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(B) Given the differential equation $\log _{e}\left(\frac{d y}{d x}\right)=3 x+4 y$,we can write it as $\frac{d y}{d x}=e^{3 x+4 y}=e^{3 x} \cdot e^{4

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$-\frac{1}{4}$

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(B) Given the differential equation $\log _{e}\left(\frac{d y}{d x}\right)=3 x+4 y$,we can write it as $\frac{d y}{d x}=e^{3 x+4 y}=e^{3 x} \cdot e^{4 y}$.Separating the variables,we get $e^{-4 y} d y=e^{3 x} d x$.Integrating both sides,$\int e^{-4 y} d y=\int e^{3 x} d x$,which gives $-\frac{1}{4} e^{-4 y}=\frac{1}{3} e^{3 x}+C$.Using the initial condition $y(0)=0$,we substitute $x=0$ and $y=0$: $-\frac{1}{4} e^{0}=\frac{1}{3} e^{0}+C \Rightarrow -\frac{1}{4}=\frac{1}{3}+C \Rightarrow C=-\frac{1}{4}-\frac{1}{3}=-\frac{7}{12}$.Thus,$-\frac{1}{4} e^{-4 y}=\frac{1}{3} e^{3 x}-\frac{7}{12}$.Multiplying by $-12$,we get $3 e^{-4 y} = 7 - 4 e^{3 x}$,so $e^{-4 y} = \frac{7 - 4 e^{3 x}}{3}$.Taking the reciprocal,$e^{4 y} = \frac{3}{7 - 4 e^{3 x}}$,so $4 y = \log _{e} \left(\frac{3}{7 - 4 e^{3 x}}\right)$.For $x = -\frac{2}{3} \log _{e} 2$,we have $e^{3 x} = e^{3 \left(-\frac{2}{3} \log _{e} 2\right)} = e^{-2 \log _{e} 2} = e^{\log _{e} (2^{-2})} = 2^{-2} = \frac{1}{4}$.Substituting this into the equation for $4y$: $4 y = \log _{e} \left(\frac{3}{7 - 4(1/4)}\right) = \log _{e} \left(\frac{3}{7 - 1}\right) = \log _{e} \left(\frac{3}{6}\right) = \log _{e} \left(\frac{1}{2}\right) = -\log _{e} 2$.Therefore,$y = -\frac{1}{4} \log _{e} 2$. Comparing this with $y = \alpha \log _{e} 2$,we get $\alpha = -\frac{1}{4}$.

Let $y=y(x)$ be the solution of the differential equation $\log _{e}\left(\frac{d y}{d x}\right)=3 x+4 y$,with $y(0)=0$. If $y\left(-\frac{2}{3} \log _{e} 2\right)=\alpha \log _{e} 2$,then the value of $\alpha$ is equal to:

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