Let y = y (x) be solution of the differential equation _ e...

Given,

$\log_{e}\left(\dfrac{dy}{dx}\right) = 3x + 4y$

$⇒ \dfrac{dy}{dx} = {e}^{3x} \cdot {e}^{4y}$

$⇒ ∫ {e}^{- 4y}dy = ∫ {e}^{3x}dx$

$⇒ \dfrac{{e}^{- 4y}}{- 4} = \dfrac{{e}^{3x}}{3} + C$

Given,

$y\left(0\right) = 0$

So,

$- \dfrac{1}{4} - \dfrac{1}{3} = C ⇒ C = - \dfrac{7}{12}$

So, the particular solution is

$\dfrac{{e}^{- 4y}}{- 4} = \dfrac{{e}^{3x}}{3} - \dfrac{7}{12}$

$⇒ {e}^{- 4y} = \dfrac{4{e}^{3x} - 7}{- 3}$

$⇒ {e}^{4y} = \dfrac{3}{7 - 4{e}^{3x}} ⇒ 4y = \ln \left(\dfrac{3}{7 - 4{e}^{3x}}\right)$

$4y = ℓn\left(\dfrac{3}{6}\right)$ when $x = - \dfrac{2}{3}ℓn2$

$⇒ y = \dfrac{1}{4}ℓn\left(\dfrac{1}{2}\right)$

$⇒ y = - \dfrac{1}{4}ℓn2$

So, $\alpha = - \dfrac{1}{4}$