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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"

User Richard Hubley
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19 votes
19 votes

Answer:

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

User Miw
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