Ex 3. Second Order Liner ODE with Constant Coefficients

Calculus 2



Weiqi Zhou

Form

The Characteristic Polynomial

  • Consider the map ， is a differential operator

  • Homogeneous equation:
    Non-homogeneous equation

  • Notice that for any ，.
    The polynomial is called the characteristic polynomial of the operator

The Approach

  • If the characteristic polynomial admits two distinct roots ，then both solve the characteristic equation

  • are linearly independent

  • Consequently any solution can be written as , for some

The Approach

  • If the characteristic polynomial admits a double root ，then is a solution

  • Suppose the other linearly independent solution is ，then



The Approach

  • Plug it back into the original equation to get


  • which is


  • Plug and in to get

The Approach

  • The origianl equation now transforms into

  • Apparently , thus

  • Th erefore and are linearly independent solutions

  • Consequently any solution can be written as , for some

The Solution

  • If admits two distinct roots ，then soluions are

  • If admits a double root ，then solutions are

Example

  • Find the solution of that satisfies
    , and compute

  • The characteristic polynomial is , it has roots

  • a solution takes the form

Example

  • Solve to get

  • Clearly

  • Therefore is the Fibonacci sequence

Exercises

  • Solve

  • Solve

  • Solve

  • Find the solution of that satisfies

Non-homogeneous Cases

Producing a particular solution is difficult in general, we consider only the case when

where is a constant, and is a polynomial of order

The Approach

  • Suppose that the solution is ，take

  • Plug it into LHS of the euqation to get


  • With ，it further simplifies into

The Approach

  • being a solution implies , thus


  • Obviously , hence


  • Therefore is also a polynomial，there are three cases then

The Approach



  • If (i.e., is not a root of the characteristic polynomial)，then is of order

  • If , and (i.e., is simple root of the characteristic polynomial)，then is of order

  • If (i.e., is a double root of the characteristic polynomial) ，then is of order

Example

  • Solve



  • is a simple root of the characteristic polynomial，there is a quadratic polynomial that satisfies

  • Suppose , then
    i.e., ，

Example

  • is a particular solution of the original equation，where is arbitrary

  • The solution to the associated homogeneous euqation is , where are constants

  • Absorbing into we obtain

Exercises

  • Solve

  • Solve

  • Solve s

Exercises

  • Solve

  • Solve

  • Find the solution of that satisfies

A Particualr Case

For

or

we may simply solve ,
and then take the real/imaginary part of the solution

Exercise

Solve

Summary

  • The General Form

  • The Characteristic Polynomial

  • Solving a Homogeneous Equation

  • Solving a Non-homogeneous Equation