Laplace Transform x cos (ax)

Laplace Transform x cos ax shahbaz ahmed June 2024 Introduction to Laplace Transform Let f(x) be defined for 0 ≤ x < ∞ and let s denote an arbitrary real variable. The Laplace transform of f (x), designated by either L or F (s), is R∞ L {f (x)}=F (s)= 0 e−sx f (x)dx for all values of s for which the improper integral converges. Convergence occur when the limit RR limR−→∞ 0 e−sx f (x)dx exists. Find L {x cos ax} Solution Since R∞ L {f (x)}=F (s)= 0 e−sx f (x)dx Putting f (x) = x cos ax R∞ L {x cos ax} = F (s) = 0 e−sx x cos ax dx 1 RR L {x cos ax} = F (s) = limR−→∞ 0 e−sx x cos ax dx RR L {x cos ax} = limR−→∞ 0 e−sx x cos ax dx d(e−sx ) = e−sx d(−sx) = −se−sx dx − 1s d(e−sx ) = e−sx dx RR L {x cos ax} = − 1s limR−→∞ 0 x cos ax d(e−sx ) RR −sL {x cos ax} = limR−→∞ 0 x cos ax d(e−sx ) RR 1 ) − limR−→∞ 0 e−sx d(x cos ax) −sL {x cos ax} = (x cos ax)( esx RR −sL {x cos ax} = (x cos ax)( e1sx ) − limR−→∞ 0 e−sx (cos ax − ax sin ax) dx RR RR −sL {x cos ax} = (x cos ax)( e1sx )−limR−→∞ 0 e−sx cos ax dx+a limR−→∞ 0 e−sx x sin ax dx Again − 1s d(e−sx ) = e−sx dx Also putting RR limR−→∞ 0 e−sx cos ax dx = s s2 +a2 RR x sin ax d(e−sx ) R R −sx 1 s a 1 −sL {x cos ax} = (x cos ax)( esx )− s2 +a d(x sin ax)] 2 − s [(x sin ax)( esx )−limR−→∞ 0 e R R −sx 1 s a 1 −sL {x cos ax} = (x cos ax)( esx )− s2 +a (sin ax+ 2 − s [(x sin ax)( esx )−limR−→∞ 0 e −sL {x cos ax} = (x cos ax)( e1sx ) − s s2 +a2 − as limR−→∞ 0 ax cos ax ]dx 2 s a 1 a a a −sL {x cos ax} = (x cos ax)( e1sx )− s2 +a 2 − s (x sin ax)( esx )+ s s2 +a2 + s L {x cos ax} −sL {x cos ax} − a2 L {x cos ax} s 1 = (x cos ax)( esx )− s s2 +a2 1 − as (x sin ax)( esx )+ a a s s2 +a2 (−s − a2 )L {x cos ax} s −( s 2 +a2 −( s 2 +a2 −( s 2 +a2 s s s + s s2 +a2 − a2 s(s2 +a2 ) 1 1 = (x cos ax)( esx ) − as (x sin ax)( esx ) 2 a a s 1 1 R )L {x cos ax}+ s2 +a 2 − s(s2 +a2 ) = limR−→∞ [(x cos ax)( esx )− s (x sin ax)( esx )]0 2 a a s 1 1 )L {x cos ax}+ s2 +a 2 − s(s2 +a2 ) = limR−→∞ [(R cos aR)( esR )− s (R sin aR)( esR )] )L {x cos ax} + s s2 +a2 − a2 s(s2 +a2 ) 2 = limR−→∞ R [(cos aR) esR − as (sin aR)] Using L’Hospital’s Rule limR−→∞ ( eRsR ) = limR−→∞ 1 ResR =0 Also sinaR and cos aR are finite numbers for all values of R So −( s 2 +a2 −( s 2 +a2 −( s 2 +a2 s s s )L {x cos ax} + s s2 +a2 )L {x cos ax} = a2 s(s2 +a2 ) )L {x cos ax} = a2 −s2 s(s2 +a2 ) 2 2 × −L {x cos ax} = − s(ss 2−a +a2 ) 1 L {x cos ax} = s2 −a2 s(s2 +a2 ) L {x cos ax} = s2 −a2 (s2 +a2 )2 × − a2 s(s2 +a2 ) − s s2 +a2 s s2 +a2 s s2 +a2 (s > 0) Applications Find area under the curve y = e−6x x cos 4x Solution Here s=6,a=4 Putting in the formula 2 2 2 2 −a Area= (ss2 +a 2 )2 −4 Area= (662 +4 2 )2 = 5 676 3 =0