Problem 10 - Calculus: Graphical, Numerical, Algebraic

Question

In Exercises \(1-10\) , use separation of variables to solve the initial value problem. Indicate the domain over which the solution is valid. \(\frac{d y}{d x}=\frac{4 \sqrt{y} \ln x}{x}\) and \(y=1\) when \(x=e\)

Step-by-Step Solution

Answer

The solution to the differential equation is \(y = [2x \ln{x} - 2x + 1 + 2e]^2\), which is valid for all \(x > 0\), and \(y \geq 0\).

Step 1: Separate the Variables

The given differential equation is \(\frac{d y}{d x}=\frac{4 \sqrt{y} \ln x}{x}\).\n We first rearrange the equation to get all terms involving \(y\) on one side of the equation and those involving \(x\) on the other side.\n Multiply both sides by \(x\) and divide by \(\sqrt{y}\) to get the equation in the separated form: \(\frac{1}{\sqrt{y}} dy = 4 \ln{x} dx \).

Step 2: Integrate Both Sides

Now, integrate both sides of the equation: \( \int \frac{1}{\sqrt{y}} dy = 4 \int \ln{x} dx \).\n This results to \(2 \sqrt{y} = 4(x \ln{x} - x) + C \), where C is the constant of integration.

Step 3: Apply the Initial Condition

Given that \(y=1\) when \(x=e\), substitute these values into the equation to find the constant: \(2 \sqrt{1} = 4(e \ln{e} - e) + C\), this simplifies to: \(2 = 4(1 - e) + C\). Solving for \(C\) yields \(C = 2 + 4e\).\n Insert this value of \(C\) back into the equation results to: \(2 \sqrt{y} = 4(x \ln{x} - x) + 2 + 4e\).

Step 4: Solve for y

Finally, solve for \(y\) by squaring both sides of the equation: \(y = [2x \ln{x} - 2x + 1 + 2e]^2\).\n This is the solution of the differential equation which is valid for all \(x>0\), and \(y \geq 0\).

Key Concepts

Separation of Variables

Separation of variables is a method used to solve differential equations. It is particularly useful for solving equations where the variables involved can be separated onto two sides of the equation. In the exercise, we have \( \frac{dy}{dx} = \frac{4 \sqrt{y} \ln x}{x} \), and our goal is to isolate all terms involving \(y\) on one side and all terms involving \(x\) on the other.
To achieve this, we multiply both sides by \(x\) and then divide by \(\sqrt{y}\). This gives us:

• \( \frac{1}{\sqrt{y}} dy = 4 \ln{x} dx \)

This separation lets us integrate each side independently to find a solution. It's a powerful technique because it turns a differential equation into two simple integrals.
Separation of variables works well when the structure of the differential equation naturally allows for such division. It's essential to think about what transformations might be needed to separate the variables fully.

Differential Equations

Differential equations involve derivatives and describe various phenomena such as motion, heat, or waves. They show how a rate of change in one variable is related to other variables. In our problem, the original differential equation is \( \frac{dy}{dx} = \frac{4 \sqrt{y} \ln x}{x} \).
To solve it, the task was to transform this into an integratable format by separating the variables, resulting in an equation with one function of \(y\) and one of \(x\) on each side:

• \( \frac{1}{\sqrt{y}} dy = 4 \ln{x} dx \)

These integrals: \( \int \frac{1}{\sqrt{y}} dy \) and \( 4 \int \ln{x} dx \) are then solved to find an expression relating \(y\) and \(x\) without derivatives.
This approach simplifies the complexity of the problem, enabling us to find solutions that describe how one variable changes with respect to another. Differential equations like this one are fundamental to calculus and are used extensively in fields such as physics, engineering, and economics.

Initial Value Problem

An Initial Value Problem (IVP) specifies the value of a solution and its derivatives at a given point. This helps find a unique solution from a family of solutions derived from solving the differential equation. In our problem, the condition given is \(y = 1\) when \(x = e\).
With the solution to our separated and integrated equation, \(2 \sqrt{y} = 4(x \ln{x} - x) + C\), we substitute the initial values to determine \(C\):

• \(2 = 4(e \ln{e} - e) + C\)
• Simplifying, \(C = 2 + 4e\)

Using this constant, the unique solution becomes \(y = [2x \ln{x} - 2x + 1 + 2e]^2\), valid for \(x > 0\).
This IVP aligns one specific solution to our differential equation, ensuring it's not just generally correct but precisely matches the scenario defined by the initial condition. These conditions are crucial, especially in real-world applications, to ensure solutions accurately reflect the modeled scenarios.