrow y = 8, 2$

$∴$ Required area = $\int_{- 2}^{4} ((x + 4) - \frac{x^{2}}{2}) d x$

= ${[\frac{x^{2}}{2} + 4 x - \frac{x^{3}}{6}]}_{- 2}^{4} = [8 + 16 - \frac{32}{3} - 2 + 8 - \frac{8}{6}]$

= 30 – 12 = 18.

Q 26 :
If the solution y(x) of the given differential equation $(e^{y} + 1) \cos x d x + e^{y} \sin x d y = 0$ passes through the point $(\frac{π}{2}, 0)$ , then the value of $e^{y (\frac{π}{6})}$ is equal to _________. [2024]

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(3)

Givem, $(e^{y} + 1) \cos x d x + e^{y} \sin x d y = 0$

$\Rightarrow d (e^{y} \sin x) + \cos x d x = 0 \Rightarrow e^{y} \sin x + \sin x = C$

The curve passes through $(\frac{π}{2}, 0)$

$\Rightarrow C = 2 \Rightarrow e^{y} \sin x + \sin x = 2$

$\Rightarrow (e^{y (\frac{π}{6})} \frac{1}{2} + \frac{1}{2}) = 2 \Rightarrow e^{y (\frac{π}{6})} + 1 = 4 \Rightarrow e^{y (\frac{π}{6})} = 3$ .

Q 27 :
Let $α | x | = | y | e^{x y - β}$ , $α, β \in N$ be the solution of the differential equation xdy – ydx + xy (xdy + ydx) = 0, y(1) = 2. Then $α + β$ is equal to __________. [2024]

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(4)

We have, $α | x | = | y | e^{x y - β}$

and $\frac{x d y - y d x}{y^{2}} + \frac{x y (x d y + y d x)}{y^{2}} = 0$

$\Rightarrow - d (\frac{x}{y}) + \frac{x}{y} d (x y) = 0 \Rightarrow \int d (x y) = \int \frac{d (\frac{x}{y})}{x / y}$

$\Rightarrow x y = \ln | \frac{x}{y} | + \ln c \Rightarrow x y = \ln (| \frac{x}{y} | \cdot c)$ ... (i)

$∵$ y(1) = 2, i.e., x = 1, y = 2

$∴ From (i), 2 = \ln (| \frac{1}{2} | c) \Rightarrow c = 2 e^{2}$

$∴ x y = \ln (| \frac{x}{y} | \cdot 2 e^{2})$

$\Rightarrow e^{x y} = \frac{| x |}{| y |} \cdot 2 e^{2} \Rightarrow 2 | x | = | y | e^{x y - 2}$

Also, solution of equation is $α | x | = | y | e^{x y - β}$ (Given)

On comparing, $α = 2, β = 2$ . Hence, $α + β = 4$ .

Q 28 :
If x = x(t) is the solution of the differential equation $(t + 1) d x = (2 x + {(t + 1)}^{4}) d t$ , x(0) = 2, then x(1) equals __________. [2024]

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(14)

We have, $(t + 1) d x = (2 x + {(t + 1)}^{4}) d t$

$\Rightarrow \frac{d x}{d t} = \frac{2 x}{t + 1} + {(t + 1)}^{3}; P = \frac{- 2}{t + 1}, Q = {(t + 1)}^{3}$

So, $I . F . = e^{\int P d t} = e^{\int \frac{- 2}{t + 1} d t} = e^{- 2 \log (t + 1)} = {(t + 1)}^{- 2}$

The Solution of given differential equation is given by $x (I . F .) = \int Q \cdot (I . F .) d t + C$

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

$\Rightarrow \frac{x}{{(t + 1)}^{2}} = \int (t + 1) d t + C \Rightarrow x = {(t + 1)}^{2} [\frac{t^{2}}{2} + t + C]$

Putting x(0) = 2, we get

$2 = {(1)}^{2} (0 + 0 + C) \Rightarrow C = 2$

$∴ x = {(t + 1)}^{2} (\frac{t^{2}}{2} + t + 2)$

At $t = 1, x = {(2)}^{2} (\frac{1}{2} + 1 + 2) = 4 (\frac{7}{2}) = 14$ .

Q 29 :
If $\frac{d x}{d y} = \frac{1 + x - y^{2}}{y}$ , x(1) = 1, then 5x(2) is equal to __________. [2024]

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(5)

We have, $\frac{d x}{d y} = \frac{1 + x - y^{2}}{y} \Rightarrow \frac{d x}{d y} - \frac{x}{y} = \frac{1 - y^{2}}{y}$

Now, $I . F . = e^{- \int \frac{1}{y} d y} = e^{- \log y} = \frac{1}{y}$

So, solution is given by,

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

$\Rightarrow x = - 1 - y^{2} + C y$

$∵ x (1) = 1 \Rightarrow 1 = - 1 - 1 + C \Rightarrow C = 3$

$\Rightarrow x = - 1 - y^{2} + 3 y$

$∴$ 5x(2) = 5(–1 – 4 + 6) = 5.

Q 30 :
If the solution of the differential equation (2x + 3y – 2)dx + (4x + 6y – 7)dy = 0, y(0) = 3, is $α x + β y + 3 \log_{e} | 2 x + 3 y - γ | = 6$ , then $α + 2 β + 3 γ$ is equal to __________. [2024]

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(29)

(2x + 3y – 2)dx + (4x + 6y – 7)dy = 0

$\frac{d y}{d x} = \frac{- (2 x + 3 y - 2)}{(4 x + 6 y - 7}$ ... (i)

Let 2x + 3y = t

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

Putting in (i), we get

$\frac{1}{3} [\frac{d t}{d x}] - \frac{2}{3} = \frac{- (t - 2)}{2 t - 7}$

$\Rightarrow \frac{1}{3} [\frac{d t}{d x}] = \frac{2}{3} - \frac{t - 2}{2 t - 7} = \frac{4 t - 14 - 3 t + 6}{3 (2 t - 7)}$

$\Rightarrow \frac{d t}{d x} = \frac{t - 8}{2 t - 7} \Rightarrow d x = \frac{2 t - 7}{t - 8} d t$

$\Rightarrow \int d x + c = \int 2 d t + 9 \int \frac{d t}{t - 8}$

$\Rightarrow x + c = 2 t + 9 \ln | t - 8 |$

$\Rightarrow 2 (2 x + 3 y) + 9 \ln | 2 x + 3 y - 8 | = x + c$

which is the solution of given differential equation.

Now, $y (0) = 3 \Rightarrow 2 (9) + 9 \ln | 1 | = c \Rightarrow c = 18$

Putting the value of c in (i), we get

$4 x + 6 y + 9 \ln | 2 x + 3 y - 8 | = x + 18$

$\Rightarrow 3 x + 6 y + 9 \ln | 2 x + 3 y - 8 | = 18$

$\Rightarrow x + 2 y + 3 \ln | 2 x + 3 y - 8 | = 6$

$∴ α = 1 . β = 2, γ = 8$

Therefore, $α + 2 β + 3 γ = 1 + 2 (2) + 3 (8) = 1 + 4 + 24 = 29$ .

Topic Question Set

Q 21 :
The temperature T(t) of a body at time t = 0 is 160º F and it decreases continuously as per the differential equation $\frac{d T}{d t} = - K (T - 80)$ , where K is positive constant. If T(15) = 120º F, then T(45) is equal to [2024]

Q 22 :
Let the solution y = y(x) of the differential equation $\frac{d y}{d x} - y = 1 + 4 \sin x$ satisfy y( $π$ ) = 1. Then $y (\frac{π}{2}) + 10$ is equal to __________. [2024]

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Q 24 :
Let f be a differentiable function in the interval (0, $\infty$ ) such that f(1) =1 and $\lim_{t \to x} \frac{t^{2} f (x) - x^{2} f (t)}{t - x} = 1$ for each x > 0. Then 2f(2) + 3f(3) is equal to __________. [2024]

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Q 25 :
Let y = y(x) be the solution of the differential equation $\frac{d y}{d x} + \frac{2 x}{{(1 + x^{2})}^{2}} y = x e^{\frac{1}{(1 + x^{2})}}$ ; y(0) = 0. Then the area enclosed by the curve $f (x) = y (x) e^{- \frac{1}{(1 + x^{2})}}$ and the line y – x = 4 is __________. [2024]

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Q 26 :
If the solution y(x) of the given differential equation $(e^{y} + 1) \cos x d x + e^{y} \sin x d y = 0$ passes through the point $(\frac{π}{2}, 0)$ , then the value of $e^{y (\frac{π}{6})}$ is equal to _________. [2024]

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Q 27 :
Let $α | x | = | y | e^{x y - β}$ , $α, β \in N$ be the solution of the differential equation xdy – ydx + xy (xdy + ydx) = 0, y(1) = 2. Then $α + β$ is equal to __________. [2024]

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Q 28 :
If x = x(t) is the solution of the differential equation $(t + 1) d x = (2 x + {(t + 1)}^{4}) d t$ , x(0) = 2, then x(1) equals __________. [2024]

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Q 29 :
If $\frac{d x}{d y} = \frac{1 + x - y^{2}}{y}$ , x(1) = 1, then 5x(2) is equal to __________. [2024]

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Q 30 :
If the solution of the differential equation (2x + 3y – 2)dx + (4x + 6y – 7)dy = 0, y(0) = 3, is $α x + β y + 3 \log_{e} | 2 x + 3 y - γ | = 6$ , then $α + 2 β + 3 γ$ is equal to __________. [2024]

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Topic Question Set

Q 21 : The temperature T(t) of a body at time t = 0 is 160º F and it decreases continuously as per the differential equation dTdt = - K (T - 80), where K is positive constant. If T(15) = 120º F, then T(45) is equal to [2024]

Q 22 : Let the solution y = y(x) of the differential equation dydx–y=1+4 sin x satisfy y(π) = 1. Then y(π2)+10 is equal to __________. [2024]

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Q 23 : Let y = y(x) be the solution of the differential equation (x+y+2)2dx=dy, y(0) = –2. Let the maximum and minimum values of the function y = y(x) in [0,π3] be α and β, respectively. If (3α+π)2+β2=γ+δ3, γ, δ∈Z, then γ+δ equals __________. [2024]

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Q 24 : Let f be a differentiable function in the interval (0, ∞) such that f(1) =1 and limt→xt2f(x)–x2f(t)t–x=1 for each x > 0. Then 2f(2) + 3f(3) is equal to __________. [2024]

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Q 25 : Let y = y(x) be the solution of the differential equation dydx+2x(1+x2)2y=xe1(1+x2); y(0) = 0. Then the area enclosed by the curve f(x)=y(x)e–1(1+x2) and the line y – x = 4 is __________. [2024]

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Q 26 : If the solution y(x) of the given differential equation (ey+1) cos xdx+ey sin xdy = 0 passes through the point (π2,0), then the value of ey(π6) is equal to _________. [2024]

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Q 27 : Let α|x|=|y|exy–β, α, β∈N be the solution of the differential equation xdy – ydx + xy (xdy + ydx) = 0, y(1) = 2. Then α + β is equal to __________. [2024]

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Q 28 : If x = x(t) is the solution of the differential equation (t+1)dx=(2x+(t+1)4)dt, x(0) = 2, then x(1) equals __________. [2024]

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Q 29 : If dxdy=1+x–y2y, x(1) = 1, then 5x(2) is equal to __________. [2024]

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Q 30 : If the solution of the differential equation (2x + 3y – 2)dx + (4x + 6y – 7)dy = 0, y(0) = 3, is αx+βy+3 loge |2x+3y–γ|=6, then α+2β+3γ is equal to __________. [2024]

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Q 21 :
The temperature T(t) of a body at time t = 0 is 160º F and it decreases continuously as per the differential equation $\frac{d T}{d t} = - K (T - 80)$ , where K is positive constant. If T(15) = 120º F, then T(45) is equal to [2024]

Q 22 :
Let the solution y = y(x) of the differential equation $\frac{d y}{d x} - y = 1 + 4 \sin x$ satisfy y( $π$ ) = 1. Then $y (\frac{π}{2}) + 10$ is equal to __________. [2024]

Q 24 :
Let f be a differentiable function in the interval (0, $\infty$ ) such that f(1) =1 and $\lim_{t \to x} \frac{t^{2} f (x) - x^{2} f (t)}{t - x} = 1$ for each x > 0. Then 2f(2) + 3f(3) is equal to __________. [2024]

Q 25 :
Let y = y(x) be the solution of the differential equation $\frac{d y}{d x} + \frac{2 x}{{(1 + x^{2})}^{2}} y = x e^{\frac{1}{(1 + x^{2})}}$ ; y(0) = 0. Then the area enclosed by the curve $f (x) = y (x) e^{- \frac{1}{(1 + x^{2})}}$ and the line y – x = 4 is __________. [2024]

Q 26 :
If the solution y(x) of the given differential equation $(e^{y} + 1) \cos x d x + e^{y} \sin x d y = 0$ passes through the point $(\frac{π}{2}, 0)$ , then the value of $e^{y (\frac{π}{6})}$ is equal to _________. [2024]

Q 27 :
Let $α | x | = | y | e^{x y - β}$ , $α, β \in N$ be the solution of the differential equation xdy – ydx + xy (xdy + ydx) = 0, y(1) = 2. Then $α + β$ is equal to __________. [2024]

Q 28 :
If x = x(t) is the solution of the differential equation $(t + 1) d x = (2 x + {(t + 1)}^{4}) d t$ , x(0) = 2, then x(1) equals __________. [2024]

Q 29 :
If $\frac{d x}{d y} = \frac{1 + x - y^{2}}{y}$ , x(1) = 1, then 5x(2) is equal to __________. [2024]

Q 30 :
If the solution of the differential equation (2x + 3y – 2)dx + (4x + 6y – 7)dy = 0, y(0) = 3, is $α x + β y + 3 \log_{e} | 2 x + 3 y - γ | = 6$ , then $α + 2 β + 3 γ$ is equal to __________. [2024]