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)=

 The required solution is

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

We have the following formulas of Laplace inverse transform:

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

By the method of partial fractions, we have

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

So, from equation (ii), we get

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

Thus, the required solution is