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Find a formula involving integrals for a particular solution of the differential equation $$ y^{\prime \prime \prime}-3 y^{\prime \prime}+3 y^{\prime}-y=g(t) $$ If \(g(t)=t^{-2} e^{t},\) determine \(Y(t)\)

Short Answer

Expert verified
The purpose of using the convolution theorem in this exercise is to rewrite the Laplace transform of the product of two functions as the integral of products of their inverse Laplace transforms, which facilitates finding the particular solution Y(t) for the given differential equation.

Step by step solution

01

Rewrite the given differential equation

We are given the differential equation: $$ y^{\prime \prime \prime}-3 y^{\prime \prime}+3 y^{\prime}-y=g(t) $$ where \(g(t)=t^{-2} e^{t}\)
02

Take Laplace transform of the equation

Taking the Laplace transform on both sides of the equation, we get: $$ \Lap[ y^{\prime \prime \prime}] - 3\Lap[y^{\prime \prime}] + 3\Lap[y^{\prime}] - \Lap[y] = \Lap[g(t)] $$ Using the Laplace transform properties, we can rewrite the equation as: $$ s^3Y(s) - 3 s^2 Y(s) + 3s Y(s) - Y(s) = \lap[ t^{-2} e^{t}] $$
03

Find the Laplace transform of a particular solution using convolution theorem

Using the convolution theorem, we can rewrite the equation as: $$ ( s^3 - 3 s^2 + 3s - 1) Y(s) = \frac{ e^s \Gamma(0,s) } {s^2} $$ Now, we need to find the inverse Laplace transform to get \(Y(t)\): $$ y(t) = L^{-1}\left( \frac{ e^s \Gamma(0,s) } {s^2(s^3 - 3 s^2 + 3s - 1)}\right) $$
04

Use convolution theorem and inverse Laplace transforms to determine \(Y(t)\)

Using convolution theorem, we have: $$ y(t) = \int_0^t L^{-1}\left( \frac{ e^s \Gamma(0,s) } {s^2} \right) L^{-1}(s^3 - 3 s^2 + 3 s - 1)ds $$ For convenience, let \(u(s) = L^{-1}\left( \frac{ e^s \Gamma(0,s) } {s^2} \right)\) and \(v(s) = L^{-1}(s^3 - 3 s^2 + 3 s - 1)\). Then: $$ y(t) = \int_0^t u(s) \cdot v(t-s) ds $$ Now, we need to find the inverse Laplace transforms of \(u(s)\) and \(v(s)\) and then calculate the integral. Note that we may need to use partial fraction decomposition or other Laplace transform properties and techniques to find the inverse Laplace transforms. After finding the inverse Laplace transforms and calculating the integral, we will have our particular solution \(Y(t)\).

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Most popular questions from this chapter

determine whether the given set of functions is linearly dependent or linearly independent. If they are linearly dependent, find a linear relation among them. $$ f_{1}(t)=2 t-3, \quad f_{2}(t)=t^{3}+1, \quad f_{3}(t)=2 t^{2}-t, \quad f_{4}(t)=t^{2}+t+1 $$

Find the general solution of the given differential equation. $$ y^{\prime \prime \prime}-3 y^{\prime \prime}+3 y^{\prime}-y=0 $$

Find the solution of the given initial value problem and plot its graph. How does the solution behave as \(t \rightarrow \infty ?\) $$ y^{\mathrm{iv}}+y=0 ; \quad y(0)=0, \quad y^{\prime}(0)=0, \quad y^{\prime \prime}(0)=-1, \quad y^{\prime \prime \prime}(0)=0 $$

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).

Find the solution of the given initial value problem and plot its graph. How does the solution behave as \(t \rightarrow \infty ?\) $$ y^{\mathrm{iv}}-4 y^{\prime \prime \prime}+4 y^{\prime \prime}=0 ; \quad y(1)=-1, \quad y^{\prime}(1)=2, \quad y^{\prime \prime}(1)=0, \quad y^{\prime \prime \prime}(1)=0 $$

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