In 19–30 use Theorem 7.4.2 to evaluate the

In Problems 1 – 8 use Theorem 7.4.1 to evaluate the given Laplace transform. \(\mathscr{L}\left\{t e^{-10 t}\right\}\) Text Transcription: L{te^-10t}

The charge q(t) on a capacitor in an LC-series circuit is given by \(\frac{d^{2} q}{d t^{2}}+q=1-4 ?(t-\quad)+6 ?(t-3)\), q(0) = 0, \(q^{\prime}(0)=0\) Text Transcription: d^2 q/dt^2+q=1-4 ?(t-quad)+6 ?(t-3) q^prime 0=0

In Problems 1 – 8 use Theorem 7.4.1 to evaluate the given Laplace transform. \(\mathscr{L}\{t \cos 2 t\}\) Text Transcription: L{t cos 2t}

In Problems 1 – 8 use Theorem 7.4.1 to evaluate the given Laplace transform. \(\mathscr{L}\left\{t^{3} e^{t}\right\}\) Text Transcription: L{t^3e^t}

In Problems 1 – 8 use Theorem 7.4.1 to evaluate the given Laplace transform. \(\mathscr{L}\left\{t^{2} \sinh t\right\}\) Text Transcription: L{t^2 sinh t}

In Problems 1 – 8 use Theorem 7.4.1 to evaluate the given Laplace transform. \(\mathscr{L}\left\{t^{2} \cos t\right\}\) Text Transcription: L{t^2 cos t}

In Problems 1 – 8 use Theorem 7.4.1 to evaluate the given Laplace transform. \(\mathscr{L}\left\{t e^{2 t} \sin 6 t\right\}\) Text Transcription: L{te^2t sin 6t}

In Problems 1 – 8 use Theorem 7.4.1 to evaluate the given Laplace transform. \(\mathscr{L}\left\{t e^{-3 t} \cos 3 t\right\}\) Text Transcription: L{te^-3t cos 3t}

In Problems 9–14 use the Laplace transform to solve the given initial-value problem. Use the table of Laplace transforms in Appendix III as needed. \(y^{\prime}-y=t e^{t} \sin t\), y(0) = 0 Text Transcription: y^prime - y = te^t sin t

In Problems 9–14 use the Laplace transform to solve the given initial-value problem. Use the table of Laplace transforms in Appendix III as needed. \(y^{\prime}+y=t \sin t\), y(0) = 0 Text Transcription: y^prime + y = t sin t

In Problems 1 – 8 use Theorem 7.4.1 to evaluate the given Laplace transform. \(\mathscr{L}\{t \sinh 3 t\}\) Text Transcription: L{t sinh 3t}

In Problems 1–8 use Theorem 7.4.1 to evaluate the given Laplace transform. \(y^{\prime \prime}+16 y=f(t)\), y(0) = 0, \(y^{\prime}(0)=1\) where \(f(t)=\left\{\begin{array}{lr} \cos 4 t, & 0 \leq t< \\ 0, & t \geq \end{array}\right. \) Text Transcription: f(t)={{array} lr cos 4t, & 0 leq t<0, & t geq {array}.

In Problems 9–14 use the Laplace transform to solve the given initial-value problem. Use the table of Laplace transforms in Appendix III as needed. \(y^{\prime \prime}+9 y=\cos 3 t\), y(0) = 2, \(y^{\prime}(0)=5\) Text Transcription: y^{prime prime}+9 y=cos 3t y^prime 0=5

In Problems 9–14 use the Laplace transform to solve the given initial-value problem. Use the table of Laplace transforms in Appendix III as needed. \(y^{\prime \prime}+y=\sin t\), y(0) = 1, \(y^{\prime}(0)=-1\) Text Transcription: y^{prime prime}+y=\sin t y^prime 0=-1

y(t) of Problem 14 for \(0 \leq t<3 \pi\) Text Transcription: 0 leq t<3 pi

In some instances the Laplace transform can be used to solve linear differential equations with variable monomial coefficients. In Problems 17 and 18 use Theorem 7.4.1 to reduce the given differential equation to a linear first-order DE in the transformed function \(Y(s)=\mathscr{L}\{y(t)\}\) Solve the first order DE for Y(s) and then find \(y(t)=\mathscr{L}^{-1}\{Y(s)\}\). \(t y^{\prime \prime}-y^{\prime}=2 t^{2}\), y(0) = 0 Text Transcription: Y(s)=mathscr L {y(t)} y(t)=mathscr L ^{-1}{Y(s)} t y^{prime prime}-y^prime=2 t^2

In some instances the Laplace transform can be used to solve linear differential equations with variable monomial coefficients. In Problems 17 and 18 use Theorem 7.4.1 to reduce the given differential equation to a linear first-order DE in the transformed function \(Y(s)=\mathscr{L}\{y(t)\}\) Solve the first order DE for Y(s) and then find \(y(t)=\mathscr{L}^{-1}\{Y(s)\}\). \(2 y^{\prime \prime}+t y^{\prime}-2 y=10\), \(y(0)=y^{\prime}(0)=0\) Text Transcription: Y(s)=mathscr L {y(t)} y(t)=mathscr L ^{-1}{Y(s)} 2 y^prime prime+ty^prime-2 y=10 y(0)=y^prime 0=0

y(t) of Problem 13 for \(0 \leq t<2 \pi\) Text Transcription: 0 leq t<2 pi

In Problems 9–14 use the Laplace transform to solve the given initial-value problem. Use the table of Laplace transforms in Appendix III as needed. \(y^{\prime \prime}+y=f(t)\), y(0) = 1, \(y^{\prime}(0)=0\) where \(f(t)=\left\{\begin{array}{lr} 1, & 0 \leq t<\quad / 2 \\ \sin t, & t \geq / 2 \end{array}\right. \) Text Transcription: y^{\prime \prime}+y=f(t) y^{\prime}(0)=0 f(t)={{array} lr 1, & 0 leq t<quad / 2 sin t, & t geq 2 {array}

In Problems 19–30 use Theorem 7.4.2 to evaluate the given Laplace transform. Do not evaluate the integral before transforming. \(\mathscr{L}\left\{1 * t^{3}\right\}\) Text Transcription: mathscr L {1 * t^3}

In Problems 19–30 use Theorem 7.4.2 to evaluate the given Laplace transform. Do not evaluate the integral before transforming. \(\mathscr{L}\left\{t^{2} * t e^{t}\right\}\) Text Transcription: mathscr L {t^2 * te^t}

In Problems 19–30 use Theorem 7.4.2 to evaluate the given Laplace transform. Do not evaluate the integral before transforming. \(\mathscr{L}\left\{e^{-t} * e^{t} \cos t\right\}\) Text Transcription: mathscr L {e^-t * e^t cos t}

In Problems 19–30 use Theorem 7.4.2 to evaluate the given Laplace transform. Do not evaluate the integral before transforming. \(\mathscr{L}\left\{e^{2 t} * \sin t\right\}\) Text Transcription: mathscr L {e^2t * sin t}

In Problems 19–30 use Theorem 7.4.2 to evaluate the given Laplace transform. Do not evaluate the integral before transforming. \(\mathscr{L}\left\{\int_{0}^{t} e^{\tau} d \tau\right\}\) Text Transcription: mathscr L {int_0^t e^tau d tau}

In Problems 19–30 use Theorem 7.4.2 to evaluate the given Laplace transform. Do not evaluate the integral before transforming. \(\mathscr{L}\left\{\int_{0}^{t} \cos \tau d \tau\right\}\) Text Transcription: mathscr L {int_0^t cos tau d tau}

In Problems 19–30 use Theorem 7.4.2 to evaluate the given Laplace transform. Do not evaluate the integral before transforming. \(\mathscr{L}\left\{\int_{0}^{t} \tau \sin \tau d \tau\right\}\) Text Transcription: mathscr L {int_0^t tau sin tau d tau}

In Problems 19–30 use Theorem 7.4.2 to evaluate the given Laplace transform. Do not evaluate the integral before transforming. \(\mathscr{L}\left\{\int_{0}^{t} e^{-\tau} \cos \tau d \tau\right\}\) Text Transcription: mathscr L {int_0^t e^{-tau} cos tau d tau}

In Problems 19–30 use Theorem 7.4.2 to evaluate the given Laplace transform. Do not evaluate the integral before transforming. \(\mathscr{L}\left\{\int_{0}^{t} \tau e^{t-\tau} d \tau\right\}\) Text Transcription: mathscr L {int_0^t tau e^{t-tau} d tau}

In Problems 31–34 use (8) to evaluate the given inverse transform. \(\mathscr{L}^{-1}\left\{\frac{1}{s(s-1)}\right\}\) Text Transcription: mathscr L^-1{1/s(s-1)}}

In Problems 19–30 use Theorem 7.4.2 to evaluate the given Laplace transform. Do not evaluate the integral before transforming. \(\mathscr{L}\left\{t \int_{0}^{t} \sin \tau d \tau\right\}\) Text Transcription: mathscr L {t int_0^t sin tau d tau}

In Problems 31–34 use (8) to evaluate the given inverse transform. \(\mathscr{L}^{-1}\left\{\frac{1}{s^{2}(s-1)}\right\}\) Text Transcription: \mathscr{L}^{-1}\left\{\frac{1}{s^{2}(s-1)}\right\}

In Problems 19–30 use Theorem 7.4.2 to evaluate the given Laplace transform. Do not evaluate the integral before transforming. \(\mathscr{L}\left\{\int_{0}^{t} \sin \tau \cos (t-\tau) d \tau\right\}\) Text Transcription: mathscr L {int_0^t sin tau cos (t-tau) d tau}

In Problems 31–34 use (8) to evaluate the given inverse transform. \(\mathscr{L}^{-1}\left\{\frac{1}{s^{3}(s-1)}\right\}\) Text Transcription: mathscr L^-1{1/s^3(s-1)}}

In Problems 31–34 use (8) to evaluate the given inverse transform. \(\mathscr{L}^{-1}\left\{\frac{1}{s(s-a)^{2}}\right\}\) Text Transcription: mathscr L^-1{1/s(s-a)^2}}

In Problems 19–30 use Theorem 7.4.2 to evaluate the given Laplace transform. Do not evaluate the integral before transforming. \(\mathscr{L}\left\{t \int_{0}^{t} \tau e^{-\tau} d \tau\right\}\) Text Transcription: mathscr L {t int_0^t tau e^{-tau} d tau}

The table in Appendix III does not contain an entry for \(\mathscr{L}^{-1}\left\{\frac{8 k^{3} s}{\left(s^{2}+k^{2}\right)^{3}}\right\}\). (a) Use (4) along with the results in (5) to evaluate this inverse transform. Use a CAS as an aid in evaluating the convolution integral. (b) Reexamine your answer to part (a). Could you have obtained the result in a different manner? Text Transcription: mathscr L^-1 {8 k^3 s}/(s^2+k^2)^3}

Use the Laplace transform and the results of Problem 35 to solve the initial-value problem \(y^{\prime \prime}+y=\sin t+t \sin t\), y(0) = 0, \(y^{\prime}(0)=0\) Use a graphing utility to graph the solution. Text Transcription: y^prime prime+y=sin t+t sin t y^prime(0)=0

In Problems 37–46 use the Laplace transform to solve the given integral equation or integrodifferential equation. \(f(t)+\int_{0}^{t}(t-\tau) f(\tau) d \tau=t\) Text Transcription: f(t)+int_0^t(t-tau) f(tau) d tau=t

In Problems 37–46 use the Laplace transform to solve the given integral equation or integrodifferential equation. \(f(t)=2 t-4 \int_{0}^{t} \sin \tau f(t-\tau) d \tau\) Text Transcription: f(t)=2 t-4 int_0^t sin tau f(t-tau) d tau

In Problems 37–46 use the Laplace transform to solve the given integral equation or integrodifferential equation. \(f(t)=t e^{t}+\int_{0}^{t} \tau f(t-\tau) d \tau\) Text Transcription: f(t)=t e^t+int_0^t tau f(t-tau) d tau

In Problems 37–46 use the Laplace transform to solve the given integral equation or integrodifferential equation. \(f(t)+2 \int_{0}^{t} f(\tau) \cos (t-\tau) d \tau=4 e^{-t}+\sin t\) Text Transcription: f(t)+2 int_0^t f(tau) cos (t-tau) d tau=4 e^-t+sin t

In Problems 37–46 use the Laplace transform to solve the given integral equation or integrodifferential equation. \(f(t)+\int_{0}^{t} f(\tau) d \tau=1\) Text Transcription: f(t)+int_0^t f(tau) d tau=1

In Problems 37–46 use the Laplace transform to solve the given integral equation or integrodifferential equation. \(t-2 f(t)=\int_{0}^{t}\left(e^{\tau}-e^{-\tau}\right) f(t-\tau) d \tau\) Text Transcription: t-2 f(t)=int_0^t (e^{tau}-e^{-tau})f(t-tau) d tau

In Problems 37–46 use the Laplace transform to solve the given integral equation or integrodifferential equation. \(f(t)=1+t-\frac{8}{3} \int_{0}^{t}(\tau-t)^{3} f(\tau) d \tau\) Text Transcription: f(t)=1+t-8/3 int_0^t (tau-t)^3 f(tau) d tau

In Problems 37–46 use the Laplace transform to solve the given integral equation or integrodifferential equation. \(f(t)=\cos t+\int_{0}^{t} e^{-\tau} f(t-\tau) d \tau\) Text Transcription: f(t)=cos t+int_0^t e^{-tau} f(t-tau) d tau

In Problems 37–46 use the Laplace transform to solve the given integral equation or integrodifferential equation. \(y^{\prime}(t)=1-\sin t-\int_{0}^{t} y(\tau) d \tau, \quad y(0)=0\) Text Transcription: y^prime(t)=1-sin t-int_0^t y(tau) d tau, quad y(0)=0

In Problems 37–46 use the Laplace transform to solve the given integral equation or integrodifferential equation. \(\frac{d y}{d t}+6 y(t)+9 \int_{0}^{t} y(\tau) d \tau=1\), y(0) = 0 Text Transcription: dy/dt+6 y(t)+9 int_0^t y(tau) d tau=1

In Problems 47 and 48 solve equation (10) subject to i(0) = 0 with L, R, C, and E(t) as given. Use a graphing utility to graph the solution for \(0 \leq t \leq 3\). L = 0.1 h, \(R=3 \Omega\), C = 0.05 f, \(E(t)=100[\mathscr{U}(t-1)-\mathscr{U}(t-2)]\) Text Transcription: R=3 Omega E(t)=100[mathscr U (t-1)-mathscr U(t-2)]

In Problems 47 and 48 solve equation (10) subject to i(0) = 0 with L, R, C, and E(t) as given. Use a graphing utility to graph the solution for \(0 \leq t \leq 3\). L = 0.005 h, \(R=1 \Omega\), C = 0.02 f, \(E(t)=100[t-(t-1) \mathscr{U}(t-1)]\) Text Transcription: R=1 Omega E(t)=100[t-(t-1) mathscr U(t-1)]

In Problems 49–54 use Theorem 7.4.3 to find the Laplace transform of the given periodic function.

In Problems 49–54 use Theorem 7.4.3 to find the Laplace transform of the given periodic function.

In Problems 49–54 use Theorem 7.4.3 to find the Laplace transform of the given periodic function.

In Problems 49–54 use Theorem 7.4.3 to find the Laplace transform of the given periodic function.

In Problems 49–54 use Theorem 7.4.3 to find the Laplace transform of the given periodic function.

In Problems 49–54 use Theorem 7.4.3 to find the Laplace transform of the given periodic function.

In Problems 55 and 56 solve equation (15) subject to i(0) = 0 with E(t) as given. Use a graphing utility to graph the solution for \(0 \leq t<4\) in the case when L = 1 and R = 1. E(t) is the sawtooth function in Problem 51 with amplitude 1 and b = 1. Text Transcription: 0 \leq t<4

In Problems 55 and 56 solve equation (15) subject to i(0) = 0 with E(t) as given. Use a graphing utility to graph the solution for \(0 \leq t<4\) in the case when L = 1 and R = 1. E(t) is the meander function in Problem 49 with amplitude 1 and a = 1. Text Transcription: 0 \leq t<4

In Problems 57 and 58 solve the model for a driven spring/ mass system with damping \(m \frac{d^{2} x}{d t^{2}}+\beta \frac{d x}{d t}+k x=f(t)\), x(0) = 0, \(x^{\prime}(0)=0\) where the driving function f is as specified. Use a graphing utility to graph x(t) for the indicated values of t. \(m=\frac{1}{2}\), \(\beta=1\), k = 5, f is the meander function in Problem 49 with amplitude 10, and \(a=\pi\), \(0 \leq t<2 \pi\). Text Transcription: m d^2x/dt^2}+beta dx/dt+k x=f(t) x^prime 0=0 m=1/2 beta=1 a=pi 0 leq t<2 pi

Discuss how Theorem 7.4.1 can be used to fin \(\mathscr{L}^{-1}\left\{\ln \frac{s-3}{s+1}\right\}\). Text Transcription: mathscr L^-1{ln {s-3/s+1}}

In Section 6.4 we saw that \(t y^{\prime \prime}+y^{\prime}+t y=0\) is Bessel’s equation of order v = 0. In view of (22) of that section and Table 6.4.1 a solution of the initial-value problem \(t y^{\prime \prime}+y^{\prime}+t y=0\), y(0) = 1, \(y^{\prime}(0)=0\), is \(y=J_{0}(t)\). Use this result and the procedure outlined in the instructions to Problems 17 and 18 to show that \(\mathscr{L}\left\{J_{0}(t)\right\}=\frac{1}{\sqrt{s^{2}+1}}\). [Hint: You might need to use Problem 46 in Exercises 7.2.] Text Transcription: t y^prime prime+y^prime+t y=0 y^prime 0=0 y=J_0t mathscr L {J_0t}=1/sqrt{s^2+1}}

In Problems 57 and 58 solve the model for a driven spring/ mass system with damping \(m \frac{d^{2} x}{d t^{2}}+\beta \frac{d x}{d t}+k x=f(t)\), x(0) = 0, \(x^{\prime}(0)=0\) where the driving function f is as specified. Use a graphing utility to graph x(t) for the indicated values of t. m = 1, \(\beta=2\), k = 1, f is the square wave in Problem 50 with amplitude 5, and \(a=\pi\), (0 \leq t<4 \pi\). Text Transcription: m d^2x/dt^2}+beta dx/dt+k x=f(t) x^prime 0=0 beta=2 a=pi 0 leq t<4 pi

Solve the integral equation \(f(t)=e^{t}+e^{t} \int_{0}^{t} e^{-\tau} f(\tau) d \tau\) Text Transcription: f(t)=e^t+e^t int_0^t e^{-tau} f(tau) d tau

(a) Laguerre’s differential equation \(t y^{\prime \prime}+(1-t) y^{\prime}+n y=0) is known to possess polynomial solutions when n is a nonnegative integer. These solutions are naturally called Laguerre polynomials and are denoted by \(L_{n}(t)\). Find \(y=L_{n}(t)\), for n 0, 1, 2, 3, 4 if it is for n = 0, 1, 2, 3, 4 if it is known that \(L_{n}(0)=1). (b) Show that \(\mathscr{L}\left\{\frac{e^{t}}{n !} \frac{d^{n}}{d t^{n}} t^{n} e^{-t}\right\}=Y(s)\). where \(Y(s)=\mathscr{L}\{y\}\), and \(y=L_{n}(t)\) is a polynomial solution of the DE in part (a). Conclude that \(L_{n}(t)=\frac{e^{t}}{n !} \frac{d^{n}}{d t^{n}} t^{n} e^{-t}\), n = 0, 1, 2, . . . This last relation for generating the Laguerre polynomials is the analogue of Rodrigues’ formula for the Legendre polynomials. See (33) in Section 6.4. Text Transcription: t y^prime prime+(1-t) y^prime +n y=0 L_n t y=L_n t L_n 0=1 mathscr L {{e^t}/n ! d^n/d t^n} t^n e^-t}=Y(s) Y(s)=mathscr L y L_n(t)={e^t}/n ! {d^n/d t^n} t^n e^-t

Use the Laplace transform as an aide in evaluating the improper integral \(\int_{0}^{\infty} t e^{-2 t} \sin 4 t d t\). Text Transcription: int_0^infty t e^-2t sin 4 t dt

(a) Show that the square wave function E(t) given in Figure 7.4.4 can be written E(t)=\sum_{k=0}^{\infty}(-1)^{k} \vartheta t(t-k)\) (b) Obtain (14) of this section by taking the Laplace transform of each term in the series in part (a). Text Transcription: E(t)=sum_k=0^infty(-1)^k vartheta t(t-k)

The Laplace transform \(\mathscr{L}\left\{e^{-t^{2}}\right\}) exists, but without finding it solve the initial-value problem \(y^{\prime \prime}+y=e^{-t^{2}}\), y(0) = 0, \(y^{\prime}(0)=0\). Text Transcription: mathscr L {e^-t^2} y^prime prime+y=e^{-t^2} y^prime 0=0

If we assume that \(\mathscr{L}\{f(t) / t\}\) exists and \(\mathscr{L}\{f(t)\}=F(s)\), then \(\mathscr{L}\left\{\frac{f(t)}{t}\right\}=\int_{s}^{\infty} F(u) d u\). Use this result to find the Laplace transform of the given function. The symbols a and k are positive constants. (a) \(f(t)=\frac{\sin a t}{t}\) (b) \(f(t)=\frac{2(1-\cos k t)}{t}\) Text Transcription: mathscr L f(t) / t mathscr L ft=F(s) mathscr L {f(t)/t}=int_s^infty F(u) d u f(t)=sin a t/t f(t)=2(1-cos k t)/t

Transform of the Logarithm Because f(t) = Int has an infinite discontinuity at t = 0 it might be assumed that \(\mathscr{L}\{\ln t\}) does not exist; however, this is incorrect. The point of this problem to guide you through the formal steps leading to the Laplace transform of f(t) = Int, t 0 (a) Use integration by parts to show that \(\mathscr{L}\{\ln t\}=s \mathscr{L}\{t \ln t\}-\frac{1}{s}\). (b) If \\(\mathscr{L}\{\ln t\}=Y(s)\), use Theorem 7.4.1 with n = 1 to show that part (a) becomes \(s \frac{d Y}{d s}+Y=-\frac{1}{s}\) Find an explicit solution of the foregoing differential equation. (c) Finally, the integral definition of Euler’s constant (sometimes called the Euler-Mascheroni constant) is \(\gamma=-\int_{0}^{\infty} e^{-t} \ln t d t\), where \(\gamma=0.5772156649\) …. Use \(Y(1)=-\gamma\) in the solution in part (b) to show that \(\mathscr{L}\{\ln t\}=-\frac{\gamma}{s}-\frac{\ln s}{s}\), s 0 Text Transcription: mathscr L ln t mathscr L ln t=s mathscr L t ln t-1/s mathscr L ln t=Y(s) s dY/ds+Y=-1/s gamma=-int_0^infty e^-t ln t dt gamma=0.5772156649 Y(1)=-gamma mathscr L ln t=-gamma/s-ln s/s

In this problem you are led through the commands in Mathematica that enable you to obtain the symbolic Laplace transform of a differential equation and the solution of the initial-value problem by finding the inverse transform. In Mathematica the Laplace transform of a function y(t) is obtained using LaplaceTransform [y[t], t, s]. In line two of the syntax we replace LaplaceTransform [y[t], t, s] by the symbol Y. (If you do not have Mathematica, then adapt the given procedure by finding the corresponding syntax for the CAS you have on hand.) Consider the initial-value problem \(y^{\prime \prime}+6 y^{\prime}+9 y=t \sin t\), y(0) = 2, \(y^{\prime}(0)=-1\) Load the Laplace transform package. Precisely reproduce and then, in turn, execute each line in the following sequence of commands. Either copy the output by hand or print out the results. diffequat y[t] 6y[t] 9y[t] t Sin[t] transformdeq LaplaceTransform [diffequat, t, s] /. {y[0] 2, y[0] 1, LaplaceTransform [y[t], t, s] Y} soln Solve[transformdeq, Y]//Flatten Y Y/.soln InverseLaplaceTransform[Y, s, t] Text Transcription: y^prime prime+6 y^prime+9 y=t sin t y^prime 0=-1

Appropriately modify the procedure of Problem 68 to find a solution o \(y^{\prime \prime \prime}+3 y^{\prime}-4 y=0\), y(0) = 0, \(y^{\prime}(0)=0\), \(y^{\prime \prime}(0)=1\) Text Transcription: y^prime prime prime+3 y^prime-4 y=0 y^prime 0=0 y^prime prime 0=1