If the solution of the differential equation ( 2x + 3y - 2 ) dx + (...

Given: $( 2x + 3y - 2 ) dx + ( 4x + 6y - 7 ) dy = 0$

$⇒ \dfrac{dy}{dx} = \dfrac{- \left(2x + 3y - 2\right)}{\left(4x + 6y - 7\right)} . . . \left(i\right)$

Let, $2x + 3y - 2 = t$

$⇒ 2 + 3\dfrac{dy}{dx} = \dfrac{dt}{dx} . . . \left(ii\right)$

Also, $4x + 6y - 4 = 2t$

$⇒ 4x + 6y - 7 = 2t - 3 . . . \left(iii\right)$

Using \left(i\right), \left(ii\right)\text{and}\left(iii\right)$,</p><p>$⇒ \dfrac{1}{3}\left(\dfrac{dt}{dx} - 2\right) = \dfrac{- t}{2t - 3}$</p><p>$⇒ \dfrac{dt}{dx} - 2 = \dfrac{- 3t}{2t - 3}$</p><p>$⇒ \dfrac{dt}{dx} = \dfrac{- 3t}{2t - 3} + 2$</p><p>$⇒ \dfrac{dt}{dx} = \dfrac{- 3t + 4t - 6}{2t - 3}$</p><p>$⇒ \dfrac{dt}{dx} = \dfrac{t - 6}{2t - 3}$</p><p>$⇒ ∫ \dfrac{2t - 3}{t - 6}dt = ∫ dx$</p><p>$⇒ ∫ \left(\dfrac{2t - 12}{t - 6} + \dfrac{9}{t - 6}\right) \cdot dt = x + c$</p><p>$⇒ 2t + 9\log ( t - 6 ) = x + c$</p><p>$⇒ 2 ( 2x + 3y - 2 ) + 9\log ( 2x + 3y - 8 ) = x + c . . . \left(iv\right)$</p><p>It is given that$y ( 0 ) = 3$.</p><p>$⇒ 2 ( 0 + 9 - 2 ) + 9\log ( 0 + 9 - 8 ) = 0 + c$</p><p>$⇒ 14 + 9\log ( 1 ) = c$</p><p>$⇒ c = 14$</p><p>Using \left(iv\right)$,

$⇒ 4x + 6y - 4 + 9\log ( 2x + 3y - 8 ) = x + 14$

$⇒ 3x + 6y + 9\log ( 2x + 3y - 8 ) = 18$

$⇒ x + 2y + 3\log ( 2x + 3y - 8 ) = 6$

Comparing with $\alphax + \betay + 3\log_{e} | 2x + 3y - \gamma | = 6$.

$⇒ \alpha = 1, \beta = 2, \gamma = 8$

$⇒ \alpha + 2\beta + 3\gamma = 1 + 4 + 24$

$⇒ \alpha + 2\beta + 3\gamma = 29$