91Ó°ÊÓ

In solving first-order linear differential equations, an important tool is the integrating factor. This is a special function that, when multiplied by the differential equation, transforms it into a form that is easier to solve. If you have an equation in the form:

• \( \frac{dy}{dx} + P(x)y = Q(x) \)

the integrating factor, \( \mu(x) \), is derived using the formula:

• \( \mu(x) = e^{\int P(x) \, dx} \)

For the given exercise, \( P(x) = \frac{1}{x} \), so the integrating factor becomes \( \mu(x) = e^{\ln|x|} = |x| \). In cases where we're dealing with positive x-values, this simplifies neatly to \( x \). By multiplying the entire differential equation by this integrating factor, we transform the left side into the derivative of a product, making it much easier to solve through integration. The process of finding and applying the integrating factor is a crucial step that often unlocks the path to the solution.

An initial value problem is a challenge where not only do you solve the differential equation, but you also find a solution that meets a specified condition at a certain point. Typically, the condition is stated like \( y = y_0 \) when \( x = x_0 \). This extra piece of information is what sets initial value problems apart. It ensures that you find a particular solution from a family of solutions.
In this exercise, the initial condition is \( y = 2 \) when \( x = \frac{\pi}{4} \). This information is plugged in after finding the general solution. Substituting the values into the solved equation allows us to find the specific constant \( C \) of integration. This little additional step tailors the wide variety of potential solutions to the exact one needed.

Integration by parts is a technique for evaluating certain integrals where other methods like substitution might not work. It is particularly useful when dealing with a product of functions, as was the case in the given problem with the integral \( \int x \sin 2x \, dx \). Integration by parts is based on the formula:

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

Here, you typically select \( u \) to be a function that simplifies when differentiated, and \( dv \) to be a function that doesn't get more complicated upon integration.
For our integral:

• Choose \( u = x \) and \( dv = \sin 2x \, dx \).
• This gives \( du = dx \) and \( v = -\frac{1}{2} \cos 2x \).

Plugging these back into the formula results in a simpler expression. Integration by parts is often used iteratively, like in this problem, to eventually reduce the integral to a form that can be easily solved. It’s a powerful technique, especially in applied calculus.