Key Concepts and Formulas

• Separable Differential Equations: A differential equation of the form $g(y) dy = f(x) dx$ can be solved by integrating both sides.
• Exponential and Logarithmic Properties: $e^{\ln x} = x$, $\ln e^x = x$, $e^{a+b} = e^a e^b$, $\ln(a/b) = \ln a - \ln b$, $a\ln b = \ln b^a$.
• Integration of Exponential Functions: $\int e^{ax} dx = \frac{1}{a} e^{ax} + C$.

Step-by-Step Solution

Step 1: Transform the Equation and Separate Variables

Our goal is to isolate $\frac{dy}{dx}$ and express the equation in a form suitable for separation of variables.

We start with the given differential equation: $\log_e \left(\frac{dy}{dx}\right) = 3x + 4y$

Using the definition of logarithms, we rewrite the equation in exponential form: $\frac{dy}{dx} = e^{3x + 4y}$

Next, we use the property of exponents to separate the terms involving $x$ and $y$: $\frac{dy}{dx} = e^{3x} \cdot e^{4y}$

Now, we separate the variables by dividing both sides by $e^{4y}$ and multiplying both sides by $dx$: $\frac{dy}{e^{4y}} = e^{3x} dx$

Rewriting the left side using the property $\frac{1}{a^n} = a^{-n}$: $e^{-4y} dy = e^{3x} dx$

Step 2: Integrate Both Sides

Our goal is to integrate both sides of the separated differential equation with respect to their respective variables.

We integrate both sides of the equation: $\int e^{-4y} dy = \int e^{3x} dx$

Using the integration formula $\int e^{ax} dx = \frac{1}{a} e^{ax} + C$, we obtain: $\frac{e^{-4y}}{-4} = \frac{e^{3x}}{3} + C$

where $C$ is the constant of integration.

Step 3: Apply Initial Condition to Find the Constant of Integration

Our goal is to determine the value of the constant of integration, $C$, using the given initial condition $y(0) = 0$.

We substitute $x = 0$ and $y = 0$ into the integrated equation: $\frac{e^{-4(0)}}{-4} = \frac{e^{3(0)}}{3} + C$ $\frac{1}{-4} = \frac{1}{3} + C$

Solving for $C$: $C = -\frac{1}{4} - \frac{1}{3} = -\frac{3}{12} - \frac{4}{12} = -\frac{7}{12}$

Step 4: Express the Particular Solution

Our goal is to express the solution with $y$ as a function of $x$.

Substituting the value of $C$ back into the integrated equation: $\frac{e^{-4y}}{-4} = \frac{e^{3x}}{3} - \frac{7}{12}$

Multiplying both sides by $-4$: $e^{-4y} = -\frac{4}{3}e^{3x} + \frac{28}{12} = -\frac{4}{3}e^{3x} + \frac{7}{3}$ $e^{-4y} = \frac{7 - 4e^{3x}}{3}$

Taking the reciprocal of both sides: $e^{4y} = \frac{3}{7 - 4e^{3x}}$

Taking the natural logarithm of both sides: $4y = \ln\left(\frac{3}{7 - 4e^{3x}}\right)$

Dividing by 4: $y = \frac{1}{4}\ln\left(\frac{3}{7 - 4e^{3x}}\right)$

Step 5: Evaluate y at the Given x-value

Our goal is to evaluate $y$ when $x = -\frac{2}{3}\log_e 2$.

First, we find $e^{3x}$: $3x = 3\left(-\frac{2}{3}\log_e 2\right) = -2\log_e 2 = \log_e 2^{-2} = \log_e \frac{1}{4}$ $e^{3x} = e^{\log_e \frac{1}{4}} = \frac{1}{4}$

Now, we substitute this value into the expression for $y$: $y = \frac{1}{4}\ln\left(\frac{3}{7 - 4\left(\frac{1}{4}\right)}\right) = \frac{1}{4}\ln\left(\frac{3}{7 - 1}\right) = \frac{1}{4}\ln\left(\frac{3}{6}\right) = \frac{1}{4}\ln\left(\frac{1}{2}\right)$

Step 6: Determine the Value of $\alpha$

Our goal is to find the value of $\alpha$ such that $y\left(-\frac{2}{3}\log_e 2\right) = \alpha \log_e 2$.

We have: $y = \frac{1}{4}\ln\left(\frac{1}{2}\right) = \frac{1}{4}\ln(2^{-1}) = \frac{1}{4}(-\ln 2) = -\frac{1}{4}\ln 2$

Comparing this with $\alpha \log_e 2$, we have: $\alpha \log_e 2 = -\frac{1}{4}\log_e 2$

Therefore: $\alpha = -\frac{1}{4}$

Common Mistakes & Tips

• Remember to include the constant of integration after integrating.
• Be careful with algebraic manipulations, especially when dealing with fractions and negative signs.
• Use logarithm and exponential properties correctly.

Summary

We solved the given differential equation by separating variables, integrating both sides, and using the initial condition to find the constant of integration. We then evaluated the solution at the given $x$-value and determined the value of $\alpha$.

The final answer is $\boxed{\text{-}{1 \over 4}}$, which corresponds to option (A).