91Ó°ÊÓ

A first-order linear differential equation is an equation of the form \( y' + p(x)y = q(x) \). In this case, \( y' \) represents the first derivative of \( y \) with respect to \( x \), \( p(x) \) is a function of \( x \), and \( q(x) \) is another function of \( x \). These equations are important because they model many real-world processes such as cooling of objects, growth and decay problems, or even electrical circuits.
To solve a first-order linear differential equation, the common approach is to first rewrite it in the standard form. For example, given \( y' = xy - 5x \), you can rewrite it as \( y' - xy = -5x \). This involves treating the left-hand side to align with the general form of a linear equation.
You may notice that these types of equations can be linear or non-linear. The term linear here refers specifically to \( y \) and its derivatives, meaning there are no powers or products of \( y \) or \( y' \) in this equation.

The integrating factor is a powerful tool used to solve first-order linear differential equations. It simplifies the left side of the equation to become a total derivative, making integration straightforward.
To find the integrating factor, \( \,\mu(x) \), compute \( e^{\int p(x) \, dx} \). For the equation in our example \( y' - xy = -5x \), \( p(x) = -x \), so \( \,\mu(x) = e^{\int -x \, dx} = e^{-\frac{x^2}{2}} \). This clever trick transforms the equation into a form where the product rule is easy to apply and integrate.
Once you multiply through by the integrating factor, it turns the equation into \( (e^{-\frac{x^2}{2}}y)' = -5xe^{-\frac{x^2}{2}} \). The benefit here is that the left side can easily be integrated in terms of \( x \), simplifying the complexity of solving the differential equation.

Initial conditions are a crucial aspect of solving differential equations. They allow us to find specific solutions that fit particular scenarios or real-world situations.
After solving a differential equation, the solution often includes an arbitrary constant, \( C \), that represents a family of solutions. Initial conditions, like \( y(0) = 4 \) in our problem, help us determine this constant. By substituting the initial condition into the solution, we solve for \( C \) and narrow down the solution to one specific curve from the family.
In the example, once we found the general solution \( y = 5 + Ce^{\frac{x^2}{2}} \), the initial condition provides \( 4 = 5 + C \cdot e^0 \), simplifying to \( C = -1 \). Verify that this solution satisfies the original equation and the initial condition to ensure it is accurate and complete.