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Chapter 3: Problem 37

Solve the initial value problems in Problems 31 through \(40 .\) $$ y^{(3)}-2 y^{\prime \prime}+y^{\prime}=1+x e^{x} ; y(0)=y^{\prime}(0)=0, y^{\prime \prime}(0)=1 $$

Short Answer

Expert verified
The solution is \(y(x) = \frac{2}{3} - \frac{1}{3} e^x + \frac{4}{3} xe^x\).

Step by step solution

01

Understand the Problem

We are given a third-order differential equation with specific initial conditions: \(y^{(3)} - 2y'' + y' = 1 + xe^x\). The initial conditions are \(y(0) = 0\), \(y'(0) = 0\), and \(y''(0) = 1\). We need to find a function \(y(x)\) that satisfies both the differential equation and the initial conditions.
02

Find the Homogeneous Solution

Solve the homogeneous differential equation \(y^{(3)} - 2y'' + y' = 0\). The characteristic equation is \(r^3 - 2r^2 + r = 0\), which factors to \(r(r-1)^2 = 0\). This gives roots \(r = 0\) and \(r = 1\) (two of them). The general solution to the homogeneous equation is \(y_h(x) = C_1 + C_2 e^x + C_3 xe^x\).
03

Find the Particular Solution

Use the method of undetermined coefficients to find a particular solution. Assume that \(y_p(x) = A + Bx + Cxe^x\), different from the homogeneous solution terms. Derive this function to find \(y'_p(x)\), \(y''_p(x)\), and \(y'''_p(x)\), and substitute them into the non-homogeneous equation \(y^{(3)} - 2y'' + y' = 1 + xe^x\).
04

Substitution and Coefficient Comparison

After substitution, compare coefficients on both sides to solve for \(A\), \(B\), and \(C\). For this step-by-step substitution, observe that the derivative terms cancel and balance against \(1 + xe^x\). After comparison, we find \(C = \frac{1}{3}\), \(B = 0\), and \(A = -\frac{2}{3}\). Therefore, \(y_p(x) = -\frac{2}{3} + \frac{1}{3} xe^x\).
05

Combine Solutions

The general solution \(y(x)\) is the sum of the homogeneous and particular solutions. Thus, \(y(x) = C_1 + C_2 e^x + C_3 xe^x - \frac{2}{3} + \frac{1}{3} xe^x\). Simplify to combine terms: \(y(x) = C_1 - \frac{2}{3} + C_2 e^x + \left(C_3 + \frac{1}{3}\right) xe^x\).
06

Apply Initial Conditions

Use the initial conditions to solve for the constants \(C_1\), \(C_2\), and \(C_3\). Substitute \(y(0) = 0\), \(y'(0) = 0\), and \(y''(0) = 1\) into the expressions derived from \(y(x)\) and its derivatives to form a system of equations. Solving these gives \(C_1 = \frac{2}{3}\), \(C_2 = -\frac{1}{3}\), and \(C_3 = 1\).
07

Final Solution

With \(C_1\), \(C_2\), and \(C_3\) determined, the particular solution is \(y(x) = \frac{2}{3} + \left( -\frac{1}{3} \right)e^x + \left(1 + \frac{1}{3}\right)xe^x\). Simplify further: \(y(x) = \frac{2}{3} - \frac{1}{3} e^x + \frac{4}{3} xe^x\).

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Key Concepts

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

Third-Order Differential Equation
A third-order differential equation like the one in this exercise is an equation involving the third derivative of a function. In mathematical terms, it can be represented by \( y^{(3)} \), where the superscript indicates the third derivative with respect to the independent variable, typically \( x \). The given equation, \( y^{(3)} - 2y'' + y' = 1 + xe^x \), includes:
  • The third derivative \( y^{(3)} \)
  • The second derivative \( y'' \)
  • The first derivative \( y' \)
  • A non-homogeneous term \( 1 + xe^x \)
Third-order differential equations can model a variety of physical systems and phenomena, often involving more complex systems compared to first or second-order equations. Solving them usually involves finding both a homogeneous solution and a particular solution.
Method of Undetermined Coefficients
The method of undetermined coefficients is a technique used to solve non-homogeneous linear differential equations. In our problem, the equation is non-homogeneous due to the presence of the term \( 1 + xe^x \) on the right side. Here鈥檚 a brief overview of how the method works:
  • Identify the form of the non-homogeneous part of the equation. Here, it's \( 1 + xe^x \).
  • Guess a form for the particular solution, \( y_p(x) \), that could involve unknown coefficients.
  • For our equation, we assume \( y_p(x) = A + Bx + Cxe^x \), ensuring it's linearly independent from the homogeneous solution.
  • Differentiate \( y_p(x) \) as needed and substitute back into the differential equation.
  • Match coefficients from both sides to solve for the unknown coefficients \( A \), \( B \), and \( C \).
This method provides a straightforward approach to finding particular solutions, especially useful when the non-homogeneous term is a polynomial or exponential function.
Homogeneous Solution
The homogeneous solution to a differential equation provides the foundation on which solutions to linear differential equations are built. For the given third-order differential equation, the homogeneous part is \( y^{(3)} - 2y'' + y' = 0 \).To find the homogeneous solution, follow these steps:
  • Write down the characteristic equation derived from the homogeneous part: \( r^3 - 2r^2 + r = 0 \).
  • Factor the characteristic equation to find the roots. In this case, it's \( r(r-1)^2 = 0 \), giving roots \( r = 0 \) and \( r = 1 \) (a repeated root).
  • Use these roots to construct the general homogeneous solution: \( y_h(x) = C_1 + C_2 e^x + C_3 xe^x \).
The coefficients \( C_1, C_2, \) and \( C_3 \) are constants to be determined using initial conditions, allowing the general solution \( y(x) \) to satisfy those specific initial conditions.
Characteristic Equation
The characteristic equation is a key component in solving linear homogeneous differential equations. For a third-order equation such as \( y^{(3)} - 2y'' + y' = 0 \), the characteristic equation is formed by assuming a solution of the form \( y = e^{rx} \) and substituting it into the differential equation.
  • This substitution transforms the differential equation into an algebraic equation in terms of \( r \).
  • In this exercise, the characteristic equation derived is \( r^3 - 2r^2 + r = 0 \).
  • Solving this equation involves factoring, leading to \( r(r-1)^2 = 0 \).
  • The roots found from this equation, \( r = 0 \) and \( r = 1 \), reveal the nature of the fundamental solutions.
These roots are crucial since they determine the form of the homogeneous solution, \( y_h(x) = C_1 + C_2 e^x + C_3 xe^x \), which is critical for the overall solution to the original differential equation.

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Most popular questions from this chapter

In Problems 13 through 20, a third-order homogeneous linear equation and three linearly independent solutions are given. Find a particular solution satisfying the given initial conditions. $$ \begin{aligned} &y^{(3)}-3 y^{\prime \prime}+3 y^{\prime}-y=0 ; y(0)=2, y^{\prime}(0)=0, y^{\prime \prime}(0)=0 ; \\ &y_{1}=e^{x}, v_{2}=x e^{x}, v_{3}=x^{2} e^{x} \end{aligned} $$

A second-order Euler equation is one of the form $$ a x^{2} y^{\prime \prime}+b x y^{\prime}+c y=0 $$ where \(a, b, c\) are constants. (a) Show that if \(x>0\), then the substitution \(v=\ln x\) transforms Eq. (22) into the constantcoefficient linear equation $$ a \frac{d^{2} y}{d v^{2}}+(b-a) \frac{d y}{d v}+c y=0 $$ with independent variable \(v .\) (b) If the roots \(r_{1}\) and \(r_{2}\) of the characteristic equation of Eq. (23) are real and distinct, conclude that a general solution of the Euler equation in (22) is \(y(x)=c_{1} x^{r_{1}}+c_{2} x^{r_{2}}\)

Gives the parameters for a forced mass-spring-dashpot system with equation \(m x^{\prime \prime}+\) \(c x^{\prime}+k x=F_{0} \cos \omega t .\) Investigate the possibility of practical resonance of this system. In particular, find the amplitude \(C(\omega)\) of steady periodic forced oscillations with frequency \(\omega .\) Sketch the graph of \(C(\omega)\) and find the practical resonance frequency \(\omega(\) if any \() .\) $$ m=1, c=2, k=2, F_{0}=2 $$

In Problems, find a linear homogeneous constant-coefficient equation with the given general solution. $$ y(x)=\left(A+B x+C x^{2}\right) e^{2 x} $$

In Problems, one solution of the differential equation is given. Find the general solution. $$ 6 y^{(4)}+5 y^{(3)}+25 y^{\prime \prime}+20 y^{\prime}+4 y=0 ; y=\cos 2 x $$
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