91Ó°ÊÓ

Initial conditions are values provided to specify the solution of a differential equation at a particular point. In this problem, we're given the initial condition \( y(1)=4 \). This means that when \( x = 1 \), the value of \( y \) is 4. Initial conditions are essential because they help us determine the particular solution from the general solution of a differential equation. Without initial conditions, we would only find a family of solutions rather than a specific one. To apply an initial condition:

• Solve the differential equation to get the general solution.
• Substitute the initial values of \(x\) and \(y\) into the general solution.
• Solve for the constant of integration, often denoted as \(C\).

In our example, after solving the differential equation, we used \(y(1) = 4\) to find the specific value of \(C\), giving us the final particular solution.

The method of separation of variables is used to solve differential equations where the variables can be separated on opposite sides of the equation. This approach involves three basic steps:
1. Rearrange the equation to isolate all terms involving \( y \) on one side and \( x \) on the other.
2. Integrate both sides separately.
3. Combine the results to find the general solution.
In the given problem, we started with: \[ \frac{dy}{dx} = \frac{\ln x}{\sqrt{x y}} \]
We rearranged it to: \[ \sqrt{y} dy = \frac{\ln(x)}{\sqrt{x}} dx \]
By doing this, we separated \(y\) and \(x\) on different sides of the equation. We could then integrate both sides to continue toward finding the solution. Separation of variables only works when the differential equation can be rearranged into this form. If not, other methods, such as integrating factor or substitution, might be needed.

Integration by parts is a technique used to integrate products of functions. The formula for integration by parts is: \[ \int u dv = uv - \int v du \]
In this problem, we applied integration by parts to: \[ \int \frac{\ln(x)}{\sqrt{x}} dx \]
We chose \( u = \ln(x) \) and \( dv = x^{-1/2} dx \). Then, we found that \( du = x^{-1} dx \) and \( v = 2x^{1/2} \). Applying the integration by parts formula gave us:
\[ \int \frac{\ln(x)}{\sqrt{x}} dx = 2 \sqrt{x} \ln(x) - 2 \int x^{-1/2} dx \]
The second integral was straightforward, and we found:
\[ 2 \sqrt{x} \ln(x) - 4 \sqrt{x} \]
Integration by parts is useful when integrating products of functions that do not simplify easily. It's often used in combination with other techniques for particularly challenging integrals.

Solving a differential equation involves finding the relationship between the dependent and independent variables that satisfies the equation. Here's a summary of the steps we followed:

• Identified the differential equation: \( \frac{dy}{dx} = \frac{\ln x}{\sqrt{x y}} \)
• Separated variables: \( \sqrt{y} dy = \frac{\ln(x)}{\sqrt{x}} dx \)
• Integrated both sides: \( \int \sqrt{y} dy = \int \frac{\ln(x)}{\sqrt{x}} dx \)
• Used integration by parts for the right-hand side.
• Combined the results: \( \frac{2}{3} y^{3/2} = 2 \sqrt{x} \ln(x) - 4 \sqrt{x} + C \)
• Applied the initial condition \( y(1) = 4 \) to find \( C \).
• Substituted \( C \) back into the equation.

Finally, we got the particular solution:
\( \frac{2}{3} y^{3/2} = 2 \sqrt{x} \ln(x) - 4 \sqrt{x} + \frac{28}{3} \).
By understanding and following these steps, solving differential equations becomes a systematic process. Breaking it down helps you tackle even the most complicated problems.