91Ó°ÊÓ

A first order linear differential equation is a type of equation that can be written in the form: \( y' + P(x)y = Q(x) \). In this form, \( y' \) represents the derivative of \( y \) with respect to \( x \), \( P(x) \) and \( Q(x) \) are functions of \( x \) alone. This type of equation describes processes where the rate of change of a quantity is somehow linked to the quantity itself.
The equation from the exercise: \( 2xy' + y = 2\sqrt{x} \) can be simplified into this standard form by dividing each term by 2: \( xy' + \frac{1}{2}y = \sqrt{x} \). Here, \( P(x) = \frac{1}{2x} \) and \( Q(x) = x^{-1/2} \). Recognizing this structure allows us to apply specific methods, like the integrating factor method, to find solutions.

The integrating factor is a powerful tool used to solve first order linear differential equations. The goal of using an integrating factor is to transform a difficult equation into one that's easier to solve by integration. For a first order linear equation \( y' + P(x)y = Q(x) \), the integrating factor \( \mu(x) \) is given by: \[ \mu(x) = e^{\int P(x) \, dx} \]

• In the exercise, we identified \( P(x) = \frac{1}{2x} \).
• Calculating the integral, \( \int \frac{1}{2x} \, dx = \frac{1}{2}\ln x \).

Thus, the integrating factor becomes \( \mu(x) = e^{\frac{1}{2}\ln x} = \sqrt{x} \). Multiplying the entire differential equation by this integrating factor alters it, making the left-hand side an exact derivative. This lets us reframe the equation in such a way that we can directly integrate both sides to find the particular solution.
This is a critical step that greatly simplifies solving the differential equation.

To find a particular solution, we use the integrating factor to integrate the modified equation. When we multiply through by the integrating factor \( \sqrt{x} \), the differential equation becomes: \( \sqrt{x}y' + \frac{1}{2}y = x^0 \). The left-hand side can now be expressed as the derivative of \( \sqrt{x}y \) with respect to \( x \): \( \frac{d}{dx}(\sqrt{x}y) = 1 \).
We integrate both sides:

• The left side integrates to: \( \sqrt{x}y \).
• The right side integrates to: \( x + C \), where \( C \) is an integration constant.

Solving for \( y \), we get the particular solution: \( y_p = \frac{x + C}{\sqrt{x}} = \sqrt{x} + \frac{C}{\sqrt{x}} \). Finding this particular solution is crucial as it describes a specific circumstance of our differential equation.

A homogeneous equation in the context of a first order linear differential equation is when \( Q(x) = 0 \). For the homogeneous part of our original equation \( xy' + \frac{1}{2}y = 0 \), we solve it separately to get the general solution. The homogeneous equation has the form: \( y' + P(x)y = 0 \). We find its solution by integrating:

• This gives us \( y_h = C e^{-\int P(x) \, dx} \).
• For our problem, \( P(x) = \frac{1}{2x} \) which integrates to \( \frac{1}{2}\ln x \).
• Therefore, \( y_h = C e^{-\frac{1}{2}\ln x} = C\sqrt{x} \).

This solution represents the complementary function to the particular solution. Together, they form the general solution of the differential equation, written as the sum of the homogeneous and particular solutions.