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Let's start with the homogeneous part of our differential equation. Can anyone remind me of the steps we take?
We need to solve y′′ + y = 0 first.
Exactly! By solving this equation, we find the general solution. What do we find when we solve the characteristic equation?
The roots are ±i, which gives us the solutions involving sine and cosine.
Correct! So, what is the general solution to the homogeneous equation?
It's C₁ cos(x) + C₂ sin(x).
Perfect! Remember, this is crucial for the next steps.
Now that we have our homogeneous solution, what comes next?
We calculate the Wronskian!
Good! The Wronskian helps us in finding the coefficients for the particular solution. Can anyone write down the formula for the Wronskian?
W(x) = y₁y₂' - y₂y₁'.
Exactly! With our solutions y₁ and y₂, what do we get as W(x)?
It simplifies to 1 since cos²(x) + sin²(x) = 1.
Great job! Remember this because we will use it to find u₁ and u₂.
Let’s move on to computing u₁ and u₂. Can anyone recall the formula we use for these?
u₁ = -∫(y₂g(x)/W(x)) dx and u₂ = ∫(y₁g(x)/W(x)) dx.
Absolutely right! Now we plug in what we have. What’s our g(x) for this problem?
It's tan(x).
Right again! So how do we set up u₁ and u₂?
For u₁, we have -∫(sin(x)tan(x)) dx, and for u₂, ∫(cos(x)tan(x)) dx.
Excellent! Now let's take these integrals one at a time.
Now that we have u₁ and u₂, how do we construct the particular solution?
We use the formula yₚ = u₁y₁ + u₂y₂.
That's right! And what is our final expression?
It simplifies to -ln|sec(x) + tan(x)| cos(x).
Exactly! We also need to combine this with the homogeneous solution for our final answer. Who can tell me the general solution?
y(x) = C₁ cos(x) + C₂ sin(x) - ln|sec(x) + tan(x)| cos(x).
Spot on! You've all done an excellent job understanding the variation of parameters.
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The advanced example demonstrates how to solve the differential equation using the variation of parameters method, detailing each step from solving the homogeneous equation to constructing the particular solution, alongside the final general solution.
In this section, we delve into an advanced example of applying the variation of parameters to solve a non-homogeneous differential equation given by
y′′ + y = tan(x), for 0 < x < π. The procedure begins with solving the corresponding homogeneous equation, yielding the general solution in terms of trigonometric functions. Next, we compute the Wronskian to facilitate the determination of functions u₁(x) and u₂(x) which are essential for deriving the particular solution. After finding these functions through integration, we construct the particular solution and combine it with the general solution to arrive at the full expression. This example not only illustrates the method in action but emphasizes the importance of each step in arriving at the solution.
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Solve the differential equation:
π
y′′+y =tanx, 0<x<2
The section begins with a differential equation that we need to solve. This equation is a non-homogeneous second-order linear differential equation, where the left side contains the second derivative of y and a first-power term of y, while the right side is a tangent function. The domain specified here is between 0 and π/2, indicating a range within which we are looking for solutions.
Think of this as a problem in physics where you want to understand how a pendulum swings under the influence of a varying external force (representing the tan(x) function). The values of x are limited to a particular range, just like how you might only observe the pendulum's motion within a specific timeframe.
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Step 1: Solve the homogeneous equation:
y′′+y =0⇒r2+1=0⇒r =±i
So, the general solution to the homogeneous equation is:
y (x)=C cosx+C sinx
Thus, y (x)=cosx, y (x)=sinx
In this step, we first solve the homogeneous part of the equation, which is y'' + y = 0. We find the characteristic equation, which gives us complex roots (±i). This indicates that the solutions will be oscillatory. Therefore, the general solution to the homogeneous part is expressed using cosine and sine functions, implying that the system's natural response is sinusoidal in nature.
Imagine you are watching a swinging pendulum. The natural motion of the pendulum corresponds to the cosine and sine functions, which describe how it swings back and forth. This represents the system's response when there are no external forces acting upon it.
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Step 2: Compute the Wronskian:
W(x)=(cid:12) (cid:12)
(cid:12)
(cid:12)−cos
sin x
x
(cid:12)
(cid:12)cos
x
sin x
(cid:12)(cid:12)
=cos2x+sin2x=1
The Wronskian is a determinant used to check the linear independence of the solutions we found earlier (cos x and sin x). In this step, we calculate the Wronskian and find that it equals 1. A non-zero Wronskian indicates that the solutions are linearly independent, confirming we can use them to construct the general solution to the non-homogeneous equation.
Think of the Wronskian as a tool that helps us determine if two paths (the solutions) are unique and can coexist without overlapping. If you have two rivers that flow differently without merging, they are analogous to the independent solutions we derived.
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Step 3: Compute u (x) and u (x):
1′ 2′
y g sin2x
u =− 2 =−sinx·tanx=−
1′ W cosx
y g sinx
u = 1 =cosx·tanx=
2′ W 1
Next, we compute the functions u1(x) and u2(x) that will help in constructing the particular solution. These functions are derived from the non-homogeneous term (tan x) and the Wronskian we calculated. By substituting the required components into the formulas for u1 and u2, we obtain expressions that account for how the input function (tan x) influences our overall solution.
Imagine trying to fit different pieces into a puzzle. Here, u1 and u2 are like special pieces that adjust the shape of our overall solution to perfectly match the contour defined by the external force represented by tan x.
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Now integrate:
Z sin2x Z (cid:18)1−cos2x(cid:19) Z Z
tud =− dx=− dx=− secxdx+ cosxdx
u =− dx=− ln|secx+tanx|+sinx
Z
u = sinxdx=−cosx
After deriving u1 and u2, we proceed to integrate these functions. The integration is necessary to find the specific functions u1(x) and u2(x) which will be utilized in constructing the particular solution. It involves using standard integration techniques to find these functions, which can sometimes get complicated depending on the expressions we have.
Think of integration as baking a cake. Each layer (integral) contributes to the final flavor of your cake (the complete solution). If you can’t mix the right ingredients (function components) properly, your cake will not turn out as expected!
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Step 4: Construct the particular solution:
y =u y +u y =(sinx−ln|secx+tanx|)cosx−cosx·sinx
p 1 1 2 2
With u1 and u2 now integrated, we substitute them back into the assumed form of the particular solution. This involves multiplying the functions we computed by the homogeneous solutions and combining them to form a complete expression that satisfies the original differential equation.
Imagine combining different colors of paint to create a new shade. Each u function we integrated reflects a unique color, and together they create a solution—a new shade—that accurately reflects the effects of the external forces acting on our system.
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Final solution:
y(x)=C cosx+C sinx−ln|secx+tanx|cosx
The final step is to express the complete solution to the differential equation by combining the homogeneous solution with the particular solution we derived. The result illustrates how the system behaves under the influence of both its natural tendencies and the external forces described by the tan function.
The final equation is like a complete story of a character's journey. The first part features the natural characteristics of the character (homogeneous solution), while the latter part describes how external events (the force of tan x) influence their journey and change their behavior.
Learn essential terms and foundational ideas that form the basis of the topic.
Key Concepts
Homogeneous Solution: The solution to the corresponding homogeneous equation, providing the baseline behavior of the system.
Wronskian: A determinant that helps us ascertain the independence of solutions and is crucial for calculating the coefficients in the variation of parameters method.
Particular Solution: The solution that accounts for the non-homogeneous part of the equation and is derived based on u₁ and u₂.
General Solution: The total solution that combines the homogeneous solution with the particular solution, forming a complete answer.
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Example illustrating the use of the Wronskian in calculating u₁ and u₂ for the differential equation.
Demonstration of constructing the general solution from the specific components of homogeneous and particular solutions.
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To find the solution's heart, solve def' equations, it's a good start.
Imagine a knight named 'Wron', who always ensured that his soldiers were independent. Every time he found a solution, he called it 'homogeneous', giving it the power to stand by itself against the forces of equations.
HUP (H - Homogeneous, U - Wronskian, P - Particular) - A way to remember the key steps: find the Homogeneous solution, compute the Wronskian, and construct the Particular solution.
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Review the Definitions for terms.
Term: Homogeneous Equation
Definition:
A differential equation in which the right-hand side equals zero.
Term: Wronskian
Definition:
A determinant used to determine the linear independence of solutions to a differential equation.
Term: Particular Solution
Definition:
A specific solution to a non-homogeneous differential equation that satisfies both the differential equation and any given initial or boundary conditions.
Term: General Solution
Definition:
The complete set of solutions to a differential equation, including constants related to initial conditions.