Question

Solve the following differential equation:

(D² – 2D – 15)y = 0 given that `("d"y)/("d"x)` = 0 and `("d"^2y)/("d"x^2)` = 2 when x = 0

Answer 1

The auxiliary equation is

m² – 2m + 15 = 0

m² + 3m – 5m – 15 = 0

m(m + 3) – 5(m + 3) = 0

(m + 3)(m – 5) = 0

m = – 3, 5

Roots are real and different

∴ The complementary function is

Ae^m₁x + Be^m₂x

C.F = Ae^–3x + Be^5x

∴ The general solution is

y = (Ae^–3x + Be^5x)  ........(1)

`("d"y)/("d"x)` = Ae^-3x (-3) + Be^5x (5)

`("d"y)/("d"x)` = – 3Ae^–3x + 5Be^5x   ........(2)

`("d"^2y)/("d"x^2)` = 9Ae^–3x + 25Be^5x  .......(3)

When x = 0, `("d"y)/("d"x)` = 0

3 Ae° + 5Be° = 0

– 3A + 5B = 0  ........(4)

When x = 0, `("d"^2y)/("d"x^2)` = 2

Equation (3)

⇒ 9Ae° + 25Be° = 2

9A + 25B = 2  ........(5)

Solving equation (4) and (5)

Equation (4) x 3 ⇒ – 9A + 15B = 0
Equation (5)      ⇒    9A  + 25B = 2
                                          40B = 2

∴ B = `2/40`

B = `1/20`

Substitute B = `1/20` in equation (4)

` 3"A" + 5(1/20)` = 0

` 3"A" + 1/4` = 0

` -3"A" = (-1)/4`

A = `1/12`

Substitute the value of A = `1/12` and B = `1/20` in equation (1)

y = `1/12 "e"^(-3x) + 1/20 "Be"^(5x)`