If the solution of the differential equation (2x + 3y – 2)dx + (4x + 6y – 7)dy = 0, y(0) = 3, is , then is equal to __________. [2024]

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Solution

Q.
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]

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

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