Question

Solve the following differential equation:

(D² – 3D + 2)y = e^3x which shall vanish for x = 0 and for x = log 2

Answer 1

(D² – 3D + 2)y = e^3x

The auxiliary equation is

m² – 3m + 2 =0

(m – 1)(m – 2) = 0

m = 1, 2

Roots are real and different

The complementary function is

C.F = Ae^m₁x + Be^m₂x

C.F = A^x + Be^2x 

P.I = `1/("D"^2 - 3"D" + 2) "e"^(3x)`

= `1/((3)^2 - 3(3) + 2) "e"^(3x)`

P.I = `"e"^(3x)/2`

The general solution is P = C.F + P.I

y = `"Ae"^x + "Be"^(2x) + "e"^((3x)/2)`   .......(1)

When x = 0, y = 0

`"Ae"^0 + "Be"^0 + "e"^0/2` = 0

`"A" + "B" + 1/2` = 0

`"A" + "B" = (-1)/2`

⇒ (12)

When x = log 2, y = 0

`"Ae"^(log2) + "BE"^(2log2) + ("e"^(3xlog2))/2` = 0

`"Ae"^(log2) + "BE"^(2log2) + "e"^(log2)/2` = 0

`2"A" + 4"B" + 8/2` = 0

2A + 4B + 4 = 0

2A + 4B = – 4  .........(3)

Solving equation (2) and (3)

Equation (2) × 2

⇒ 2A + 2B = – 1

Equation (3)

⇒ 2A + 4B = – 4
   (–)     (–)      (+)
         – 2B  =   3 

∴ B = `(-3)/2`

Substitute the value of B = `(-3)/2` in equation (2)

`"A" - 3/2 = - 1/2`

A = `(-1)/2 + 3/2`

A = `2/2` = 1

∴ A = 1

Substitute A = 1 and B = `(-3)/2` in equation (1)

y = `"e"^x - 3/2 "e"^(2x) + "e"^(3x)/2`