Question

a) Find the general solution of the differential equation Y'(B) + 2y(s) = (1)3 8>0. b) Find the inverse Laplace transform y(t) = --!{Y(s)}, where Y(s) is the solution of part (a). c) Use Laplace transforms to find the solution of the initial value problem ty"(t) – ty' (t) + y(t) = te", y(0) = 0, y(0) = 1, fort > 0. You may use the above results if you find them helpful. (Correct solutions obtained without Laplace transform methods may receive only partial credit.)

Answer 1

solution : (a) The given differential equation is Y'(s) + 5 Y(s) =(S-1)3,570, Here integrating factor is as des es multiplying both side by on both sides! Therefore multiplying I. F. and then integrating we get S2 Y(s).st -ds to where i ist - -( s de a constant (31) te (-) )ک y (6) .5 = 4 + 2 (3-1/2 te - YCS) - st 2 285-102t es? 05-12 where c is (2) an artitrary constant This is the general given equation solution de of the CamScanner

Now taking inverse Laplace transför- mation on both sidest we get F (409) = F" [ +32 + es'] yety ="[*]+CLOC] + "[«s?] [since it is linear] tel? (32) [:ct is linear transforming Fl" [$]+{"L}+] + { ["[06-0577 + 1" [5+1] +42". [1] + c !'[^^] =ét + et + E6592] +38' C++i) + 8(t) + r® Lt) [[ (sh) = din) (t)] where I 1 a, tão. yet = 2e + II (smje] + 5'641) + < S (1) + C JM%C4). d (t)=40, +70 Hence CS Scanned with CamScanner

The given problem is ty" – ty' (t) tyt) - tet, (0) 208 A 76(0)21 Taking Laplace transformation on both side of (3) we get, ehtynth{-ty!} +24%} = { ret} or; - 444"} + +1473 +2243 = -14tet} ...[LMSF (9) = 60'+'f++)] - + = (- ) Lý L, 13 29f"CH 3 := 5" L48 1493 - Sapcu) - 7 thy snýc h ::.- Fton] or - [s? 5- %60)=y0] + fe [sý - 900)] = h - 740) – 4'(0);! 7-1 [s5 - 1] + [17:] + 1 = 0, - 25 dy + 7 ts a Is ty