Let $y=y(x)$ be solution of the differential equation $\log _{e}\left(\frac{d y}{d x}\right)=3 x+4 y$, with $y(0)=0$
If $\mathrm{y}\left(-\frac{2}{3} \log _{\mathrm{e}} 2\right)=\alpha \log _{\mathrm{e}} 2$, then the value of $\alpha$ is equal to:
Correct Option: 1
$\frac{d y}{d x}=e^{3 x} \cdot e^{4 y} \Rightarrow \int e^{-4 y} d y=\int e^{3 x} d x$
$\frac{\mathrm{e}^{-4 \mathrm{y}}}{-4}=\frac{\mathrm{e}^{3 \mathrm{x}}}{3}+\mathrm{C} \Rightarrow-\frac{1}{4}-\frac{1}{3}=\mathrm{C} \Rightarrow \mathrm{C}=-\frac{7}{12}$
$\frac{\mathrm{e}^{-4 \mathrm{y}}}{-4}=\frac{\mathrm{e}^{3 \mathrm{x}}}{3}-\frac{7}{12} \Rightarrow \mathrm{e}^{-4 \mathrm{y}}=\frac{4 \mathrm{e}^{3 \mathrm{x}}-7}{-3}$
$e^{4 y}=\frac{3}{7-4 e^{3 x}} \Rightarrow 4 y=\ln \left(\frac{3}{7-4 e^{3 x}}\right)$
$4 \mathrm{y}=\ell \mathrm{n}\left(\frac{3}{6}\right)$ when $\mathrm{x}=-\frac{2}{3} \ell \mathrm{n} 2$
$y=\frac{1}{4} \ell \mathrm{n}\left(\frac{1}{2}\right)=-\frac{1}{4} \ell \mathrm{n} 2$
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