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Chapter 8: Problem 23

\(z^{\prime \prime}+x z^{\prime}+z=x^{2}+2 x+1\)

Short Answer

Expert verified
The general solution of the equation \(z^{\prime \prime}+x z^{\prime}+z=x^{2}+2 x+1\) will be in the form of a quadratic function due to the nature of the non-homogeneous term. It consists of a homogeneous solution and a particular solution.

Step by step solution

01

Finding the Homogeneous Solution

The homogeneous form of the given differential equation is \(z^{\prime \prime}+x z^{\prime}+z=0\). Solving this differential equation will provide the complementary function which is part of the general solution. This can be solved using standard methods for solving second-order homogeneous differential equations, such as the characteristic equation method or the undetermined coefficients method. In this case, because the coefficients are not constant, special methods like reduction of order or variation of parameters might be required.
02

Finding the Particular Solution

Next, a particular solution to the original (non-homogeneous) equation needs to be found. This can often be accomplished by assuming the function z is of the same form as the non-homogeneous term on the right-hand side of the equation, which is a quadratic polynomial in this case. The coefficients of the polynomial are then determined by substitifying this assumed form and its derivatives into the original differential equation and equating coefficients.
03

Forming the General Solution

The general solution to the original differential equation is the sum of the complementary function found from the homogeneous equation and the particular solution. This solution will contain two arbitrary constants, which can be determined if initial or boundary conditions are given.

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Key Concepts

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

Homogeneous Solution
To understand differential equations, it's essential to grasp what a homogeneous solution is. When dealing with a second-order differential equation like the one in our exercise, the homogeneous solution arises from solving the equation where all non-homogeneous terms (like constants or functions on the right side of the equation) are set to zero.

In our example, this shifts the equation to:
  • \[ z'' + x z' + z = 0 \]
This part of the equation represents how the system behaves without any outside influences (the right side of the equation).

Finding the homogeneous solution involves solving this equation often using a method tailored to the specific equation. For equations with non-constant coefficients, techniques like reduction of order or variation of parameters can be effective. By solving this homogeneous equation, we find what is known as the 'complementary function', which offers a view of the system鈥檚 natural dynamics.
Particular Solution
The particular solution is the key to addressing the 'non-homogeneity' of a differential equation. It focuses on fitting the equation to match the effects of non-zero terms, like the polynomial on the right-hand side of our original equation:
  • \[ x^2 + 2x + 1 \]
To find the particular solution, we typically assume that the solution will take the form of the non-homogeneous terms. Since our equation鈥檚 right side is a quadratic polynomial, we assume a similar polynomial form for our particular solution, such as:
  • \[ z_p = Ax^2 + Bx + C \]
We then determine the coefficients \( A, B, \) and \( C \) by substituting \( z_p \) and its derivatives back into the original equation and matching coefficients. This step ensures that the particular solution accounts for the specific influences that make the equation non-homogeneous.
General Solution
Upon solving the homogeneous and particular solutions, we can construct the general solution. This solution encompasses both the natural behavior of the system (via the homogeneous solution) and the response to external factors (via the particular solution).

The general solution is formed by summing these two components:
  • \[ z = z_h + z_p \]
Where \( z_h \) is the homogeneous solution, and \( z_p \) is the particular solution.

This combination gives a full descriptive representation of the system, capturing both static and dynamic responses. Additionally, the general solution includes two arbitrary constants, typically denoted as \( C_1 \) and \( C_2 \). These constants are vital for tailoring the general solution to specific scenarios, using initial or boundary conditions if they are provided. By adjusting \( C_1 \) and \( C_2 \), the solution can be fine-tuned to match any particular conditions presented within the problem domain.

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Most popular questions from this chapter

\(x(1-x) y^{\prime \prime}+(1-3 x) y^{\prime}-y=0\)

Show that \(x^{\prime} J_{\nu}(x)\) satisfies the equation \(x y^{\prime \prime}+(1-2 v) y^{\prime}+x y=0, \quad x>0\) and use this result to find a solution for the equation \(x y^{\prime \prime}-2 y^{\prime}+x y=0, \quad x>0\)

$$x^{2} y^{\prime \prime}+x y^{\prime}+x^{2} y=0$$

The equation $$\left(1-x^{2}\right) y^{\prime \prime}-2 x y^{\prime}+n(n+1) y=0$$ where \(n\) is an unspecified parameter, is called Legendre's equation. This equation occurs in applications of differential equations to engineering systems in spherical coordinates. (a) Find a power series expansion about \(x=0\) for a solution to Legendre's equation. (b) Show that for \(n\) a nonnegative integer, there exists an \(n\) th-degree polynomial that is a solution to Legendre's equation. These polynomials, up to a constant multiple, are called Legendre polynomials. (c) Determine the first three Legendre polynomials (up to a constant multiple).

In applying the method of Frobenius, the following recurrence relation arose: \(a_{k+1}=15^{7} a_{k} /(k+1)^{9}\) \(k=0,1,2, \ldots\) (a) Show that the coefficients are given by the formula \(a_{k}=15^{7 k} a_{0} /(k !)^{9}, k=0,1,2, \ldots\) (b) Use the formula obtained in part (a) with \(a_{0}=1\) to compute \(a_{5}, a_{10}, a_{15}, a_{20},\) and \(a_{25}\) on your computer or calculator. What goes wrong? (c) Now use the recurrence relation to compute \(a_{k}\) for \(k=1,2,3, \ldots, 25,\) assuming \(a_{0}=1 .\) (d) What advantage does the recurrence relation have over the formula?
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