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Chapter 4: Problem 40

Solve the initial-value problems. $$ 3 y^{\prime \prime}+4 y^{\prime}-4 y=0, \quad y(0)=2, \quad y^{\prime}(0)=-4. $$

Short Answer

Expert verified
The specific solution for the given initial-value problem is \(y(x) = 2e^{-2x}\).

Step by step solution

01

Find the characteristic equation of the differential equation

To solve the given second-order linear homogeneous differential equation, first, we need to find its characteristic equation. The differential equation is \(3y'' + 4y' - 4y = 0\), so the characteristic equation will be: \[3r^2 + 4r - 4 = 0\]
02

Solve the characteristic equation for r

Now we need to solve the quadratic equation for r: \[3r^2 + 4r - 4 = 0\] Applying the quadratic formula, with \(a = 3, b = 4\), and \(c = -4\): \[r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\] \[r = \frac{-4 \pm \sqrt{4^2 - 4(3)(-4)}}{2(3)}\] \[r = \frac{-4 \pm \sqrt{64}}{6}\] \[r = \frac{-4 \pm 8}{6}\] We get two distinct roots: \[r_1 = \frac{(-4 + 8)}{6} = \frac{4}{6} = \frac{2}{3}\] \[r_2 = \frac{(-4 - 8)}{6} = -2\]
03

Write the general solution of the differential equation

Since we have two distinct roots, the general solution of the differential equation is: \[y(x) = c_1 e^{\frac{2}{3}x} + c_2 e^{-2x}\]
04

Apply the initial condition y(0)=2

Now, we will apply the initial condition \(y(0) = 2\): \[2 = c_1 e^{\frac{2}{3}(0)} + c_2 e^{-2(0)}\] \[2 = c_1(1) + c_2(1)\] \[2 = c_1 + c_2\]
05

Apply the initial condition y'(0)=-4

First, find the derivative of \(y(x)\): \[y'(x) = \frac{2}{3}c_1 e^{\frac{2}{3}x} - 2c_2 e^{-2x}\] Now, apply the initial condition \(y'(0) = -4\): \[-4 = \frac{2}{3}c_1 e^{\frac{2}{3}(0)} - 2c_2 e^{-2(0)}\] \[-4 = \frac{2}{3}c_1(1) - 2c_2(1)\] \[-4 = \frac{2}{3}c_1 - 2c_2\]
06

Solve the system of equations for c1 and c2

Now we have a system of linear equations: \[\begin{cases} c_1 + c_2 = 2 \\ \frac{2}{3} c_1 - 2c_2 = -4 \end{cases}\] We can solve this system using substitution or elimination. We will use substitution: From the first equation: \(c_1 = 2 - c_2\) Replace \(c_1\) in the second equation: \[\frac{2}{3}(2 - c_2) - 2 c_2 = -4\] Solve for \(c_2\): \[\frac{4}{3} - \frac{2}{3}c_2 - 2 c_2 = -4\] \[- \frac{2}{3} c_2 - 2 c_2 = -\frac{16}{3}\] \[-\frac{8}{3}c_2 = -\frac{16}{3}\] \[c_2 = 2\] Now, find \(c_1\): \[c_1 = 2 - c_2 = 2 - 2 = 0\]
07

Write the specific solution for the initial-value problem

Now that we have the values of \(c_1\) and \(c_2\), we can write the specific solution for the initial-value problem: \[y(x) = 0\cdot e^{\frac{2}{3}x} + 2\cdot e^{-2x}\] \[y(x) = 2 e^{-2x}\] Thus, the specific solution for the given initial-value problem is \(y(x) = 2e^{-2x}\).

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Key Concepts

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

Second Order Differential Equations
Second-order differential equations are equations that involve the second derivative of a function, typically denoted as \(y''\). In many physical systems, they describe phenomena such as oscillations and wave propagation. They can be expressed generally in the form:
  • \(ay'' + by' + cy = f(x)\)
where \(a\), \(b\), and \(c\) are constants, and \(f(x)\) is a function of \(x\). The specific problem we addressed is a homogeneous second-order differential equation because \(f(x) = 0\). Solving these equations often requires finding a characteristic equation that helps identify the type of solutions that fit the system's dynamics.
Characteristic Equation
The characteristic equation is a crucial component when solving linear homogeneous differential equations. It is derived from the differential equation by assuming a solution of the form \(y = e^{rx}\), which leads to an algebraic equation in terms of \(r\). For the given problem, the characteristic equation is:
  • \(3r^2 + 4r - 4 = 0\)
Solving this polynomial equation provides the roots \(r\), which significantly aid in constructing the general solution of the differential equation. The nature of the roots (real and distinct, repeated, or complex) determines the form of the solution.
Homogeneous Differential Equations
A homogeneous differential equation is one in which all terms are dependent on the unknown function and its derivatives. The equation \(3y'' + 4y' - 4y = 0\) is homogeneous because there is no term independent of \(y\). The solutions to a homogeneous second-order differential equation often involve combinations of exponential functions, specifically:
  • \(y(x) = c_1 e^{r_1 x} + c_2 e^{r_2 x}\)
Here, \(r_1\) and \(r_2\) are the roots of the characteristic equation, and \(c_1\), \(c_2\) are determined through initial conditions or boundary values. These solutions describe how the state of a system evolves over time.
Quadratic Formula
The quadratic formula is a mathematical method for solving quadratic equations of the form \(ax^2 + bx + c = 0\). It provides the solution for \(x\) in terms of the coefficients \(a\), \(b\), and \(c\):
  • \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)
In the context of differential equations, it is used to find the roots of the characteristic equation, which guide the form of the solution. For our particular problem, this formula revealed the roots \(r_1 = \frac{2}{3}\) and \(r_2 = -2\), leading to the general solution.
Systems of Linear Equations
In solving initial-value problems, systems of linear equations often arise when applying initial conditions. For our problem, after defining the general solution, the constants \(c_1\) and \(c_2\) were found by solving:
  • \(c_1 + c_2 = 2\)
  • \(\frac{2}{3}c_1 - 2c_2 = -4\)
These equations were solved using techniques like substitution, resulting in specific values for \(c_1\) and \(c_2\) that satisfy the initial conditions \(y(0) = 2\) and \(y'(0) = -4\). This method highlights how initial conditions can determine the unique solution to a homogeneous differential equation.

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Most popular questions from this chapter

Solve the initial-value problems in Exercises $$ y^{\prime \prime}+6 y^{\prime}+9 y=27 e^{-6 x}, \quad y(0)=-2, \quad y^{\prime}(0)=0 $$

Find the general solution of each of the differential equations. $$ y^{\mathrm{iv}}+y=0. $$

Solve the initial-value problems. $$ y^{\prime \prime}+6 y^{\prime}+13 y=0, \quad y(0)=3, \quad y^{\prime}(0)=-1 $$

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