Question

Solve (D² – 3D + 2)y = e^4x given y = 0 when x = 0 and x = 1

Answer 1

(D² – 3D + 2) y = e^4x

The auxiliary equation is m² – 3m + 2 = 0

(m – 1)(m – 2) = 0

m = 1, 2

The roots are real and different

C.F = Ae^m₁x + Be^m₁x

C.F = Ae^x + Be^2x 

P.I = `1/(("D"^2 - 3"D" + 2)) "e"^(4x)`

= `1/([(4)^2 - 3(4) + 2]) "e"^(4x)`

= `"e"^(4x)/((16 - 12 + 2))`

 `"e"^(4x)/6`

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

y = `"Ae"^x + "Be"^(2x) + "e"^(4x)/6`  ......(1)

When x = 0, y = 0

Equation (1) ⇒ 0 = `"Ae"^0 + "Be"^0 + "e"^0/6`

0 = `"A"(1) + "B"(1) + 1/6`

A + B = `0 - 1/6`

A + B = `- 1/6`   ........(2)

When x = 1, y = 0

Equation (1) ⇒ 0 = `"Ae"^1 + "Be"^2+ "e"^4/6` = 0

`"Ae" + "Be"^2 + "e"^4/6` = 0  ......(3)

From (2)

B = `(-1)/6 - A"`

Substitute B = `(-1)/6 - "A"` in equation (3)

`"Ae" + "e"^2 ((-1)/6 - "A") + "e"^4/6` = 0

`"Ae" - "e"^2/6 - "A"^2 + "e"^4/6` = 0

`"Ae" - "Ae"^2 = ("e"^2 - "e"^4)/6`

`"A"("e" - "e"^2) = (("e"^2 - "e"^4))/6`

A = `(("e"^2 - "e"^4))/(6("e" - "e"^2))`

`("e"^2(1 - "e"^2))/(6"e"(1 - "e")) = ("e"^2(1 + "e")(1 - "e"))/(6"e"(1 - "e"))`

A = `("e"(1 + "e"))/6`

⇒ A = `("e" + "e"^2)/6`

Then B = `- 1/6 - "A"`

B = `- 1/6 - (("e" + "e"^2)/6) = (-1 - "e" - "e"^2)/6`

B = `(-("e"^2 + "e" + 1))/6`

Substitute the values of A and B in equation (1)

y = `(("e" + "e"^2)/6) "e"^x - (("e"^2 + "e" + 1))/6 ("e"^(2x)) + "e"^(4x)/6`

y = `(("e"^2 + 2)"e"^x - ("e"^2 + "e" + 1)"e"^(2x) + "e"^(4x))/6`

⇒ 6y = `("e"^2 + "e")"e"^x - ("e"^2 + "e" + 1)"e"^(2x) + "e"^(4x)`