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b a d (s+1) S-1 5+1 st-2s +1 Example 3: Find the solution to the 4th order IVP y'' - 2y" + y = 0, y(0) = 1, y'0) = -1, y"0) = 0, and y'" (0)=2. Solution: Step 1: Take the LT of this IVP: (s*Y(s) - sy(0) - say'(0) - sy"(0) - y" (0)) - 2 (s2Y(s) - sy(0) - y'(0)) + Y($) = 0 so (s* – 252 + 1)Y(s) – 53 +52 - 2 - 2 = 0 which leads to Y(s) = 52-52-28+4 Step 2: Write st-2s+1 Y(s) in terms of known LTs: Because s4 - 2s2 +1 = (s2 – 1)2 = (s - 1)(s + 1)2 the partial fraction decomposition of Y(s) has form Y(s) = + $9-52 +4 ++ To complete this step, (S-1) 3-1 (3-1) +86-132 + 5* + 6+12 a(s-1)(s+1)2 +b(s+1)2 +c(5–192(s+1)+d(s+1)2 – (a+c)s2+(a+b+c+d)s2+(-a+2b-c-2d)s+(-a+b+c+d) 3°(8+1)(8+2) 2.Equating like powers of sin $(s+1)(8+2) the numerator: =1, _=-2, and _=4. These can be separated into 2 pairs of equations: 2b-2d=-1 and 2b+2d=3, which mean b=_ and d and a+c=1 and a-c=-5/2, which mean ar and ca Thus, y(s) = Step 3: Find ILT of y(s): y(t) =- Wow! That was a fourth-order linear DE. In fact, this method works for any linear DE with constant coefficients.

Answer 1

here, 2 e et + 1 t. e² +7 et + toe-t s?-52-25+4. a + b. d 5-25+) S- (8-112 (5+1) St12 (2+0)33+ (atbac+d)s+(-a+2b-c-3d) 54 (-atbreid) (S-1)2(S+1) -Q+2b-c-2d=-2 the numerator : a+c=1, a+b-c+d=-1, & -a+b+c+d=4 Now solving 26-2d=-1 & 2b+2d=3 => b=1 & derella d = 1 a+c=1 Solving & a-(=-5/ we get a- C=7 9 → Y(s)= - 3 I ? + + 405-1) 2651)2' 465+1) (5+1) since I leat)= & L (*) - 9 » Y(t) = -3 1 1 4 2 "(t) = -3 et + 1 tet 2 -3 > + (5+1) - Now s-a sht + + Lone} 크 나 4 Larry I be 6-t t te-t 4 Y 2