The Laplace Transform

The Laplace Transform Solved by Shahbaz Ahmed June 2024 Introduction 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. Q1 L {sin ax} Solution Since 1 R∞ L {f (x)}=F (s)= 0 e−sx f (x)dx Putting f (x) = sin ax R∞ L {sin ax}=F (s)= 0 e−sx sin axdx Since d(e−sx ) = e−sx d(−sx) = −se−sx dx − 1s d(e−sx ) = e−sx dx R∞ L {sin ax}=F (s)=− 1s 0 sin ax d(e−sx ) R∞ L {sin ax} = − 1s 0 sin ax d(e−sx )   R∞ L {sin ax} = − 1s [sin ax][ e1sx ] − 0 e−sx d(sin ax)   R∞ L {sin ax} = − 1s [sin ax][ e1sx ] − 0 e−sx cos ax d(ax)   R∞ L {sin ax} = − 1s [sin ax][ e1sx ] − a 0 e−sx cos ax dx R∞ L {sin ax} = − 1s [sin ax][ e1sx ] + as 0 e−sx cos ax dx R∞ L {sin ax} = − 1s [sin ax][ e1sx ] + as 0 cos ax (e−sx dx) R∞ L {sin ax} = − 1s [sin ax][ e1sx ] + as 0 cos ax (− 1s )d(e−sx ) R∞ L {sin ax} = − 1s [sin ax][ e1sx ] − sa2 0 cos ax d(e−sx )   R∞ L {sin ax} = − 1s [sin ax][ e1sx ] − sa2 (cos ax)( e1sx ) − 0 e−sx d(cos ax)   R∞ L {sin ax} = − 1s [sin ax][ e1sx ]− sa2 (cos ax)( e1sx ) − 0 e−sx (− sin ax) d(ax)   R∞ L {sin ax} = − 1s [sin ax][ e1sx ]− sa2 (cos ax)( e1sx ) − 0 e−sx (−a sin ax) dx   R∞ L {sin ax} = − 1s [sin ax][ e1sx ]− sa2 (cos ax)( e1sx ) + a 0 e−sx sin ax dx 2 R ∞ L {sin ax} = − 1s [sin ax][ e1sx ] − sa2 (cos ax)( e1sx ) − as2 0 e−sx sin ax dx 2 L {sin ax} = − 1s [sin ax][ e1sx ] − L {sin ax} + a2 s2 L {sin ax} a 1 s2 (cos ax)( esx ) − = − 1s [sin az][ e1sx ] − a2 s2 L {sin ax} a 1 s2 (cos ax)( esx ) a2 s2 )L {sin ax} = − 1s [sin ax][ e1sx ] − sa2 (cos ax)( e1sx )  1 R 2 2 1 a 1 )L {sin ax} = − [sin ax][ ] − (cos ax)( ) ( s s+a 2 s esx s2 esx 0  1  2 2 1 a 1 a ( s s+a 2 )L {sin ax} = − s [sin aR][ esR ] − s2 (cos aR)( esR ) + s2  1 2 2 1 a 1 ( s s+a )L {sin ax} = lim 2 R−→∞ − s [sin aR][ esR ] − s2 (cos aR)( esR ) + (1 + a s2  Since sin aR and cos aR are finite numbers for any value of R and 1 limR−→∞ [ esR ] = 0,hence 2 2 ( s s+a 2 )L {sin ax} = L {sin ax} = a s2 L {sin ax} = a s2 +a2 × a s2 s2 s2 +a2 for ( s > 0) Application of Laplace Transform Q2 Find area under the curve y = e−4x sin 5x for 0 ≤ x < ∞ Solution Here s = 4, a = 5 3 Putting in the formula F (s) = a s2 +a2 F (4) = 5 42 +52 = 5 41 4