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

Answer:

Explanation:

Given that:

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

where;

$g'(0) = \dfrac{1}{m}$     and $mg"+kg = 0$

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

$\dfrac{d}{dt} \int ^{b(t)}_{a(t)} \ f (t,s) \ ds = f(t,b(t) ) * \dfrac{d}{dt}b(t) - f(t,a(t)) *\dfrac{d}{dt}a(t) + \int ^{b(t)}_{a(t)}\dfrac{\partial}{\partial t} f(t,s) \ dt$

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

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

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

$y'' = \dfrac{1}{m} f'(t) + g"(0) f(t) + \int^t_o g"'(t-s) f(s)ds --- (1)$

$Since \ \ mg" (t) +kg (t) = 0 \\ \\ \implies g" (t) = -\dfrac{k}{m} g(t) --- (111) \\ \\ put \ t \ =0 \ we \ get;\\g" (0) = - \dfrac{k}{m } g(0) \\ \\ g"(0) = 0 \ \ \ \ ( because \ g(0) =0)$

$Now \ differentiating \ equation (111) \ with \ respect \ to \ t \\ \\ g"'(t) = -\dfrac{k}{m}g(t) \\ \\ replacing \ it \ into \ equation \ (1) \\ \\ y" = \dfrac{1}{m}f' (t) + 0 + \int ^t_o \dfrac{-k}{m}g' (t-s) f(s) \ ds \\ \\ y" = \dfrac{1}{m}f' (t) - \dfrac{k}{m} \int ^t_o g' (t-s) \ f(s) \ ds \\ \\ y" = \dfrac{1}{m}f'(t) - \dfrac{k}{m}y \\ \\ my" = f'(t)-ky \\ \\ \implies \mathbf{ my" +ky = f'(t)}$