Question

Show that y = '(t - s)f(s)ds is a solution to my" + ky = f(t). Use g' (0) = 1/m and mg" + kg = 0. 6.1) Derive y' 6.2) Using g(0) = 0 and derive y"

Answer 1

Step-by-step explanation:

Given that:

`y = \int^t_og'(t-s) f(s) ds \ \text{is solution to } \ my`

where;

`g'(0) = (1)/(m)` and
`mg`

`\text{Using Leibniz Formula to prove the above equation:}`

`(d)/(dt) \int ^(b(t))_(a(t)) \ f (t,s) \ ds = f(t,b(t) ) * (d)/(dt)b(t) - f(t,a(t)) *(d)/(dt)a(t) + \int ^(b(t))_(a(t))(\partial)/(\partial t) f(t,s) \ dt`

So,
`y = \int ^t_0 g' (t-s) f(s) \ ds`

`\text{By differentiation with respect to t;}`

`y' = g'(o) f(t) (d)/(dt)t- 0 + \int^(t)_(0)g'' (t-s) f(s) ds \\ \\ y' = (1)/(m)f(t) + \int ^t_0 g'' (ts) f(s) \ ds`

`y'' = (1)/(m) f'(t) + g`

`Since \ \ mg`

`Now \ differentiating \ equation (111) \ with \ respect \ to \ t \\ \\ g`