Linear Algebra and Differential Equations

Sujin Khomrutai, Ph.D.

Department of Math & Computer Science Chulalongkorn University

Lecture 1

1 Introduction

2 Some Aspects

3 First Order ODEs: A Review

4 Second Order Linear Equations

Definition

A differential equation(or DE) is an equation involving a function and some of its derivatives.

EX.

1 y⁰ = 2xy is a DE. (variable x, a function y(x).)

2 x²+u² = 4 is nota DE. No derivative!

3 t²+ (y⁰⁰)² = 4 is a DE. (variablet, a function y(t).)

4 ux+ 5uy = 1 is a DE. (variablesx,y, a functionu(x,y).)

Definition

A system of DEsis a group of equations involving two or more functions and some of their derivatives.

EX.

1

(u⁰ = 2u+v

v⁰=−u+ 4v. ⇒ a system of DEs. (functions u,v)

2

(x+ 2y= 2

−x+y = 0. ⇒not a system of DEs. No derivatives!

3







u= 3u−v v⁰ =u+w w⁰=−1

⇒ a system of DES. (functions u,v,w)

Definition In a DE,

1 only one variable⇒ ordinary differential equation (or ODE).

2 two or more variables⇒ partial differential equations(or PDE).

System of ODEs = 1 variable, at least 2 functions

System of PDEs= at least 2 variables, at least 2 functions.

EX.

1 y⁰ = 2xy is an ODE.

2 u_xx+u_yy = 1 is a PDE.

3

(u⁰ = 2u+v

v⁰=−u+ 4v is a system of ODEs.

Goal. We study DEs and sytems of DEs.

Variable. t= time, or x= a space variable.

Functions. y (ODE), vectors~x= (x₁, . . . ,x_n) (Systems of ODES) Derivatives.

y⁰ = dy

dt, y⁰⁰ = d²y

dt², . . . ,y⁽ⁿ⁾= dⁿy dtⁿ, . . . and

~

x⁰ = (x₁⁰,x₂⁰, . . . ,x_n⁰), ~x⁰⁰= (x₁⁰⁰,x₂⁰⁰, . . . ,x_n⁰⁰), . . .

Definition (Order)

For a DE or system of DEs, the highest order of derivative appeared is called the order.

EX.

1 y⁰⁰+ 2y⁰=t³ ⇒ order = 2. (ODE)

2 y⁽⁴⁾−y⁰⁰+ (y⁰)⁵= 0 ⇒ order = 4. (ODE)

3

(x₁⁰ = 2x1−x2

x₂⁰ =−x₁+x2

⇒ order = 1. (system of ODEs)

4 u_t+u(u_xxt)⁵ = 0⇒ order = 3. (PDE)

Definition (Solutions)

A solution to a DE is a function that the equation becomes true when we substitute the function and its derivatives into the equation.

EX.

1 y⁰ = 2t ⇒ y(t) =t².

2 y⁰ = 2y ⇒ y(t) =e^2t.

3 y⁰+ 3y = 6 ⇒y(t) =e^−3t+ 2.

4 y⁰⁰−3y⁰+ 2y = 0 ⇒ y(t) =C1e^t+C2e^2t.

5 y⁽⁴⁾ = 1 ⇒ y(t) = ₂₄^t4.

Definition (Particular & General solutions)

A solution to a DE that contains no arbitrary constants is called a particular solution.

A solution to a DE that contains some arbitrary constants and represents all possible particular solution is called ageneral solution.

EX.Consider the ODE y⁰ = 2t.

y₁(t) =t² and y₂(t) =t²+ 3 are solutions. They are particular solutions.

To find a general solution, we need integration y⁰= 2t ⇒

Z

y⁰dt = Z

2tdt which gives

y(t)−y(0) =t² ⇒ y(t) =t²+C. where C =y(0) is an arbitrary constant.

Definition (Linear vs Nonlinear)

An ODE is called alinear ODEif it can be expressed as any⁽ⁿ⁾+· · ·+a2y⁰⁰+a1y⁰+a0y =f(t).

Otherwise, it is called nonlinear ODE.

EX.

1 y⁰⁰+ 2ty⁰−3y =e^t ⇒linear.

2 ty⁽⁴⁾= 3y⁰⁰+ sint ⇒ linear.

3 y⁰+y² = 0⇒ nonlinear.

4 y⁰⁰⁰+y⁰⁰+ 2y⁰+ 4y =x² ⇒ linear. (Observe: variable isx.)

Definition (Initial Value Problems)

An initial value problem (or IVP) is a problem that has a DE together with some conditions for the solutions at some t₀ (or somex₀).

Solving IVP⇒ Particular solution.

EX.Solve the initial value problem

y⁰ = 3t+ 2, y(0) = 0.

Sol. We integrate

y⁰ = 3t+ 2 ⇒ Z

y⁰dt = Z

(3t+ 2)dt then

y(t)−y(0) = 3t² 2 + 2t Using y(0) = 0, we obtain

y(t) = 3t² 2 + 2t.

Definition

A first order ODE is an equation of the form F(t,y,y⁰) = 0.

If it can be put into the form

y⁰=f(t)g(y) it is calledseparable.

If it can be put into the form

y⁰+p(t)y =q(t) it is a linear equation.

How to solve separable eqns.?

To solve a separable equation, we write dy

dt =f(t)g(y) ⇒ 1

g(y)dy =f(t)dt.

Then integrate

Z 1 g(t)dy =

Z

f(t)dt.

Solution can be implicitly defined.

EX.Solve the ODE

dy dt = t

y. Sol. We write

ydy=tdt.

Integrate Z

ydy = Z

tdt ⇒ y(t)²

2 −y(0)² 2 = t²

2. Setting C =y(0)², we get

y(t)²=t²+C. This is a general solution.

EX.Sole the ODE

dy

dx = 3x² y−siny. Sol. We express

dy

dx = 3x²

y−siny ⇒ (y−siny)dy = 3x²dx.

Then integrate Z

(y−siny)dy = Z

3x²dx ⇒ y²

2 + cosy =x³+C. The solution is implicit!

EX.Sole the ODE

y⁰ =xy. Sol. We write

y⁰ =xy ⇒ 1

ydy =xdx.

Integrate

Z 1 ydy =

Z

xdx ⇒ lny(t) = x² 2 +C. So

y(t) =e^x

2

2+C =De^x

2 2.

Definition

For a linear equation

y⁰+p(t)y =q(t), theintegrating factor is the function

I(t) =e

Rp(t)dt.

Multiplying the ODE with I(t) and integrating, then y(t) = 1

I(t) Z

I(t)q(t)dt+C

.

EX.Solve the ODE

y⁰+ 3t²y = 6t².

Sol. This is a linear equation: p(t) = 3t² andq(t) = 6t². The integrating factor

I(t) = exp Z

3t²dt =e^t3. Plug into the formula

y(t) = 1 e^t3

Z

e^t36t²dt+C

∴ y(t) =e^−t3

2e^t3+C

= 2 +Ce^−t3.

x²y⁰+xy = 1 (x>0), y(1) = 2.

Sol. Variable isx! Divide by x² to bring coefficient of y⁰ to 1:

y⁰+ 1 xy = 1

x².

This is a linear equation: p(x) = 1/x and q(x) = 1/x². Integrating factor

I(x) = exp Z 1

xdx =e^lnx =x.

Plug into the formula y(x) = 1

x Z

x· 1

x²dx+C

= lnx+C

x .

Definition

A second order ODE is an equation of the form F(t,y,y⁰,y⁰⁰) = 0.

If the ODE can be put into the form

a(t)y⁰⁰+b(t)y⁰+c(t)y =f(t) or

y⁰⁰+p(t)y⁰+q(t)y =r(t) it is called a second order linear ODE.

Ifa,b,c are constants, the ODE is calledconstant coefficients.

Iff(t) = 0 or r(r) = 0, the ODE is calledhomogeneous.

EX.Consider the ODE

3y⁰⁰+ 2y⁰+ 4y=t²+ 1.

This is a second order ODE.

It is linear, constant coefficients, and nonhomogeneous.

EX.Consider the ODE

5y⁰⁰+ (tant)y⁰+y²= 0.

This is a second order ODE.

It is nonlinear because of the term y².

EX.Consider the ODE

ty⁰⁰+ (t+ 1)y⁰+t²y = 0.

This is a second order ODE

a(t) =t, b(t) =t+ 1, c(t) =t², f(t) = 0.

It is linear, non-constant coefficients, and homogeneous.

We can divide by t to get another form y⁰⁰+ t+ 1

t y⁰+ty = 0.

So p(t) = (t+ 1)/t andq(t) =t.

Definition

If a second order ODE is supplied with two conditions of the form y(t₀) =k, y⁰(t₀) =l

it is called an initial value problem (or IVP).

Thus an IVP for a second order linear ODE is (IVP)

(y⁰⁰+p(t)y⁰+q(t)y =f(t) y(t0) =k, y⁰(t0) =l. Usually but not always, we choose t0 = 0.

Theorem

Consider an IVP of a second order linear ODE (IVP)

(y⁰⁰+p(t)y⁰+q(t)y =f(t) y(t₀) =k, y⁰(t₀) =l.

If p(t),q(t),r(t)are continuous functions on an interval I , then the IVP has a solution and it is unique.

The solution depends on the initial conditions k,l .

• The theorem can be applied to determine the longest intervalI so that the problem has a unique solution.

(IVP)

((t−1)y⁰⁰+ty⁰+ (t+ 1)y = 0 y(2) = 2, y⁰(2) = 3.

has a unique solution.

Sol. The conclusion does not involve the initial conditions!

Divide by (t−1) so the coefficient of y⁰⁰ becomes 1 y⁰⁰+ t

t−1y⁰+t+ 1 t−1y = 0.

The functions p(t) =t/(t−1),q(t) = (t+ 1)/(t−1), and r(t) = 0 are continuous on (−∞,1)∪(1,∞).

The interval containing the initial time t0 = 2 is (1,∞) so the longest

EX.Find the solution to the IVP (IVP)

(y⁰⁰+p(t)y⁰+q(t)y= 0 y(t₀) = 0, y⁰(t₀) = 0.

where p(t),q(t) are continuous functions.

Sol. By the theorem there is a unique solution.

Note that if we substitutey(t) = 0, it works

0⁰⁰+p(t)0⁰+q(t)0 = 0, 0(t₀) = 0, 0⁰(t₀) = 0.

So we get the solution to the IVP to bey(t) = 0.

Theorem

If y₁,y₂ are solutions of a homogeneous ODE y⁰⁰+p(t)y⁰+q(t)y = 0, then C1y1+C2y2 is also a solution.

Note. It is not true for nonhomogeneous equation y⁰⁰+p(t)y⁰+q(t)y =r(t).

In order thatC₁y₁+C₂y₂ is a general solution y₁,y₂ have to be extra!

EX.Consider the ODE

y⁰⁰+ 3y⁰+ 2y= 0.

It can be shown that

y₁(t) =e^−t, y₂(t) =e^−2t

are solutions. Using the superposition principle, we form a solution formula y(t) =C1e^−t+C2e^−2t.

Whether this is a general solution depends on whether it can satisfy the initial conditions

y(t₀) =k, y⁰(t₀) =l, for anyk,l.

Plug the formula y(t) into the conditions

(C1y1(t0) +C2y2(t0) =k, C1y₁⁰(t0) +C2y₂⁰(t0) =l.

This system can be solved for any k,l if and only if det

y₁(t₀) y₂(t₀) y₁⁰(t₀) y₂⁰(t₀)

6= 0.

Definition

For any two functions f,g, theWronskian of f,g is the function W(f,g) = det

f(t) g(t) f⁰(t) g⁰(t)

.