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Question Number 100243 by mhmd last updated on 25/Jun/20

f i n d l a p l a c e t r a n s f o r m e o f t h e f u n c t i o n

f (t) = {(a - b t)}^{2} + {c o s}^{2} (w t) ?

h e l p m e s i r ?

Answered by mathmax by abdo last updated on 25/Jun/20

for that let find L (1), L (t^{n}) and L (\cos (wt)) and use linearity of L

L (1) = \int_{0}^{\infty} e^{- tx} dx = {[- \frac{1}{t} e^{- xt}]}_{0}^{\infty} = \frac{1}{t}

L (t^{n}) = \int_{0}^{\infty} x^{n} e^{- tx} dx =_{tx = u} \int_{0}^{\infty} \frac{u^{n}}{t^{n}} e^{- u} \frac{du}{t} = \frac{1}{t^{n + 1}} \int_{0}^{\infty} u^{n} e^{- u} du

= \frac{Γ (n + 1)}{t^{n + 1}} = \frac{n!}{t^{n + 1}}

L (\cos (wt)) = \int_{0}^{\infty} \cos (wx) e^{- tx} dx = Re (\int_{0}^{\infty} e^{iwx + tx} dx) and

\int_{0}^{\infty} e^{(- t + iw) x} dx = {[\frac{1}{- t + iw} e^{(- t + iw) x}]}_{x = 0}^{\infty} = - \frac{1}{- t + iw} = \frac{1}{t - iw} = \frac{t + iw}{t^{2} + w^{2}} \Rightarrow

L (\cos (wt)) = \frac{t}{t^{2} + w^{2}} we have f (t) = a^{2} - 2 abt + b^{2} t^{2} + \frac{1 + \cos (2 wt)}{2} \Rightarrow

L (f (t)) = a^{2} L (1) - 2 ab L (t) + b^{2} L (t^{2}) + \frac{1}{2} L (1) + \frac{1}{2} L (\cos (2 wt))

= \frac{a^{2}}{t} - \frac{2 ab}{t^{2}} + b^{2} \frac{2}{t^{3}} + \frac{1}{2 t} + \frac{1}{2} \times \frac{t}{t^{2} + {4 w}^{2}} \Rightarrow

L (f (t)) = \frac{a^{2}}{t} - \frac{2 ab}{t^{2}} + \frac{{2 b}^{2}}{t^{3}} + \frac{1}{4 t} + \frac{t}{2 (t^{2} + {4 w}^{2})}

Commented by mathmax by abdo last updated on 25/Jun/20

L (f (t)) = \frac{a^{2}}{t} - \frac{2 ab}{t^{2}} + \frac{{2 b}^{2}}{t^{3}} + \frac{1}{2 t} + \frac{t}{2 (t^{2} + {4 w}^{2})}

Answered by smridha last updated on 25/Jun/20

f (s) = L [(a^{2} - 2 a b t + b^{2} t^{2})] + \frac{1}{2} L [(1 + c o s 2 w t)]

= a^{2} L [1] - 2 a b L [t] + b^{2} L [t^{2}] + \frac{1}{2} L [1] + \frac{1}{2} L [c o s 2 w t]

= \frac{a^{2}}{s} - 2 a b . \frac{1}{s^{2}} + b^{2} . \frac{2}{s^{3}} + \frac{1}{2 s} + \frac{1}{2} . \frac{s}{(s^{2} + 4 w^{2})} .

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