Laplace Transform Assignment #1

Laplace Transform of [xn−1eax] and [e ax −ebx a−b ] shahbaz ahmed July 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. Formulas used (1)L’Hopital’s rule. If the limit (x) lim fg(x) 1 is of indeterminate type 0 0 or ∓∞ ∓∞ then, (x) = lim lim fg(x) dn (f (x)) dxn dn (g(x)) dxn Provided the derivative exists 1 (2)limR−→∞ [ e(s−a)x ]R 0 1 limR−→∞ [ e(s−a)R − 1 ] e(s−a)×0 1 limR−→∞ [ e(s−a)R − 1] 1 [ e(s−a)×∞ − 1] 1 − 1] [∞ [0 − 1] = −1 n−1 n−1 x R (3) limR−→∞ [ e(s−a)x ]R o = limR−→∞ [ e(s−a)R ] Applying the L’Hopital’s rule repeatedly n−1 (n−1)! R 1 limR−→∞ [ e(s−a)R ] = (s−a) n−1 limR−→∞ [ e(s−a)R ] = 0 R∞ (4)L {eax }=F (s)= 0 e−sx eax dx R∞ L {eax }=F (s)= 0 e−(s−a)x dx RR 1 limR−→∞ 0 d(e−(s−a)x ) L {eax }=F (s)=− s−a 1 1 limR−→∞ [ e(s−a)x ]R L {eax }=F (s)=− s−a 0 1 1 − 1] L {eax }=F (s)=− s−a limR−→∞ [ e(s−a)R 1 L {eax }=F (s)=− s−a (−1) = 1 s−a Find L {xn−1 eax }(n = 1, 2, 3, .....) (s > a) Solution Since R∞ L {f (x)}=F (s)= 0 e−sx f (x)dx Putting f (x) = xn−1 eax R∞ L {xn−1 eax }=F (s)= 0 e−sx xn−1 eax dx 2 L {xn−1 eax } = R∞ e−(s−a)x xn−1 dx RR L {xn−1 eax } = limR−→∞ 0 e−(s−a)x xn−1 dx 0 d(e−(s−a)x ) = e−(s−a)x d(−(s − a))x d(e−(s−a)x ) = −(s − a)e−(s−a)x dx 1 − s−a d(e−(s−a)x ) = e−(s−a)x dx RR 1 L {xn−1 eax } = − s−a limR−→∞ 0 xn−1 d(e−(s−a)x ) i h R R −(s−a)x n−2 xn−1 R 1 n−1 ax x dx L {x e } = − s−a limR−→∞ [ e(s−a)x ]0 − (n − 1)limR−→∞ 0 e h i R R −(s−a)x n−2 1 Rn−1 n−1 ax L {x e } = − s−a limR−→∞ e(s−a)R − (n − 1)limR−→∞ 0 e x dx Now n−1 R limR−→∞ e(s−a)R =0 L {xn−1 eax } = n−1 s−a h n−1 s−a L {xn−2 eax RR limR−→∞ 0 e−(s−a)x xn−2 dx i Or L {xn−1 eax } = Again i RR limR−→∞ 0 e−(s−a)x xn−2 dx h i R R n−2 −1 −(s−a)x ( ) lim x d(e ) L {xn−1 eax } = n−1 R−→∞ s−a s−a 0 i h R R −(s−a)x n−3 n−2 n−1 x L {xn−1 eax } = − (s−a) ]R − (n − 2)lim e x dx limR−→∞ [ e(s−a)x R−→∞ 2 0 0 h i R n−2 R n−1 R n−1 ax −(s−a)x n−3 L {x e } = − (s−a)2 limR−→∞ [ e(s−a)R ] − (n − 2)limR−→∞ 0 e x dx L {xn−1 eax } = n−1 s−a h n−2 R limR−→∞ [ e(s−a)R ]=0 L {xn−1 eax } = (n−1)(n−2) (s−a)2 L {xn−3 eax } It may be induced mathematically L {xn−1 eax } = (n−1)(n−2)(n−3)....[n−(n−1)] (s−a)n−1 L {xn−1 eax } = (n−1)! 1 (s−a)n−1 s−a = (n−1)! (s−a)n L {eax } (s > a) 3 ax bx −e } Find L { e a−b Solution Since R∞ L {f (x)}=F (s)= 0 e−sx f (x)dx Putting f (x) = eax −ebx a−b R ∞ −sx eax −ebx e ( a−b ) dx } = F (s) = a−b 0 R ax −ebx ax −ebx ∞ L { e a−b } = 0 e−sx ( e a−b ) dx R ax −ebx R 1 L { e a−b } = a−b limR−→∞ 0 [e−(s−a)x − e−(s−b)x ] dx RR RR ax −ebx 1 1 } = a−b limR−→∞ 0 e−(s−a)x dx − a−b limR−→∞ 0 e−(s−b)x dx L { e a−b ax −ebx L {e Now d(e−(s−a)x ) = e−(s−a)x d[−(s − a)x] d(e−(s−a)x ) = −(s − a)e−(s−a)x dx 1 − s−a d(e−(s−a)x ) = e−(s−a)x dx Similarly 1 − s−b d(e−(s−a)x ) = e−(s−b)x dx ax −ebx L {e a−b ax −ebx L {e a−b ax −ebx L {e a−b ax −ebx L {e a−b ax −ebx L {e a−b ax −ebx L {e a−b ax −ebx L {e a−b ax −ebx L {e a−b 1 1 } = − s−a limR−→∞ a−b RR 0 d(e−(s−a)x ) + 1 1 } = − (a−b)(s−a) limR−→∞ [ e(s−a)x ]R 0 + 1 (a−b)(s−b) 1 1 } = − (a−b)(s−a) limR−→∞ [ e(s−a)R − 1] + 1 } = − (a−b)(s−a) [−1] + }= 1 (a−b)(s−a) }= 1 [ 1 a−b s−a }= 1 s−b−(s−a) [ ] a−b (s−a)(s−b) }= 1 [ s−b−s+a) ] a−b (s−a)(s−b) − − 1 [−1] (a−b)(s−b) 1 (a−b)(s−b) 1 ] s−b 4 1 1 s−b a−b limR−→∞ RR 0 d(e−(s−b)x ) 1 limR−→∞ [ e(s−b)x ]R 0 1 (a−b)(s−b) 1 limR−→∞ [ e(s−b)R − 1] ax −ebx L {e a−b ax −ebx L {e a−b }= 1 [ a−b ] a−b (s−a)(s−b) }= 1 (s−a)(s−b) Applications (a)Find area under the curve y = x5 e−4x Solution Since L {xn−1 eax } = R∞ 0 e−(s−a)x xn−1 dx = (n−1)! (s−a)n (s > a) Putting n = 6, a = 3, s = 7 L {x5 e3x } = 5! (7−3)6 = 15 = 512 Area under the curve (b)Find area under the curve y= e−5x −e−3x −2 Solution Since ax −ebx L {e a−b }= 1 a−b limR−→∞ RR 0 [e−(s−a)x − e−(s−b)x ] dx = Also s − b − (s − a) = a − b = −2 Putting (s − a) = 5, (s − b) = 3 −5x −e−3x L {e −2 }= 1 5×3 = 1 = 15 Area under the curve 5 1 (s−a)(s−b)