Question

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 $y\left(-\frac{2}{3} \log _{e} 2\right)=\alpha \log _{e} 2$, then the value of $\alpha$ is equal to:

• A. $-\frac{1}{4}$
• B. $\frac{1}{4}$
• C. $2$
• D. $-\frac{1}{2}$

Answer 1

Answer: (A)

$\frac{d y}{d x}=e^{3 x} \cdot e^{4 y}$
$\Rightarrow \int e^{-4 y} d y=\int e^{3 x} d x$
$\frac{e^{-4 y}}{-4}=e^{3 x} 3+C$
$\Rightarrow-\frac{1}{4}-\frac{1}{3}=C$
$\Rightarrow C=-\frac{7}{12}$
$\frac{e^{-4 y}}{-4}=\frac{e^{3 x}}{3}-\frac{7}{12}$
$\Rightarrow e^{-4 y}=\frac{4 e^{3 x}-7}{-3}$
$e^{4 y}=\frac{3}{7-4 e^{3 x}}$
$\Rightarrow 4 y=\ln \left(\frac{3}{7-4 e^{3 x}}\right)$
$4 y=\ln \left(\frac{3}{6}\right)$ when $x=-\frac{2}{3} \ln 2$
$y=\frac{1}{4} \ln \left(\frac{1}{2}\right)=-\frac{1}{4} \ln 2$