8.3. Derivation of the Variation of Parameters Formula

Differentiate y (x): p y (x)=u (x)y (x)+u (x)y (x)+u (x)y (x)+u (x)y (x) p′ 1′ 1 1′ 2′ 2 2 2′

In this step, we begin to find the particular solution for the non-homogeneous differential equation. We differentiate the assumed form of our particular solution, which is expressed in terms of two functions, u₁(x) and u₂(x), multiplied by the homogeneous solutions y₁(x) and y₂(x). This differentiation allows us to manipulate the equation further to ultimately solve for the unknown functions u₁ and u₂. It highlights that we are dealing with two dynamic parts that will contribute to the change in our dependent variable y.

Think of differentiating the assumed solution as taking a snapshot of an ongoing process. Just like observing how much a car has traveled over time can be described by measuring its speed at different moments, here, we're observing how changes in the functions u₁ and u₂ affect the overall behavior of the solution y.

To simplify the derivation, we impose a constraint: u (x)y (x)+u (x)y (x)=0 1′ 1 2′ 2

To make our equations manageable, we impose a constraint between the derivatives of the functions u₁ and u₂ and the homogeneous solutions y₁ and y₂. This constraint reduces the complexity of our differentiation results and is pivotal in applying the variation of parameters method. It ensures that the changes we analyze in our solution do not contribute to interference when we substitute back into the original equation.

Imagine you are tuning an instrument. By adjusting a certain string while keeping another string still, you're optimizing the sound without creating discord. The constraint we apply functions similarly by ensuring that while we modify u₁ and u₂, they must work harmoniously together with y₁ and y₂.

Differentiate again: y (x)=u (x)y (x)+u (x)y (x)+u (x)y (x)+u (x)y (x) p′′ 1′ 1′ 1 1′′ 2′ 2′ 2 2′′

In this step, we differentiate the partially derived form of y again to obtain a second derivative. This robust differentiation is crucial as it captures more detailed behavior of our assumed particular solution and facilitates substitution into the non-homogeneous equation. At this point, we are preparing to fully integrate the results into the main equation to derive our solution for u₁ and u₂.

Consider measuring the height of a person as they grow. The first differentiation tells us the speed of growth (how fast they are growing), while the second differentiation tells us if that growth speed is increasing or tapering off. This deeper insight helps us predict future heights more accurately, just as our second derivative will aid in finding precise functions for our solution.

Now substitute y ,y ,y into the original non-homogeneous equation: p p′ p′′ y +p(x)y +q(x)y =g(x)

Now that we have our differentiated forms ready, we substitute them into the original non-homogeneous differential equation. This is where we apply all our work into one cohesive form. With this substitution, we will reveal relationships that directly lead us to forming a system of equations for u₁ and u₂. It allows us to directly link the coefficients of the non-homogeneous terms to our modified solutions.

Think of this step as putting together a puzzle where you've spent significant time on the edges and corners. Now, as you substitute everything into the full picture, you begin to see the connections and relationships between the pieces, ultimately leading you to the completed image.

After substituting and simplifying using the homogeneous equation, we get: u (x)y (x)+u (x)y (x)=g(x) 1′ 1′ 2′ 2′

After performing the necessary substitutions and simplifications, we arrive at a system of two linear equations that involve our unknown functions u₁(x) and u₂(x). This step condenses our calculations into a manageable format that allows us to use algebraic methods (like Cramer's rule) or matrix techniques for solving systems. These equations express how the particular solution is directly influenced by different factors in the non-homogeneous term g(x).

This process is similar to solving for variables in a budget. When you have multiple expenses (like rent, food, and entertainment) that must add up to a total budget, you can set up equations based on your knowns and unknowns to find out how much you can spend on each category.

Let W(x) be the Wronskian of y and y : W(x)=y (x)y (x)−y (x)y (x) 1 2′ 1′ 2

We define W(x) as the Wronskian, which is a determinant used to determine whether our two homogeneous solutions y₁ and y₂ are independent solutions. The Wronskian plays a crucial role in finding the coefficients u₁ and u₂. If W(x) is zero, the functions are not independent, meaning we cannot use them effectively in the variation of parameters method. This determinant is essential for ensuring our solution's viability and correctness.

Think of the Wronskian as a test of collaboration between two team members on a project. If they have distinct skills (independent solutions), they complement each other well. However, if both have the same skills (dependent solutions), their collaboration isn't as effective. The Wronskian tests this collaboration.

Then: u (x)=− y (x)g(x) , u (x)= y (x)g(x) 1′ W(x) 2′ W(x)

Based on our earlier system, we derive equations for u₁ and u₂ using the Wronskian. The formulas express how the particular solutions are weighted by the impact of the forcing function g(x) and the relationships captured in the Wronskian. By deriving u₁ and u₂ through these relations, we are systematically allocating the influence of g(x) across our homogeneous solutions.

Imagine you have a recipe that requires a certain proportion of different spices (influences from y solutions) to achieve a desired taste (the outcome from g). Each spice contributes differently, analogous to how u₁ and u₂ bring the right balance to solve for the particular solution.

Now integrate both: Z y (x)g(x) Z y (x)g(x) u (x)=− 2 dx, u (x)= 1 dx 1 W(x) 2 W(x)

To finalize our solution, we need to integrate the expressions derived for u₁ and u₂. These integrals provide us with the specific forms of our functional coefficients that we can utilize in the formulation of our particular solution. This step essentially transforms our relationships into solvable functions that play a critical role in determining the overall solution to the differential equation.

When you mix ingredients, like when you bake, there's a moment where just mixing them is not enough – you have to 'cook' it (integrate) to achieve the final dish. Here, ‘cooking’ through integration turns our raw functional relationships into usable forms.

Thus, the particular solution is: Z y (x)g(x) Z y (x)g(x) y (x)=−y (x) 2 dx+y (x) 1 dx p 1 W(x) 2 W(x)

After integrating, we construct the final form of the particular solution by combining u₁ and u₂ with the homogeneous solutions y₁ and y₂. This final formula is hugely significant as it gives us the particular solution to the non-homogeneous equation we started with. It marks the culmination of our systematic process and provides insight into how the solution encompasses both the natural behavior (homogeneous solution) and the imposed external effects (particular solution).

Think of this as putting on the final touches to a painting. You have the base colors (homogeneous solutions) and you mix in highlights and shadows (particular solutions) to create a complete and vibrant artwork. This step signifies bringing together all elements into a definitive form.

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