Math 240: Linear Differential Equations - Lecture Notes, Slides of Differential Equations

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Math 240: Linear Differential Equations

Ryan Blair

University of Pennsylvania

Tuesday February 15, 2011

Outline

(^1) Review

(^2) Today’s Goals

3 General Solutions

4 Results For Nonhomogeneous Equations

1 Solving D.E.s Using Auxiliary Equations

Review

(^1) The following is a general nth-order linear D.E.

an(x)

dny dxn^

• an− 1 (x)

dn−^1 y dxn−^1

• ...a 1 (x)

dy dx

• a 0 (x)y = g (x)

Review

(^1) The following is a general nth-order linear D.E.

an(x)

dny dxn^

• an− 1 (x)

dn−^1 y dxn−^1

• ...a 1 (x)

dy dx

• a 0 (x)y = g (x)

(^2) For a linear homogeneous D.E., linear combinations of solutions

are again solutions.

Today’s Goals

Today’s Goals

(^1) Construct general solutions to homogeneous and nonhomogeneous linear D.E.s (^2) Use auxiliary equations to solve constant coefficient linear homogeneous D.E.s

Today’s Goals

Linearly Independent Solutions

Theorem Let y 1 , y 2 , ..., yn be n solutions to a homogeneous linear nth-order differential equation on an interval I. The the set of solutions is linearly independent on I if and only if W (y 1 , y 2 , ..., yn) 6 = 0 for every x in the interval.

Today’s Goals

Linearly Independent Solutions

Theorem Let y 1 , y 2 , ..., yn be n solutions to a homogeneous linear nth-order differential equation on an interval I. The the set of solutions is linearly independent on I if and only if W (y 1 , y 2 , ..., yn) 6 = 0 for every x in the interval.

If the solutions y 1 , y 2 , ..., yn are linearly independent they are said to be a fundamental set of solutions. Note: There always exists a fundamental set of solutions to an nth-order linear homogeneous differential equation on an interval I.

General Solutions

General Solution

Theorem Let y 1 , y 2 , ..., yn be a fundamental set of solutions set of solutions to an nth-order linear homogeneous differential equation on an interval I. Then the general solution of the equation on the interval is

y = c 1 y 1 (x) + c 2 y 2 (x) + ... + cnyn(x) where the ci are arbitrary constants.

Results For Nonhomogeneous Equations

Superposition Principle for Nonhomogeneous Equations

Theorem Suppose ypi denotes a particular solution to the differential equation

an(x)

dny dxn^

• an− 1 (x)

dn−^1 y dxn−^1

• ...a 1 (x)

dy dx

• a 0 (x)y = gi (x)

Where i = 1, 2 , ..., k. Then yp = yp 1 + yp 2 + ... + ypk is a particular solution of

an(x)

dny dxn^

• an− 1 (x)

dn−^1 y dxn−^1

• ...a 1 (x)

dy dx

• a 0 (x)y =

g 1 (x) + g 2 (x) + ... + gk (x)

Solving D.E.s Using Auxiliary Equations

A Motivating Example

Our goal is to solve constant coefficient linear

homogeneous differential equations.

Solving D.E.s Using Auxiliary Equations

A Motivating Example

Our goal is to solve constant coefficient linear

homogeneous differential equations.

What if we guess y = emx^ as a solution to

y ′′^ + y ′^ − 6 y = 0?

What if we guess y = emx^ as a solution to

ay ′′^ + by ′^ + cy = 0?

Solving D.E.s Using Auxiliary Equations

A Motivating Example

Our goal is to solve constant coefficient linear

homogeneous differential equations.

What if we guess y = emx^ as a solution to

y ′′^ + y ′^ − 6 y = 0?

What if we guess y = emx^ as a solution to

ay ′′^ + by ′^ + cy = 0?

In this case, we get emx^ (am^2 + bm + c) = 0. There are

three possibilities for the roots of a quadratic equation.

Solving D.E.s Using Auxiliary Equations

Case 2: Repeated Roots

If am^2 + bm + c has a repeated root m 1 , then the general

solution to ay ′′^ + by ′^ + cy = 0 is

y = c 1 em^1 x^ + c 2 xem^1 x

Solving D.E.s Using Auxiliary Equations

Case 3: Complex Roots

If am^2 + bm + c has complex roots m 1 = α + i β and

m 2 = α − i β, then the general solution to

ay ′′^ + by ′^ + cy = 0 is

y = c 1 eαx^ cos(βx) + ic 2 eαx^ sin(βx)