Solve the initial value problem y 3y 10y 24y 0 y0 0 y0

Solve the initial value problem {y\"\' + 3y\" - 10y\' - 24y = 0 y(0) = 0 y\'(0) = 10 y\"(0) = 20

Solution

Given that

y\'\'\' + 3y\'\' - 10y\' - 24y = 0

The D-operator form is ,

(D³ + 3D² -10D - 24 ) y = 0

The auxialary equation is,

m³ + 3m² -10m - 24 = 0

( m - 3 ) (m + 4) (m + 2) = 0

m₁ = 3 , m₂ = -4 , m₃ = -2

The roots are real and distinct .

The solution is , y(x) = c₁ e^m₁x + c₂ e^m₂x + c₃ e^m₃x

   y(x) = c₁ e^3x + c₂ e^-4x + c₃ e^-2x.....................................1

Given that

y(0) = 0 , y\'(0) = 10 , y\'\'(0) = 20

At y(0) = 0

   y(0) = c₁ e^3.0 + c₂ e^-4.0 + c₃ e^-2.0

0 = c₁ + c₂ + c₃     [ since , e⁰ = 1 ]

   c₁ + c₂ + c₃ = 0........................................2

At y\'(0) = 10

y(x) = c₁ e^3x + c₂ e^-4x + c₃ e^-2x

y\'(x) = c₁ e^3x.3 + c₂ e^-4x.-4 + c₃ e^-2x.-2 [ since, d/dx(e^ax) = ae^x , a = constant ]

  y\'(x) = 3c₁ e^3x - 4c₂ e^-4x - 2c₃ e^-2x

  y\'(0) = 3c₁ e^3.0 - 4c₂ e^-4.0 - 2c₃ e^-2.0

10 = 3c₁ - 4c₂ - 2c₃ [ since , e⁰ = 1 ]

3c₁ - 4c₂ - 2c₃ = 10 ...................................3

At y\"(0) = 20

   y(x) = c₁ e^3x + c₂ e^-4x + c₃ e^-2x

  y\'(x) = 3c₁ e^3x - 4c₂ e^-4x - 2c₃ e^-2x

  y\'\'(x) = 3c₁ e^3x.3 - 4c₂ e^-4x.-4 - 2c₃ e^-2x.-2

y\'\'(x) = 9c₁e^3x + 16c₂e^-4x + 4c₃e^-2x

y\'\'(0) = 9c₁e^3.0 + 16c₂e^-4.0 + 4c₃e^-2.0

   20 = 9c₁ + 16c₂ + 4c₃    [ since , e⁰ = 1 ]

9c₁ + 16c₂ + 4c₃ = 20...........................................4

Solve equations 2 and 3

Multiply the equation 2 with 3

  c₁ + c₂ + c₃ = 0

   3c₁ + 3c₂ + 3c₃ = 0

3c₁ = - 3c₂ - 3c₃.........................5

From equation 3

   3c₁ - 4c₂ - 2c₃ = 10

3c₁ = 10 + 4c₂ + 2c₃ .........................6

Equating the equations 5 and 6

  - 3c₂ - 3c₃ =10 + 4c₂ + 2c₃

   7c₂ +5c₃ = -10................................7

Solve equations 3 and 4

  3c₁ - 4c₂ - 2c₃ = 10

Multiply equation 3 with 3

    9c₁ - 12c₂ - 6c₃ = 30

    9c₁ = 30 +12c₂ + 6c₃ .............................8

From equation 4

  9c₁ + 16c₂ + 4c₃ = 20

  9c₁ =20 -16c₂ - 4c₃ ..............................9

Equating the equations 8 and 9

30 +12c₂ + 6c₃ =20 -16c₂ - 4c₃

   28c₂ +10c₃= -10 ..............................10

Solve equations 7 and 10

   7c₂ +5c₃ = -10

Multiply equation 7 with 4

28c₂+ 20c_{3 =} -40

28c₂ = -40 - 20c₃................................11

From equation 10

28c₂ +10c₃= -10

28c₂ = -10 - 10c₃.........................12

Equating the equations 11 and 12

   -40 - 20c₃ = -10 - 10c₃

-10c₃ = 30

c₃ = -30/10

c₃ = -3

Substitute c₃ = -3 in    7c₂ +5c₃ = -10

7c₂ + 5(-3) = -10

7c₂ -15 = -10

7c₂ = 5

c₂ = 5/7

From equation 2

     c₁ + c₂ + c₃ = 0

c₁ + 5/7 - 3 = 0

c₁ - 16/7 = 0

c₁ = 16/7

Therefore,

c₁ = 16/7 , c₂ = 5/7 , c₃ = -3

Substitute c_{1 ,} c₂ , c₃ in equation 1

y(x) = c₁ e^3x + c₂ e^-4x + c₃ e^-2x

   = (16/7)e^3x + (5/7)e^-4x - 3e^-2x

Therefore,

The general solution is ,

y(x) =  (16/7)e^3x + (5/7)e^-4x - 3e^-2x