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Chapter 6: Problem 43

Solve the boundary-value problem, if possible. $$ y^{\prime \prime}-4 y^{\prime}+4 y=0, \quad y(0)=2, y(1)=-1 $$

Short Answer

Expert verified
The solution is \( y(t) = 2e^{2t} + \left(-\frac{1}{e^{2}} - 2\right)te^{2t} \).

Step by step solution

01

Write the characteristic equation

Given the differential equation is \( y'' - 4y' + 4y = 0 \). The characteristic equation associated with this is \( r^2 - 4r + 4 = 0 \).
02

Solve the characteristic equation

We need to factor the characteristic equation \( r^2 - 4r + 4 = 0 \). This factors to \( (r - 2)^2 = 0 \). So, we have a repeated root \( r = 2 \).
03

Write the general solution

Since we have a repeated root \( r = 2 \), the general solution of the differential equation is \( y(t) = C_1 e^{2t} + C_2 t e^{2t} \), where \( C_1 \) and \( C_2 \) are constants.
04

Apply the initial condition \( y(0) = 2 \)

Using the condition \( y(0) = 2 \), we substitute into the general solution: \( y(0) = C_1 e^{0} + C_2 \cdot 0 \cdot e^{0} = C_1 = 2 \). Thus, \( C_1 = 2 \).
05

Apply the initial condition \( y(1) = -1 \)

Now, use the second boundary condition \( y(1) = -1 \): substitute into the general solution: \( y(1) = 2e^{2\cdot 1} + C_2 \cdot 1 \cdot e^{2\cdot 1} = 2e^{2} + C_2 e^{2} = -1 \). This simplifies to \( 2e^{2} + C_2 e^{2} = -1 \).
06

Solve for \( C_2 \)

From \( 2e^{2} + C_2 e^{2} = -1 \), factor out \( e^{2} \): \( e^{2}(2 + C_2) = -1 \). Divide both sides by \( e^{2} \) to isolate \( C_2 \): \( 2 + C_2 = -\frac{1}{e^{2}} \). Therefore, \( C_2 = -\frac{1}{e^{2}} - 2 \).
07

Write the particular solution

Substitute \( C_1 = 2 \) and \( C_2 = -\frac{1}{e^{2}} - 2 \) back into the general solution. The particular solution satisfying the boundary conditions is: \( y(t) = 2e^{2t} + \left(-\frac{1}{e^{2}} - 2\right)te^{2t} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Characteristic Equation
In solving boundary-value problems involving differential equations, the characteristic equation plays a critical role. It is derived from a linear differential equation by considering solutions of the form \( y(t) = e^{rt} \). For the given differential equation
  • \( y'' - 4y' + 4y = 0 \)
we replace derivatives of \( y \) with their respective \( r \) terms. This yields the characteristic equation:
  • \( r^2 - 4r + 4 = 0 \).
The characteristic equation is quadratic and offers insight into the nature of the solutions, indicating whether they involve real distinct, repeated, or complex roots. Solving this equation gives us the roots necessary for constructing the general solution.
Differential Equation
Differential equations describe relationships involving rates of change or slopes. They are key in modeling physical processes and appear frequently in boundary-value problems, which impose conditions on the solutions. Here, the problem involves a second-order linear homogeneous differential equation:
  • \( y'' - 4y' + 4y = 0 \)
with specific conditions at the boundaries (points \( t = 0 \) and \( t = 1 \)). Solving such an equation involves transforming it into a characteristic form to discover the general solution, which then must satisfy the given boundary conditions. Understanding how to set up and solve differential equations is fundamental in mathematics and engineering.
Repeated Roots
When solving the characteristic equation \( r^2 - 4r + 4 = 0 \), we find that it can be factored as:
  • \( (r - 2)^2 = 0 \)
We have a repeated root, \( r = 2 \). Repeated roots arise when the discriminant (\( b^2 - 4ac \) in the case of a quadratic) is zero. This results in both roots of the quadratic being the same.In the context of differential equations, repeated roots require a special form for the general solution.
  • For a repeated root \( r \), the solution becomes: \( y(t) = C_1 e^{rt} + C_2 t e^{rt} \)
where \( C_1 \) and \( C_2 \) are constants determined by initial or boundary conditions.
Initial Conditions
To fully solve a boundary-value problem, initial or boundary conditions must be applied to find specific constants in the general solution. These conditions specify exact values of the function at certain points, allowing us to determine the unique solution that fits the problem.In our exercise, two initial conditions are given:
  • \( y(0) = 2 \)
  • \( y(1) = -1 \)
By applying \( y(0) = 2 \), we directly find the constant \( C_1 = 2 \) since the exponential component does not affect this equation at \( t = 0 \).Using \( y(1) = -1 \), we solve for \( C_2 \):
  • Substitute into the solution: \( y(1) = 2e^{2} + C_2 e^{2} = -1 \)
  • Solving this gives: \( C_2 = -\frac{1}{e^{2}} - 2 \)
With these constants known, the particular solution is fully determined, satisfying both the differential equation and its boundary conditions.

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Most popular questions from this chapter

For the following problems, find the solution to the initial-value problem, if possible.\(y^{\prime \prime}+4 y^{\prime}+6 y=0, y(0)=0, y^{\prime}(0)=\sqrt{2}\)

A 9-kg mass is attached to a vertical spring with a spring constant of \(16 \mathrm{~N} / \mathrm{m}\). The system is immersed in a medium that imparts a damping force equal to 24 times the instantaneous velocity of the mass. a. Find the equation of motion if it is released from its equilibrium position with an upward velocity of \(4 \mathrm{~m} / \mathrm{sec}\). b. Graph the solution and determine whether the motion is overdamped, critically damped, or underdamped.

Find the charge on the capacitor in an \(R L C\) series circuit where \(L=40 \mathrm{H}, R=30 \Omega, C=1 / 200 \mathrm{~F}\), and \(E(t)=200 \mathrm{~V}\). Assume the initial charge on the capacitor is \(7 \mathrm{C}\) and the initial current is \(0 \mathrm{~A}\).

Find the unique solution satisfying the differential equation and the initial conditions given, where \(y_{p}(x)\) is the particular solution. $$ y^{\prime \prime}-5 y^{\prime}=e^{5 z}+8 e^{-5 x}, y_{p}(x)=\frac{1}{5} x e^{5 x}+\frac{4}{25} e^{-5 x}, y(0)=-2, y^{\prime}(0)=0 $$

Two linearly independent solutions- \(y_{1}\) and \(y_{2}\) -are given that satisfy the corresponding homogeneous equation. Use the method of variation of parameters to find a particular solution to the given nonhomogeneous equation. Assume \(x>0\) in each exercise. $$ x^{2} y^{\prime \prime}+2 x y^{\prime}-2 y=3 x, y_{1}(x)=x, \quad y_{2}(x)=x^{-2} $$
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