Question

The inverse forms of the results in Problem 49 in Exercises 7.1 are ℒ−1 s − a (s − a)2 + b2 = eat cos bt and ℒ−1 b (s − a)2 + b2 = eat sin bt. Use the Laplace transform and these inverses to solve the given initial-value problem. y' + y = e−7t cos 6t, y(0) = 0y(t)= _____.

Answer 1

Answer:  The required solution is

$y(t)=\dfrac{1}{12}e^{-t}-\dfrac{1}{12}e^{-7t}\cos 6t+\dfrac{1}{12}e^{-7t}\sin 6t.$

Step-by-step explanation:  We are given to use the Laplace transform to solve the following initial-value problem :

$y^\prime+y=e^{-7t}\cos 6t~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(i)$

We have the following formulas of Laplace inverse transform:

$(i)~~L^{-1}\{\dfrac{s-a}{(s-a)^2+b^2}\}=e^{at}\cos bt,\\\\\\(ii)~~L^{-1}\{\dfrac{b}{(s-a)^2+b^2}\}=e^{at}\sin bt.$

Applying Laplace transform on both sides of equation (i), we have

$L\{y^\prime+y\}=L\{e^{-7t}\cos 6t\}\\\\\Rightarrow sY(s)-y(0)+Y(s)=\dfrac{s-(-7)}{(s-(-7))^2+6^2}\\\\\\\Rightarrow (s+1)Y(s)-0 =\dfrac{s+7}{(s+7)^2+6^2}\\\\\\\Rightarrow Y(s)=\dfrac{s+7}{(s+1)(s^2+14s+85)}~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(ii)$

By the method of partial fractions, we have

$\dfrac{s+7}{(s+1)(s^2+14s+85)}=\dfrac{A}{s+1}+\dfrac{Bs+C}{s^2+14s+85}\\\\\\\Rightarrow s+7=A(s^2+14s+85)+(Bs+C)(s+1)\\\\\Rightarrow s+7=(A+B)s^2+(14A+B+C)s+(85A+C)$

Equating the coefficients of s², s and the constant term from the above equation and solving, we get

$A=\dfrac{1}{12},~~B=-\dfrac{1}{12},~~C=\dfrac{25}{12}.$

So, from equation (ii), we get

$Y(s)=\dfrac{1}{12(s+1)}-\dfrac{s+1}{12(s^2+14s+85)}\\\\\\\Rightarrow Y(s)=\dfrac{1}{12(s+1)}-\dfrac{1}{12}\dfrac{s+7}{(s+7)^2+6^2}+\dfrac{6}{12}\dfrac{1}{(s+7)^2+6^2}$

Taking inverse Laplace transform on both sides of the above, we get

$L^{-1}\{Y(s)\}=L^{-1}\{\dfrac{1}{12(s+1)}-\dfrac{1}{12}\dfrac{s+7}{(s+7)^2+6^2}+\dfrac{1}{2}\dfrac{1}{(s+7)^2+6^2}\}\\\\\\\Rightarrow y(t)=\dfrac{1}{12}e^{-t}-\dfrac{1}{12}e^{-7t}\cos 6t+\dfrac{1}{2\times6}e^{-7t}\sin 6t\\\\\\\Rightarrow y(t)=\dfrac{1}{12}e^{-t}-\dfrac{1}{12}e^{-7t}\cos 6t+\dfrac{1}{12}e^{-7t}\sin 6t.$

Thus, the required solution is

$y(t)=\dfrac{1}{12}e^{-t}-\dfrac{1}{12}e^{-7t}\cos 6t+\dfrac{1}{12}e^{-7t}\sin 6t.$