Question

(D^3-2D^2-5D+6)y=e^3x

Answer 1

      solution

(D³-2D²-5D+6)y=e∧3x

let find the complementary factor

⇒m³-2m²-5m+6=0

 let set the factor of 1

⇒ when x-1=0

 let check and see (1)³-2(1)-5(1)+6=0

 ⇒x=1 is the factor

  using long division method

                m²-m-6

  (m-1)√(m³-2m²-5m+6)

            m³-m²

                 -m²-5m+6

                 -m²+m

                       -6m+6

                       -6m+6

                               0

     ⇒ m²-m+6=0 is also a factor

   m=-2   or    m=3

⇒ the factors are m=(1,-2,3)

  ⇒      the complementary factor is required as

  y=c₁e∧x+c₂e∧-2x+c₃e∧3x

  then let  find the  other factor

 y=e∧3x/(D³-2D²-5D+6)

 ⇒Y=e∧3x(3³-2(3)-5(3)+6)

 ⇒y=27-6-15+6

 ⇒y=12

⇒ the solution is

   y=c₁e∧x+c₂e∧-2x+c₃e∧3x+12e∧3x

Answer 2

Answer:

Step-by-step explanation: