CHARACTERISTIC EQUATIONS AND IVP
1. Find the general solution to the equation (using the characteristic polynomials).
2๐ฆโฒโฒโฒ+๐ฆโฒโฒ+2๐ฆโฒ=0โ2๐3+๐2+2๐=0
since ๐(๐) do not have a constant term and the term with the lowest power is zr, then
๐(๐)
๐=2๐2+๐+2=0
Then, ๐=0. To find the remaining roots, we can apply the quadratic formula
๐=โ๐ยฑโ๐2โ4๐๐
2๐ =โ1ยฑโ12โ4(2)(2)
2(2) =โ1ยฑโ15๐
4=โ1
4ยฑ๐โ15
4
with roots ๐=0,โ1
4+๐โ15
4,โ1
4โ๐โ15
4
Thus, the general solution is given by: ๐ฆ๐บ=๐1๐๐1๐ฅ+๐2๐๐2๐ฅ+โฏ+๐๐๐๐๐๐ฅ
๐ฆ๐บ=๐1+๐2๐โ1/4๐ฅcosโก(โ15๐ฅ/4)+๐3๐โ1/4๐ฅsinโก(โ15๐ฅ/4)
2. Using Laplace transforms, solve the IVP
๐ฆโฒโฒโ3๐ฆโฒ+2๐ฆ=20sin(2๐ก),โกโกโก๐ฆ(0)=0,๐ฆโฒ(0)=0
โ{๐ฆโฒโฒโ3๐ฆโฒ+2๐ฆ}=โ{20sinโก(2๐ก)}โ ๐
๐ 2+๐2,๐ >0
๐ 2๐(๐ )โ๐ ๐ฆ(0)โ๐ฆโฒ(0)โ3[๐ ๐(๐ )โ๐ฆ(0)]+2๐(๐ )=20(2
๐ 2+22)
Substitute the terms,
๐ 2๐(๐ )โ0โ0โ3[๐ ๐(๐ )โ0]+2๐(๐ )= 40
๐ 2+4
๐(๐ )(๐ 2โ3๐ +2)=40
๐ 2+4
๐ฆ(๐ )= 40
๐ 2+4(๐ 2โ3๐ +2)
40
(๐ 2+4)(๐ 2โ3๐ +2)=40
(๐ 2+4)(๐ โ1)(๐ โ2)=๐ด๐ +๐ต
(๐ 2+4)+๐ถ
(๐ โ1)+๐ท
(๐ โ2)
40=๐ด๐ +๐ต(๐ โ1)(๐ โ2)+๐ถ(๐ 2+4)(๐ โ2)+๐ท(๐ 2+4)(๐ โ1)
40=๐ด๐ 3โ3๐ด๐ 2+2๐ด๐ +๐ต๐ 2โ3๐ต๐ +2๐ต+๐ถ๐ 3โ2๐ถ๐ 2+4๐ถ๐ โ8๐+๐ท๐ 3โ๐ท๐ 2+4๐ท๐ โ4๐ท
