Question

Apply the appropriate La Place transform properties to get the solution to the problem. (It can only be solved by this method).  Can someone please solve it STEP BY STEP without skipping any step please? It would be really helpful. I´m lost ):

x" - 3x' = te-t x(0) = 0 x'(0) = 0

Answer 1

Fact: 1 2[fm (4) s" L[fc] - $hf(0) = 5"-25'0)... 42-22 16-1) -Sf 0) f(o) Fact! 2 (First ( First shifting shifting the onem): Leat fcul FCS-a) where F(S) = 2 [ f(t)] Fact : 3 La f (t) + ß g (H) XL [ f(t) + B L [ glt)] CS S CamScaniter Scanned da is linear

1 x' tet 3X taplace we get apply LE both side L[tet 11 38 IT 2[x") - 32[x] Lłtet 6: Faet (31) CS Scanned with CamScanner

- 3 SLExt] X00?) → (5? LEXO) - S X(0) – x'(0) L[tet fact 1 (put n=2 and n=1 n=1 + -> (52-39.) Lo [x4] = (st? Fact 2 and L[6] = and X(0) = x'(o)=0 → to = txt] CS canned with S(9-3)(8+12 am Scanner

Let us use partial fraction + + + IS (5-3) (571)? D S+1)? S 15-3) (5+1) 1 - (2 A CS-3)(8+1) + BCS) (Stu- it c (s)(5-3)(5+1) + D65) (5-3) put fo both side 1563) A I we get A = 3 - 5 put S=3 both side we get = 48 both side, we get put S= - D - T Compare the coefficient of 3 both side of ep. (2) 0 A+ B + c - -A-1 1 / 1 48 - scan16

So eg. (1) Becomes. - + + To Set 5 16(5+1) 4 (5+1)? IS S(53) (5+1)? 4818-3 + + + So eg. (0) Becomes 5 2 Ext)) = - 48(5-3) HotSth 9 (5+1)2 taking intense taplace both side get X (D) - 1/2 L ( t) L suz) we + 48 + d + L St (5+1)? 4 To 35 5 e - 1 / 2 ét I ett + t 48 t6 n! th al and first shifting theorem sh) 3t + + XU. 5 16 et hétt 48 Ang cam Scanner