Question

Let $y=y(x)$ be solution of the differential equation ${\log }_e\left(\frac{dy}{dx}\right)=3x+4y$, 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)

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