91Ó°ÊÓ

A key technique in solving differential equations is variable separation. This involves rearranging an equation so that similar variables are grouped together. In the exercise, the differential equation \( \frac{d y}{d x}=\frac{\ln x}{x y} \) is tackled using this approach. Essentially, we want all the \( y \) terms on one side and the \( x \) terms on the other. We achieve this by multiplying both sides of the equation by \( y \) and \( dx \), leading to:

• \( y \, dy = \frac{\ln x}{x} \, dx \)

By separating these variables, we prepare the equation for integration. The aim is to integrate each side independently to simplify the equation further. Successfully separating variables like this sets the foundation for solving the differential equation as each side can be approached with basic calculus techniques.
Thus, separation of variables is a fundamental and powerful technique, simplifying the process into more manageable integrals.

When faced with difficult integrals, such as \( \int \frac{\ln x}{x} \, dx \), the method of integration by parts comes to the rescue. Integration by parts is a technique derived from the product rule for differentiation and is particularly useful for functions that are products of two simpler functions. In formula terms, it states:

• \( \int u \, dv = uv - \int v \, du \)

For our problem, we let \( u = \ln x \) and \( dv = \frac{1}{x} \, dx \). This choice simplifies the derivative \( du = \frac{1}{x} \, dx \) and makes \( v = \ln x \), as \( dv \) integrates to \( v \, \). Applying the formula results in:

• \( \int \ln x \, \frac{1}{x} \, dx = \ln x \ln x - \int \ln x \, \frac{dx}{x} \)

Although the integral may seem more complex at first, basic algebra and calculus usually allow simplification. This is why integration by parts is such a valuable method for tackling integral calculus problems, providing solutions to seemingly daunting integrals.

When solving differential equations, initial conditions are critical for finding specific solutions. Using the information \( y(1) = 2 \), we determine constants necessary for solving our differential equation. Once the general solution \( \frac{y^2}{2} = \frac{(\ln x)^2}{2} + C_2 \) is derived, the initial condition helps pinpoint \( C_2 \).

• Plug \( x = 1 \) and \( y = 2 \) into the equation.
• Evaluate to find \( C_2 \): \[ 2 = 0 + C_2 \, \Rightarrow \, C_2 = 2 \]

With \( C_2 \) known, the specific solution is further simplified. Reapplying this constant back reveals \( y = \sqrt{(\ln x)^2 + 4} \). By incorporating initial conditions, we transition from a general solution to one reflecting specific values that the problem demands. Thus, initial conditions not only guide us to a suitable constant but also align the solution with the real-world context or constraints given in a problem.