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 :
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For differential equations of the form $\frac{dy}{dx} = f(ax+by)$, a substitution $u=ax+by$ can be useful. However, if the function can be written as $g(x)h(y)$, as in this case ($e^{3x}e^{4y}$), the variable separable method is the most direct approach.
The given differential equation is $\ln\left(\frac{dy}{dx}\right) = 3x+4y$.
Exponentiating both sides gives:
$\frac{dy}{dx} = e^{3x+4y} = e^{3x} \cdot e^{4y}$.
This is a variable separable differential equation.
$\frac{dy}{e^{4y}} = e^{3x} dx$.
$e^{-4y} dy = e^{3x} dx$.
Integrate both sides:
$\int e^{-4y} dy = \int e^{3x} dx$.
$\frac{e^{-4y}}{-4} = \frac{e^{3x}}{3} + C$, where C is the integration constant.
We are given the initial condition $y(0)=0$. Substitute $x=0, y=0$ to find C.
$\frac{e^{-4(0)}}{-4} = \frac{e^{3(0)}}{3} + C$.
$\frac{1}{-4} = \frac{1}{3} + C \implies C = -\frac{1}{4} - \frac{1}{3} = -\frac{3+4}{12} = -\frac{7}{12}$.
The particular solution is:
$-\frac{e^{-4y}}{4} = \frac{e^{3x}}{3} - \frac{7}{12}$.
Multiply by -12: $3e^{-4y} = -4e^{3x} + 7$.
Now we need to find the value of y when $x = -\frac{2}{3}\ln 2$.
Let $x_0 = -\frac{2}{3}\ln 2$.
$e^{3x_0} = e^{3(-\frac{2}{3}\ln 2)} = e^{-2\ln 2} = e^{\ln(2^{-2})} = 2^{-2} = \frac{1}{4}$.
Substitute this into the solution:
$3e^{-4y} = -4\left(\frac{1}{4}\right) + 7$.
$3e^{-4y} = -1 + 7 = 6$.
$e^{-4y} = \frac{6}{3} = 2$.
Take the natural logarithm of both sides:
$\ln(e^{-4y}) = \ln 2$.
$-4y = \ln 2 \implies y = -\frac{1}{4}\ln 2$.
We are given that $y\left(-\frac{2}{3}\log_e 2\right) = \alpha \log_e 2$.
Comparing the two expressions for y, we get:
$\alpha = -\frac{1}{4}$.
