Let y = y(x) be solution of the differential equation logₑ(dy/dx) = 3x + 4y, with y(0) = 0. If y(-2/

Q: Wait, why do we move the $e^{4y}$ to the left side? Can't we just integrate with respect to $x$?

Great question! In a differential equation, we need to group all terms involving $y$ with $dy$ and all terms involving $x$ with $dx$. If you tried to integrate with respect to $x$ directly, you would have an $e^{4y}$ term that depends on $x$ in a way that makes direct integration impossible. By moving $e^{4y}$ to the left, we isolate the variables, which is the standard and most powerful technique for solving first-order differential equations.

Q: What happens to the constant $C$? Do I really need it?

Absolutely! Without $C$, you are describing a whole family of curves that satisfy the differential equation. The initial condition $y(0) = 0$ is the key that selects the one specific curve that passes through the origin. If you omit $C$, you won't be able to satisfy the initial condition, and your final answer for $\alpha$ will be incorrect.

Q: Is there a shortcut for the $e^{3x}$ evaluation at $x = -\frac{2}{3} \ln 2$?

There is a very elegant way to handle it! When you substitute $x = -\frac{2}{3} \ln 2$ into $e^{3x}$, you get $e^{3(-\frac{2}{3} \ln 2)}$, which simplifies to $e^{-2 \ln 2}$. Using the property $a \ln b = \ln(b^a)$, this becomes $e^{\ln(2^{-2})}$. Since $e^{\ln(u)} = u$, this simplifies directly to $2^{-2}$, which is $\frac{1}{4}$. It's a beautiful cancellation that makes the rest of the calculation much easier!

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:

Options:

(A) $-\frac{1}{4}$

(B) $\frac{1}{4}$

(C) 2

(D) $-\frac{1}{2}$

Sigma Percentile · Accepted Answer

### The Beauty of the Logarithmic Cage

Welcome, fellow traveler on the path of mathematics. Today, we are going to dismantle a differential equation that, at first glance, might seem like a daunting fortress.

We are given the equation $\log_e\left(\frac{dy}{dx}\right) = 3x + 4y$, with the anchor condition $y(0) = 0$. Our mission is to find the value of $\alpha$ such that $y\left(-\frac{2}{3} \ln 2\right) = \alpha \ln 2$.

This is not just a calculation; it is a story of liberation and transformation.

### Phase 1: The Exponential Liberation

Look at the derivative $\frac{dy}{dx}$. It is currently imprisoned inside a natural logarithm. In the world of calculus, we cannot easily manipulate a derivative when it is trapped.

We must use the fundamental definition of the logarithm to set it free. By exponentiating both sides, we transform the equation into:

$$ \frac{dy}{dx} = e^{3x + 4y} $$

Suddenly, the derivative is free, and we have a clear path forward. The exponential function is a powerful tool, and now it is working for us.

### Phase 2: The Dance of Variables

Now, we observe the exponent: $3x + 4y$. The laws of exponents are our best friends here. We know that $e^{3x + 4y}$ is equivalent to $e^{3x} \cdot e^{4y}$.

This is the moment of clarity. We can now separate our variables. By dividing both sides by $e^{4y}$ (or multiplying by $e^{-4y}$), we get:

$$ e^{-4y} \, dy = e^{3x} \, dx $$

We have successfully separated the $y$ world from the $x$ world. This is the heart of the variable separable method, and it is the most elegant way to solve this type of problem.

### Phase 3: The Integration and the Constant of Destiny

With our variables separated, we move to the next stage: integration. We set up our integrals:

$$ \int e^{-4y} \, dy = \int e^{3x} \, dx $$

These are standard integrals, but we must be precise. Integrating $e^{-4y}$ gives us $-\frac{1}{4}e^{-4y}$, and integrating $e^{3x}$ gives us $\frac{1}{3}e^{3x}$.

And, of course, we must include the constant of integration, $C$. Our equation is now:

$$ -\frac{1}{4}e^{-4y} = \frac{1}{3}e^{3x} + C $$

This constant $C$ is the 'constant of destiny'—it defines which specific curve in the family of solutions we are working with.

### Phase 4: Locking Down the Curve

We use our initial condition $y(0) = 0$ to find $C$. Substituting $x=0$ and $y=0$ into our equation, we get:

$$ -\frac{1}{4}e^{0} = \frac{1}{3}e^{0} + C \Rightarrow -\frac{1}{4} = \frac{1}{3} + C $$

Solving for $C$, we find $C = -\frac{1}{4} - \frac{1}{3} = -\frac{7}{12}$. Now, our equation is fully defined:

$$ -\frac{1}{4}e^{-4y} = \frac{1}{3}e^{3x} - \frac{7}{12} $$

To make our lives easier, let's multiply by $-4$ to get:

$$ e^{-4y} = -\frac{4}{3}e^{3x} + \frac{7}{3} $$

### Phase 5: The Final Evaluation

We are almost there. We need to evaluate this at $x = -\frac{2}{3} \ln 2$. First, let's calculate $e^{3x}$ at this point:

$$ e^{3\left(-\frac{2}{3} \ln 2\right)} = e^{-2 \ln 2} = e^{\ln(2^{-2})} = 2^{-2} = \frac{1}{4} $$

Substituting this back into our equation, we get:

$$ e^{-4y} = -\frac{4}{3}\left(\frac{1}{4}\right) + \frac{7}{3} = -\frac{1}{3} + \frac{7}{3} = \frac{6}{3} = 2 $$

Finally, we have $e^{-4y} = 2$. Taking the natural log of both sides, $-4y = \ln 2$, which means $y = -\frac{1}{4} \ln 2$.

Comparing this to $\alpha \ln 2$, we see that **$\alpha = -\frac{1}{4}$**. You have successfully navigated the complexity and arrived at the elegant solution. Well done!

Animated Solution for Mathematics - Differential Equations: Let $y = y (x)$ be solution of the differential equation $lo g_{e} (\frac{d y}{d x}) = 3 x + 4 y$ , with $y (0) = 0$ . If $y (- \frac{2}{3} lo g_{e} 2) = α lo g_{e} 2$ , then the value of $α$ is equal to:

Select Answer:

Visualized Solution

Analyzing the Differential Equation

Exponential Transformation

Variable Separation

Setting up Integration

Integrating the Terms

Applying Initial Condition

Solving for Constant $C$

General Solution Setup

Simplifying the Expression

Evaluating at Target $x$

Simplifying the Exponent

Calculating $e^{- 4 y}$

Solving for $y$

Finding $α$

The Sigma Insight: Variable Separable Method

The Beauty of the Logarithmic Cage

Phase 1

Phase 2

Phase 3

Phase 4

Phase 5

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