Question

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

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

Answer 1

The correct option is B $-\frac{1}{4}$

Given: $\ln \left(\frac{dy}{dx}\right)=3x+4y$
$\Rightarrow \frac{dy}{dx}=e^{3x+4y}$
$\Rightarrow e^{-4y}dy=e^{3x}dx$
$\Rightarrow \int e^{-4y}dy=\int e^{3x}dx$
$\Rightarrow \frac{e^{-4y}}{-4}=\frac{e^{3x}}{3}+C$
$\because y\left(0\right)=0\Rightarrow C=-\frac{7}{12}$
$\therefore e^{-4y}=\frac{7}{3}-\frac{4e^{3x}}{3}$
$\Rightarrow e^{4y}=\frac{3}{7-4e^{3x}}$
$\Rightarrow y=\frac{1}{4}\ln \left(\frac{3}{7-4e^{3x}}\right)$
At $x=-\frac{2}{3}\ln 2,$ we have
$y(-\frac{2}{3}\ln 2)=\frac{1}{4}\ln \left(\frac{3}{6}\right)=-\frac{1}{4}\ln 2$
$\therefore \alpha =-\frac{1}{4}$