ORDINARY DIFFERENTIAL EQUATIONS

Although the differential equation in 1.1.10 has infinitely many solutions, the associatedinitial value problem has a unique solution.. 1.1.10,Example 1.1.8: Find the unique solution of

ORDINARY DIFFERENTIAL EQUATIONS

GABRIEL NAGY

Mathematics Department, Michigan State University, East Lansing, MI, 48824.

MAY 24, 2017

Summary This is an introduction to ordinary differential equations We describe the main ideas to solve certain differential equations, like first order scalar equations, second order linear equations, and systems of linear equations We use power series methods

to solve variable coefficients second order linear equations We introduce Laplace form methods to find solutions to constant coefficients equations with generalized source functions We provide a brief introduction to boundary value problems, eigenvalue- eigenfunction problems, and Fourier series expansions We end these notes solving our first partial differential equation, the heat equation We use the method of separation

trans-of variables, where solutions to the partial differential equation are obtained by solving infinitely many ordinary differential equations.

0

G NAGY – ODE May 24, 2017

Contents

2.2.4 Exercises 94

G NAGY – ODE May 24, 2017

6.2.4 The Stability of Nonlinear Systems 273

G NAGY – ODE May 24, 2017

Chapter 1 First Order Equations

We start our study of differential equations in the same way the pioneers in this field did

We show particular techniques to solve particular types of first order differential tions The techniques were developed in the eighteenth and nineteenth centuries and theequations include linear equations, separable equations, Euler homogeneous equations, andexact equations Soon this way of studying differential equations reached a dead end Most

equa-of the differential equations cannot be solved by any equa-of the techniques presented in the firstsections of this chapter People then tried something different Instead of solving the equa-tions they tried to show whether an equation has solutions or not, and what properties suchsolution may have This is less information than obtaining the solution, but it is still valu-able information The results of these efforts are shown in the last sections of this chapter

We present theorems describing the existence and uniqueness of solutions to a wide class offirst order differential equations

G NAGY – ODE may 24, 2017

1.1 Linear Constant Coefficient Equations1.1.1 Overview of Differential Equations A differential equation is an equation, wherethe unknown is a function and both the function and its derivatives may appear in theequation Differential equations are essential for a mathematical description of nature—they lie at the core of many physical theories For example, let us just mention Newton’sand Lagrange’s equations for classical mechanics, Maxwell’s equations for classical electro-magnetism, Schr¨odinger’s equation for quantum mechanics, and Einstein’s equation for thegeneral theory of gravitation We now show what differential equations look like

Example 1.1.1:

(a) Newton’s law: Mass times acceleration equals force, ma = f , where m is the particlemass, a = d2x/dt2is the particle acceleration, and f is the force acting on the particle.Hence Newton’s law is the differential equation

Remark: This is a second order Ordinary Differential Equation (ODE)

(b) Radioactive Decay: The amount u of a radioactive material changes in time as follows,

du

dt(t) = −k u(t), k > 0,where k is a positive constant representing radioactive properties of the material.Remark: This is a first order ODE

(c) The Heat Equation: The temperature T in a solid material changes in time and inthree space dimensions—labeled by x = (x, y, z)—according to the equation

Remark: This is a first order in time and second order in space PDE

(d) The Wave Equation: A wave perturbation u propagating in time t and in three spacedimensions—labeled by x = (x, y, z)—through the media with wave speed v > 0 is

on two or more independent variables, t, x, y, and z, and their partial derivatives appear inthe equations

The order of a differential equation is the highest derivative order that appears in theequation Newton’s equation in example (a) is second order, the time decay equation inexample (b) is first order, the wave equation in example (d) is second order is time and

space variables, and the heat equation in example (c) is first order in time and second order

in space variables

1.1.2 Linear Differential Equations We start with a precise definition of a first orderordinary differential equation Then we introduce a particular type of first order equations—linear equations

Definition 1.1.1 Afirst order ODE on the unknown y is

where f is given and y0 =dy

dt The equation islineariff the source function f is linear onits second argument,

Boyce-as y0+ a y = b, they get a plus sign on the left-hand side In any case, we stick here to theconvention y0= ay + b

(b) Another example of a first order linear ODE is the equation

We denote by y : D ⊂ R → R a real-valued function y defined on a domain D Such

a function is solution of the differential equation (1.1.1) iff the equation is satisfied for allvalues of the independent variable t on the domain D

Example 1.1.3: Show that y(t) = e2t−3

2 is solution of the equation

G NAGY – ODE may 24, 2017

Example 1.1.4: Find the differential equation y0 = f (y) satisfied by y(t) = 4 e2t+ 3.Solution: We compute the derivative of y,

y0= 2 y + 3,and integrate with respect to t on both sides,

Z

y0(t) dt = 2

Zy(t) dt + 3t + c, c ∈ R

The Fundamental Theorem of Calculus implies y(t) =R y0(t) dt, so we get

y(t) = 2

Zy(t) dt + 3t + c

Integrating both sides of the differential equation is not enough to find a solution y Westill need to find a primitive of y We have only rewritten the original differential equation

as an integral equation Simply integrating both sides of a linear equation does not solvethe equation

We now state a precise formula for the solutions of constant coefficient linear equations.The proof relies on a new idea—a clever use of the chain rule for derivatives

Theorem 1.1.2 (Constant Coefficients) The linear differential equation

(c) It makes sense that we have a free constant c in the solution of the differential tion The differential equation contains a first derivative of the unknown function y,

equa-so finding a equa-solution of the differential equation requires one integration Every nite integration introduces an integration constant This is the origin of the constant cabove

indefi-Proof of Theorem 1.1.2: First consider the case b = 0, so y0= a y, with a ∈ R Then,

y0 = a y ⇒ y

0

y = a ⇒ ln(|y|)0= a ⇒ ln(|y|) = at + c0,where c0∈ R is an arbitrary integration constant, and we used the Fundamental Theorem

of Calculus on the last step,R ln(|y|)0dt = ln(|y|) Compute the exponential on both sides,

y(t) = ±eat+c0 = ±ec0eat, denote c = ±ec0 ⇒ y(t) = c eat, c ∈ R

This is the solution of the differential equation in the case that b = 0 The case b 6= 0 can

be converted into the case above Indeed,

Remark: We solved the differential equation above, y0 = a y, by transforming it into atotal derivative Let us highlight this fact in the calculation we did,

ln(|y|)0 = a ⇒ (ln(|y|) − at
0= 0 ⇔ ψ(t, y(t))0= 0, with ψ = ln(|y(t)|) − at.The function ψ is called a potential function This is how the original differential equationgets transformed into a total derivative,

y0 = a y → ψ0= 0

Total derivatives are simple to integrate,

ψ0= 0 ⇒ ψ = c0, c0∈ R

So the solution is

ln(|y|) − at = c0 ⇒ ln(|y|) = c0+ at ⇒ y(t) = ±ec0 +at= ±ec0eat,

and denoting c = ±ec 0 we reobtain the formula

G NAGY – ODE may 24, 2017

Remark: We converted the original differential equation y0= 2 y + 3 into a total derivative

of a potential function ψ0 = 0 The potential function can be computed from the step

ln(|˜y|)0 = 2 ⇒ ln(|˜y|) − 2t
0

= 0,then a potential function is ψ(t, y(t)) = ln y(t) + 3

2



− 2t Since the equation is now

ψ0 = 0, all solutions are ψ = c0, with c0∈ R That is

cannot be generalized in a simple way to all linear equations with variable coefficients ever, there is a way to solve linear equations with both constant and variable coefficients—theintegrating factor method Now we give a second proof of Theorem1.1.2using this method.Second Proof of Theorem1.1.2: Write the equation with y on one side only,

How-y0− a y = b,and then multiply the differential equation by a function µ, called an integrating factor,

Eq (1.1.6), the differential equation for µ, which is simple to solve,

µ0 = −a µ ⇒ µ

0

µ = −a ⇒ ln(|µ|)
0

= −a ⇒ ln(|µ|) = −at + c0.Computing the exponential of both sides in the equation above we get

µ = ±ec0 −at= ±ec0e−at ⇒ µ = c1e−at, c1= ±ec0.Since c1is a constant which will cancel out from Eq (1.1.5) anyway, we choose the integrationconstant c0= 0, hence c1= 1 The integrating function is then

µ(t) = e−at.This function is an integrating factor, because if we start again at Eq (1.1.5), we get

e−aty0− a e−aty = b e−at ⇒ e−aty0+ e−at
0y = b e−at,

where we used the main property of the integrating factor, −a e−at = e−at
0 Now theproduct rule for derivatives implies that the left-hand side above is a total derivative,

e−aty
0

= b e−at

The right-hand side above can be rewritten as a derivative, b e−at =−b

We solve the example below following the second proof of Theorem1.1.2

Example 1.1.6: Find all solutions to the constant coefficient equation

The equation is a total derivative, ψ0 = 0, of

the potential function

ψ(t, y) =y +3

2



e−2t.Now the equation is easy to integrate,



y +32

c > 0

c = 0

c < 0

Figure 1 A few solutions

to Eq (1.1.8) for different c

We now solve the same problem above, but now using the formulas in Theorem1.1.2

G NAGY – ODE may 24, 2017

Example 1.1.7: Find all solutions to the constant coefficient equation

be interested only in solutions such that the particle is at a specific position at the initialtime Such condition is called an initial condition, and it selects a subset of solutions of thedifferential equation An initial value problem means to find a solution to both a differentialequation and an initial condition

Definition 1.1.3 Theinitial value problem (IVP) is to find all solutions y to

that satisfy the initial condition

where a, b, t0, and y0are given constants

Remark: The equation (1.1.11) is called theinitial conditionof the problem

Although the differential equation in (1.1.10) has infinitely many solutions, the associatedinitial value problem has a unique solution

Theorem 1.1.4 (Constant Coefficients IVP) Given the constants a, b, t0, y0 ∈ R, with

a 6= 0, the initial value problem

y0 = a y + b, y(t0) = y0,has the unique solution

y(t) = c eat−b

a.The initial condition determines the value of the constant c, as follows

Introduce this expression for the constant c into the differential equation in Eq (1.1.10),

Example 1.1.8: Find the unique solution of the initial value problem

Solution: All solutions of the differential equation are given by

y(t) = ce2t−3

2,where c is an arbitrary constant The initial condition in Eq (1.1.13) determines c,

1 = y(0) = c −3

2.Then, the unique solution to the initial value problem above isy(t) = 5

e3ty
0

= e3t.The exponential e3t is an integrating factor Integrate on both sides of the equation,

G NAGY – ODE may 24, 2017

1.1.6 Exercises

1.1.1.- Find the differential equation of the

form y0= f (y) satisfied by the function

y(t) = 8e5t−2

5.1.1.2.- Find constants a, b, so that

(b) Write the equations as a total

de-rivative of a function ψ, that is

y0= −4y + 2 ⇔ ψ0= 0

(c) Integrate the equation for ψ

(d) Compute y using part (c)

1.1.5.- Find all solutions of

y0= 2y + 51.1.6.- Find the solution of the IVP

y0= −4y + 2, y(0) = 5.1.1.7.- Find the solution of the IVP

dy

dt(t) = 3 y(t) − 2, y(1) = 1.1.1.8.- Express the differential equation

y0= 6 y + 1 (1.1.14)

as a total derivative of a potential tion ψ(t, y), that is, find ψ satisfying

func-y0= 6 y + 1 ⇔ ψ0= 0.Integrate the equation for the poten-tial function ψ to find all solutions y of

Eq (1.1.14)

1.1.9.- Find the solution of the IVP

y0= 6 y + 1, y(0) = 1

1.2 Linear Variable Coefficient Equations

In this section we obtain a formula for the solutions of variable coefficient linear equations,which generalizes Equation (1.1.4) in Theorem1.1.2 To get this formula we use the integrat-ing factor method—already used for constant coefficient equations in § 1.1 We also showthat the initial value problem for variable coefficient equations has a unique solution—just

as happens for constant coefficient equations

In the last part of this section we turn our attention to a particular nonlinear differentialequation—the Bernoulli equation This nonlinear equation has a particular property: it can

be transformed into a linear equation by an appropriate change in the unknown function.Then, one solves the linear equation for the changed function using the integrating factormethod The last step is to transform the changed function back into the original function.1.2.1 Review: Constant Coefficient Equations Let us recall how we solved the con-stant coefficient case We wrote the equation y0= a y + b as follows

y0= ay + b

a

.The critical step was the following: since b/a is constant, then (b/a)0= 0, hence



y + ba

0

= ay + b

a



At this point the equation was simple to solve,

(y +ab)0

(y +a

b) = a ⇒ ln y + b

a

0

= a ⇒ ln y + b

a ... data-page="27">

1.3 Separable Equations1 .3.1 Separable Equations More often than not nonlinear differential equations areharder to solve than linear equations Separable equations are an exception—they... sides of the differential equation We tried this idea

to solve linear equations, but it did not work However, it works for separable equations. Definition 1.3.1 Aseparabledifferential equation... two examples above we see that linear differential equations, with a 6= 0,are separable for b/a constant, and not separable otherwise Separable differential equationsare simple to solve We just