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8.5. Example 1

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Interactive Audio Lesson

Session 1: Finding the Homogeneous Solution

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SarahInstructor

Today, we will solve the differential equation $y'' - y = e^x$ . The first step is to find the general solution of the associated homogeneous equation, which is $y'' - y = 0$ . Can anyone tell me how we can derive the characteristic equation?

Noah

Isn't it $r^2 - 1 = 0$ ?

SarahInstructor

Correct! We factor that gives us roots $r = 1$ and $r = -1$ . So the general solution to the homogeneous equation is...

Isabella

Is it $y_h(x) = C_1 e^x + C_2 e^{-x}$ ?

SarahInstructor

Exactly! Great job. Now we have our first solution. Remember, we denote these solutions as $y_1$ and $y_2$ . Now let’s move to the next step!

Session 2: Computing the Wronskian

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RobertInstructor

Next, we need to find the Wronskian, which will help us evaluate the coefficients for the particular solution. Does anyone remember the formula for the Wronskian?

Akash

Yes, it’s $W = y_1 y_2' - y_2 y_1'$ .

RobertInstructor

That's right! Now can we define $y_1$ and $y_2$ to compute the Wronskian?

Ananya

Sure, $W = e^x(-e^{-x}) - e^{-x}(e^x) = -2$ .

RobertInstructor

Perfect! Now we have $W = -2$ . It's essential for our next steps.

Session 3: Finding the Particular Solution

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SarahInstructor

Now, onto calculating the derivatives for our particular solution! Who can recall the formulas for $u_1'$ and $u_2'$ ?

Noah

I remember! It’s $u_1' = -\frac{y_2 g(x)}{W}$ and $u_2' = \frac{y_1 g(x)}{W}$ .

SarahInstructor

Exactly! Great memory! Now substituting in our functions, what do we get for $u_1'$ ?

Isabella

So we have $u_1' = -\frac{e^{-x} imes e^x}{-2} = \frac{1}{2}$ .

SarahInstructor

Excellent! Now let's find $u_2'$ using the same steps.

Session 4: Integrating for Coefficients

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RobertInstructor

Now, to find $u_1$ and $u_2$ , we need to integrate the derivatives we calculated.

Akash

So, that means $u_1 = \int \frac{1}{2} \, dx = \frac{x}{2}$ and $u_2 = -\int \frac{e^{2x}}{2} \, dx = -\frac{e^{2x}}{4}$ ?

RobertInstructor

Exactly, nice work! Now, how do we construct the particular solution?

Ananya

We combine them: $y_p = u_1 y_1 + u_2 y_2 = \frac{x}{2} e^x - \frac{e^{2x}}{4} e^{-x}$ .

RobertInstructor

That's correct! Now we can finalize it and write down the general solution.

Session 5: Finalizing the General Solution

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SarahInstructor

Finally, let's write the general solution for our differential equation. Does anyone want to summarize what we have?

Noah

We have $y(x) = C_1 e^x + C_2 e^{-x} + \frac{x}{2} e^x - \frac{e^{x}}{4}$ .

SarahInstructor

Excellent summary! This solution combines the homogeneous and particular solutions successfully.

Isabella

I find it easier to follow the steps when we systematically break it down like this!

SarahInstructor

I'm glad to hear that! Remember, each step is important in the method of variation of parameters!

Overview

Short Summary

This section demonstrates the application of the variation of parameters method by solving a specific differential equation.

Medium Summary

In this section, a specific differential equation is solved using the variation of parameters method. By following systematic steps, including finding the homogeneous solution, computing the Wronskian, and constructing the particular solution, students can understand how to apply this powerful technique effectively.

Detailed Summary

Example 1: Solving a Second-Order Differential Equation

In this section, the method of variation of parameters is applied to solve the equation $y'' - y = e^x$ . First, we identify the homogeneous equation associated with our differential equation, which is given by $y'' - y = 0$ . The characteristic equation $r^2 - 1 = 0$ yields roots $r = \pm 1$ , resulting in the general solution for the homogeneous equation:

$y_h(x) = C_1 e^x + C_2 e^{-x}$

The next step involves calculating the Wronskian $W$ of the solutions:\n $W = e^x e^{-x} - e^{-x} e^x = -2$ \n This Wronskian is crucial for determining the functions $u_1$ and $u_2$ in the particular solution. Using the formulas for the derivatives:

$u_1' = -\frac{y_2 g(x)}{W} \quad \text{and} \quad u_2' = \frac{y_1 g(x)}{W}$

we find:

$u_1' = -\frac{e^{-x} e^x}{-2} = \frac{1}{2} \quad \text{and} \quad u_2' = \frac{e^x e^x}{-2} = -\frac{e^{2x}}{2}$

After computing the integrals:

$u_1 = \int \frac{1}{2} \, dx = \frac{x}{2} \quad \text{and} \quad u_2 = -\int \frac{e^{2x}}{2} \, dx = -\frac{e^{2x}}{4}$

the particular solution can be formed as follows:

$y_p = u_1 y_1 + u_2 y_2 = \frac{x}{2} e^x - \frac{e^{2x}}{4} e^{-x} = \frac{x}{2} e^x - \frac{e^x}{4}$

Finally, combining the general solution yields:

$y(x) = C_1 e^x + C_2 e^{-x} + \frac{x}{2} e^x - \frac{e^x}{4}$

Audio Book

Voice:

Step 1: Homogeneous Equation

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Solve: y′′−y =ex

Step 1: Homogeneous equation: y′′−y =0⇒r2−1=0⇒r =±1 So: y (x)=ex, y (x)=e−x

Detailed Explanation

In the first step, we start by rewriting the given differential equation, which is a second-order linear non-homogeneous equation. The first task is to solve the associated homogeneous equation, which is obtained by setting the right-hand side equal to zero. We solve for the roots of the characteristic equation resulting from the homogeneous form, r² - 1 = 0, which gives us the roots r = ±1. This means the two fundamental solutions to the homogeneous equation are e^x and e^(-x). Thus, the general solution to the homogeneous equation is a linear combination of these solutions.

Examples & Analogies

Think of the homogeneous equation as finding the natural behavior of a tuned musical instrument, where the two tuning frequencies correspond to the two solutions, e^x and e^(-x). Like different notes that resonate naturally, these solutions show how the system behaves without any external forces acting on it.

Step 2: Compute Wronskian

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Step 2: Compute Wronskian: W =(cid:12) (cid:12)ex e−x (cid:12) (cid:12)=−2 (cid:12)ex −e−x(cid:12)

Detailed Explanation

In this step, we compute the Wronskian, which is a determinant used to check if our solutions are linearly independent. The Wronskian W is computed for the two solutions e^x and e^(-x). The formula for the Wronskian is W = y₁y₂' - y₂y₁'. Here, upon calculating, we find that W = -2, which is a non-zero value indicating that y₁ and y₂ are linearly independent and therefore valid for further calculations.

Examples & Analogies

Imagine the Wronskian as a measure of how different your notes sound when played together on a musical scale. If you strike the two keys (e^x and e^(-x)) and they sound distinct (the Wronskian is non-zero), it means they can continuously create beautiful harmonies, just like our solutions can combine to form the complete general solution.

Step 3: Compute Derivatives

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Step 3: Compute derivatives: y g e−x·ex 1 x u =− 2 =− = ⇒u = 1′ W −2 2 1 2

y g ex·ex e2x 1 u = 1 = =− ⇒u =− e2x

Detailed Explanation

Next, we compute the derivatives needed to find the particular solution using the variation of parameters method. For both u₁ and u₂, we use the formulas obtained for their derivatives with respect to the non-homogeneous term g = e^x. This leads us to the specific forms of u₁ and u₂ after substituting in the Wronskian calculated earlier. Here, we derive expressions for u₁ and u₂ that will later be integrated to find the particular solution.

Examples & Analogies

Think of computing these derivatives like preparing ingredients for a complex dish. Just as you measure out different spices according to your recipe, we are carefully calculating how much of each function will contribute to achieving a flavor profile (the correct particular solution) that will balance the overall taste (the final solution to our differential equation).

Step 4: Particular Solution

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Step 4: Particular solution: y =u y +u y = ex− e2xe−x = ex− ex

Detailed Explanation

Finally, we construct the particular solution using the functions u₁ and u₂ that we derived previously. The particular solution yₚ is given by a combination of our independent solutions multiplied by the calculated u₁ and u₂. This results in yₚ = e^x - (1/4)e^(2x)e^(-x) which simplifies to the required form. This final step allows us to add this specific response to the overall behavior of our system, making it responsive to the external input defined by e^x.

Examples & Analogies

Picture this step as assembling the final dish after all individual components have been prepared. Just like in cooking, where you combine all the ingredients you’ve meticulously measured out to create a delicious meal, here we are combining our calculated contributions from u₁ and u₂ to form the complete particular solution that addresses the external influences defined in the original problem.

Step 5: General Solution

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So: y(x)=C ex+C e−x+ − ex

Detailed Explanation

With our particular solution in hand, we can now write the general solution of our original equation. The general solution is a combination of the homogeneous solution and the particular solution. We express this as y(x) = C₁e^x + C₂e^(-x) + e^x/4, where C₁ and C₂ are constants determined by initial conditions. This encapsulates both the natural behavior of the system and its response to external forces.

Examples & Analogies

Think of this last assembly as having both your foundational recipe (how the dish should taste naturally without any additions) and the flavorful twists you added in the form of the particular solution. Just as the final dish incorporates all elements to represent the chef’s creative expression, the general solution captures the full response of our system to various influences.

8.5. Example 1

Interactive Audio Lesson

Overview

Short Summary

Medium Summary

Detailed Summary

Example 1: Solving a Second-Order Differential Equation

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Audio Book

Detailed Explanation

Examples & Analogies

Detailed Explanation

Examples & Analogies

Detailed Explanation

Examples & Analogies

Detailed Explanation

Examples & Analogies

Detailed Explanation

Examples & Analogies

Key concepts

Examples

Memory aids

Rhyme

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Acronym

Flash Cards

Glossary

Homogeneous Equation

Particular Solution

Wronskian

Characteristic Equation

Variation of Parameters

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