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determine a suitable form for Y(\(t\)) if the method of undetermined coefficients is to be used. Do not evaluate the constants. \(y^{\prime \prime \prime}-y^{\prime}=t e^{-t}+2 \cos t\)

Short Answer

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Question: Determine a suitable form for Y(t) using the method of undetermined coefficients for the given differential equation: \(y^{\prime \prime \prime}-y^{\prime}=t e^{-t}+2 \cos t\). Answer: A suitable form for Y(t) is given by: \(Y(t) = Y_{CF}(t) + Y_P(t) = (C_1 + C_2 e^{-t} + C_3 e^{t}) + [(A t + B) e^{-t} + C \cos t + D \sin t]\), where Y(t) is the general solution to the given differential equation, Y_{CF}(t) is the complementary function, and Y_P(t) is the general particular solution.

Step by step solution

01

Determining the complementary function (CF)

First, we'll find the complementary function by solving the homogeneous equation (associated with the given differential equation): \(y^{\prime \prime \prime}-y^{\prime}=0\) This is a linear differential equation with constant coefficients, and its characteristic equation is: \(r^3 - r = 0\) \(r(r^2 - 1) = 0\) \(r(r + 1)(r - 1) = 0\) The roots are r = 0, r = -1, and r = 1. So the complementary function takes the general form: \(Y_{CF}(t) = C_1 e^{0 \cdot t} + C_2 e^{-t} + C_3 e^{t} = C_1 + C_2 e^{-t} + C_3 e^{t}\)
02

Writing the general form of the particular solution for each part of the inhomogeneous term

We'll now have to find the appropriate forms for the particular solutions for each part of the inhomogeneous term. For the exponential part \(t e^{-t}\), if we try a form of the particular solution in the form \(Y_{p1}(t) = Ate^{-t}\), this will create a problem since our second term of complementary function is \(C_2 e^{-t}\). Therefore, we need to try another form which does not include any terms from the complementary function. The new form we will try is \(Y_{p1}(t) = (A t + B) e^{-t}\), where A and B are constants to be determined. For the trigonometric part \(2 \cos t\), we can try a form of the particular solution in the form of \(Y_{p2}(t) = C \cos t + D \sin t\), where C and D are constants to be determined.
03

Combining both particular solutions to find the general particular solution

Since the inhomogeneous term has two separate functions (exponential and trigonometric), we'll simply add the particular solutions of each part to find the overall general particular solution. Thus, the general form of the particular solution is given by: \(Y_P(t) = Y_{p1}(t) + Y_{p2}(t) = (A t + B) e^{-t} + C \cos t + D \sin t\) Now, our task was only to determine a suitable form for Y(t) using the method of undetermined coefficients. We have found the general forms for the complementary function and the particular solution. We don't need to determine the constants in this case, so we can conclude our solution here.

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Key Concepts

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

Differential Equations
Differential equations are mathematical equations that describe the relationship between a function and its derivatives. They serve as a cornerstone in various fields such as physics, engineering, economics, and biology, providing a powerful tool for modeling dynamic systems.

In the given exercise, we are presented with a third-order linear differential equation with an inhomogeneous term. This kind of equation is particularly interesting because it not only dictates the rate of change of a phenomenon but also includes external forces or inputs represented by the non-homogeneous part, in this case, it's the term involving a combination of both exponential and trigonometric functions.
Complementary Function
The complementary function (CF) is a solution to the homogeneous part of a differential equation. It captures the intrinsic behavior of the system without external factors. To find it, one solves the homogeneous differential equation, which is obtained by setting the inhomogeneous part to zero.

In the step-by-step solution, we find the CF by solving the characteristic equation. The CF, in this case, comprises a combination of exponential functions, corresponding to the roots of the characteristic equation. The constants in the CF, typically denoted by C's, encapsulate initial conditions of the problem that must be satisfied, which is why they remain undetermined at this stage.
Characteristic Equation
The characteristic equation is a polynomial equation derived from a linear differential equation with constant coefficients. By solving this equation, we can find the roots that determine the form of the complementary function.

For our exercise, the characteristic equation is simplified to \(r^3 - r = 0\), with the roots revealing the nature of the solution – whether it's purely real, complex, or a combination. Here, r = 0, -1, and 1 are the roots indicating that the solution involves real exponents of t. These exponents form the basis of the exponential terms in the CF.
Particular Solution
The particular solution (PS) is the part of the total solution that specifically addresses the inhomogeneous component of the differential equation. It essentially represents the system's response to the external input or driving function.

In our exercise, the method of undetermined coefficients is used to propose a likely form for the particular solution. This method involves guessing a form similar to the inhomogeneous term, but with undetermined coefficients that will later be calculated. However, if this proposed form is already a part of the complementary solution, as it was initially for the exponential term in our problem, it needs to be modified to prevent redundancy. The generalized particular solution, therefore, combines corrected proposed forms for each unique input term.

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Most popular questions from this chapter

Find the general solution of the given differential equation. Leave your answer in terms of one or more integrals. $$ y^{\prime \prime \prime}-y^{\prime \prime}+y^{\prime}-y=\sec t, \quad-\pi / 2

We consider another way of arriving at the proper form of \(Y(t)\) for use in the method of undetermined coefficients. The procedure is based on the observation that exponential, polynomial, or sinusoidal terms (or sums and products of such terms) can be viewed as solutions of certain linear homogeneous differential equations with constant coefficients. It is convenient to use the symbol \(D\) for \(d / d t\). Then, for example, \(e^{-t}\) is a solution of \((D+1) y=0 ;\) the differential operator \(D+1\) is said to annihilate, or to be an annihilator of, \(e^{-t}\). Similarly, \(D^{2}+4\) is an annihilator of \(\sin 2 t\) or \(\cos 2 t,\) \((D-3)^{2}=D^{2}-6 D+9\) is an annihilator of \(e^{3 t}\) or \(t e^{3 t},\) and so forth. Consider the problem of finding the form of the particular solution \(Y(t)\) of $$ (D-2)^{3}(D+1) Y=3 e^{2 t}-t e^{-t} $$ where the left side of the equation is written in a form corresponding to the factorization of the characteristic polynomial. (a) Show that \(D-2\) and \((D+1)^{2},\) respectively, are annihilators of the terms on the right side of Eq. and that the combined operator \((D-2)(D+1)^{2}\) annihilates both terms on the right side of Eq. (i) simultaneously. (b) Apply the operator \((D-2)(D+1)^{2}\) to Eq. (i) and use the result of Problem 20 to obtain $$ (D-2)^{4}(D+1)^{3} Y=0 $$ Thus \(Y\) is a solution of the homogeneous equation (ii). By solving Eq. (ii), show that $$ \begin{aligned} Y(t)=c_{1} e^{2 t} &+c_{2} t e^{2 t}+c_{3} t^{2} e^{2 t}+c_{4} t^{3} e^{2 t} \\ &+c_{5} e^{-t}+c_{6} t e^{-t}+c_{7} t^{2} e^{-t} \end{aligned} $$ where \(c_{1}, \ldots, c_{7}\) are constants, as yet undetermined. (c) Observe that \(e^{2 t}, t e^{2 t}, t^{2} e^{2 t},\) and \(e^{-t}\) are solutions of the homogeneous equation corresponding to Eq. (i); hence these terms are not useful in solving the nonhomogeneous equation. Therefore, choose \(c_{1}, c_{2}, c_{3},\) and \(c_{5}\) to be zero in Eq. (iii), so that $$ Y(t)=c_{4} t^{3} e^{2 t}+c_{6} t e^{-t}+c_{7} t^{2} e^{-t} $$ This is the form of the particular solution \(Y\) of Eq. (i). The values of the coefficients \(c_{4}, c_{6},\) and \(c_{7}\) can be found by substituting from Eq. (iv) in the differential equation (i).

verify that the given functions are solutions of the differential equation, and determine their Wronskian. $$ y^{\mathrm{iv}}+y^{\prime \prime}=0 ; \quad 1, \quad t, \quad \cos t, \quad \sin t $$

Find the general solution of the given differential equation. $$ y^{\mathrm{iv}}-4 y^{\prime \prime \prime}+4 y^{\prime \prime}=0 $$

Follow the procedure illustrated in Example 4 to determine the indicated roots of the given complex number. $$ 1^{1 / 3} $$

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