Chapter 4: Problem 4
Express the given complex number in the form \(R(\cos \theta+\) \(i \sin \theta)=R e^{i \theta}\) $$ -i $$
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Consider the nonhomogeneous \(n\) th order linear differential equation $$ a_{0} y^{(n)}+a_{1} y^{(n-1)}+\cdots+a_{n} y=g(t) $$ where \(a_{0}, \ldots, a_{n}\) are constants. Verify that if \(g(t)\) is of the form $$ e^{\alpha t}\left(b_{0} t^{m}+\cdots+b_{m}\right) $$ then the substitution \(y=e^{\alpha t} u(t)\) reduces the preceding equation to the form $$ k_{0} u^{(n)}+k_{1} u^{(n-1)}+\cdots+k_{n} u=b_{0} t^{m}+\cdots+b_{m} $$ where \(k_{0}, \ldots, k_{n}\) are constants. Determine \(k_{0}\) and \(k_{n}\) in terms of the \(a^{\prime}\) 's and \(\alpha .\) Thus the problem of determining a particular solution of the original equation is reduced to the simpler problem of determining a particular solution of an equation with constant coefficients and a polynomial for the nonhomogeneous term.
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).
Express the given complex number in the form \(R(\cos \theta+\) \(i \sin \theta)=R e^{i \theta}\) $$ 1+i $$
determine whether the given set of functions is linearly dependent or linearly independent. If they are linearly dependent, find a linear relation among them. $$ f_{1}(t)=2 t-3, \quad f_{2}(t)=2 t^{2}+1, \quad f_{3}(t)=3 t^{2}+t $$
Find the solution of the given initial value problem. Then plot a graph of the solution. \(y^{\mathrm{iv}}+2 y^{\prime \prime}+y=3 t+4, \quad y(0)=y^{\prime}(0)=0, \quad y^{\prime \prime}(0)=y^{\prime \prime \prime}(0)=1\)
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