Solve the initialvalue problem y6y9y0 initial conditions y10

Solve the initial-value problem y\"+6y\'+9y=0 initial conditions y(1)=0 and y\'(1)=1

Solution

Given that y\"+6y\'+9y=0

The symbolic form is [D² + 6D +9]y = 0

The auxiliary equation is

m²+6m+9 =0

(m+3)² =0

m =-3,-3

Hence,

the complementary function is y = (c₁+c₂x)e^-3x

Given that y(1)=0,

0 = (c₁+c₂.1)e^-3.1

c₁+c₂ = 0

Also given that y\'(1)=1.

  y = (c₁+c₂x)e^-3x = c₁e^-3x + c₂xe^-3x

y\' = c₁.-3.e^-3x + c₂[ x.-3e^-3x+ e^-3x.1]

y\' = -3c₁ e^-3x -3c₂ xe^-3x+ c₂ e^-3x

Since c₁+c₂ =0, put c₂=-c₁ in above equation.

y\' = -3c₁e^-3x -3(-c₁)xe^-3x+ (-c₁)e^-3x

   y\' = -3c₁e^-3x +3c₁xe^-3x-c₁e^-3x

   y\' = 3c₁ xe^-3x - 4c₁ e^-3x

   y\'= c₁[3xe^-3x- 4e^-3x]

Now apply, y\'(1)=1

1 = c₁[3.1.e^-3.1-4 e^-3.1]

1 = c₁[3e^-3-4 e^-3]

1 = -c¹e^-3

c₁ = -e³

Then, c₂ = -c₁ = -[-e³] = e³

Hence, substitute c1 and c2 in y = (c₁+c₂x)e^-3x

   Therefore,

y = ( -e³+ e³x)e^-3x