MATH 2055

Introduction to Differential Equations

2

Instructor

Assoc. Prof. Dr. Mustafa YILMAZ

Lecture - Wednesday, Thursday 09:30-12:20PM

Office: MC670, Phone: 1548

Office Hours: WT 4:30 p.m. & by appointments

E-mail address: mustafa.yilmaz@marmara.edu.tr

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Teaching Assistant

Mr. *

Office: MC***

Office Hours:

Phone: ****

e-mail: *@marmara.edu.tr

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Course Website

http://akademik.marmara.edu.tr/mustafa.yilmaz/teaching for

Syllabus, assignments, etc...

https://www.wolframalpha.com for interactive answer check

Exams

• 2 Midterms:

• Midterm 1: will be anounced

• Midterm 2: will be anounced

• Final Exam: will be anounced

Grading Scheme & Grading Scale

• Scheme:

• 9% HW, 24% Midterm 1, 27% Midterm 2, 40% Final Exam

• Scale:

• Relative Evaluation System (Curve)

Mathematical Symbols And Abbreviations

Symbol Say Means∴ therefore therefore∵ because because∀ for all for all⇐⇒ if and only if or iff if and only if or iff∝ proportional to proportional to⇒ implies calculation on left of symbol imply those on the rightfn function function

wrt with respect to with respect toLHS Left-hand side Left-hand sideRHS right-hand side right-hand side

Mathematical Symbols And AbbreviationsSymbol Say Means

𝒅𝒅𝒅𝒅𝒅𝒅𝒅𝒅

Dee y dee x Differentiate fn y wrt x

𝑑𝑑2𝑦𝑦𝑑𝑑𝑥𝑥2

Dee 2 y dee x squaredDouble differentiate fn y wrt xSecond derivative of fn y

𝝏𝝏𝒅𝒅𝝏𝝏𝒅𝒅

,𝒅𝒅𝒅𝒅 Partial y wrt x Partial derivative of y wrt x

f’(x) f prime of x or f primeDifferentiate fn f(x) wrt x,

equivalent to 𝒅𝒅𝒅𝒅𝒅𝒅𝒅𝒅

if y=f(x)

y’ y primeDifferentiate fn y wrt x,

equivalent to 𝒅𝒅𝒅𝒅𝒅𝒅𝒅𝒅

if y=f(x)

�̇�𝒅(dot above variable x)

x dot Differentiate fn x wrt t

Mathematical Symbols And AbbreviationsSymbol Say Means

f’’(x)f double prime of x or fdouble prime

Differentiate fn f(x) wrt x twice, Second derivative of fn f(x),

equivalent to 𝒅𝒅𝟐𝟐𝒅𝒅𝒅𝒅𝒅𝒅𝟐𝟐

,𝒅𝒅𝒚𝒚 if y=f(x)

�̈�𝒅 x double dot

Differentiate fn x wrt t twice,Second derivative of fn x,

equivalent to 𝒅𝒅𝟐𝟐𝒅𝒅𝒅𝒅𝒕𝒕𝟐𝟐

ℒ Laplace Transformconverts a time domain function tocomplex-domain function byintegration from zero to infinity

Terminology

This course deals with ordinary differential equations, which are

equations that contain one or more derivatives of a function of a single

variable. Such equations can be used to model a rich variety of

phenomena of interest in the sciences, engineering, economics,

ecological studies, and other areas.

Functions: dependent variables and independent variables.

10

Concepts of differential equations

In general, a differential equation is an equation that contains an

unknown function and its derivatives. The order of a differential

equation is the order of the highest derivative that occurs in the

equation.

A function y=f(x) is called a solution of a differential equation if the

equation is satisfied when y=f(x) and its derivatives are substituted into

the equation.

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e.g.

Show that 𝑦𝑦 = 2𝑒𝑒2𝑥𝑥 is a particular solution of the ordinary differential equation:

𝑑𝑑2𝑦𝑦𝑑𝑑𝑥𝑥2

− 𝑑𝑑𝑦𝑦𝑑𝑑𝑥𝑥− 2𝑦𝑦 = 𝑒𝑒𝑥𝑥

Taking derivatives of fn𝑦𝑦𝒚 = 4𝑒𝑒2𝑥𝑥

𝑦𝑦𝒚𝒚 = 8𝑒𝑒2𝑥𝑥

Substituting into the DE, 8𝑒𝑒2𝑥𝑥 − 4𝑒𝑒2𝑥𝑥 − 2 2𝑒𝑒2𝑥𝑥 = 0

so 𝑦𝑦 = 2𝑒𝑒2𝑥𝑥 sol’n of the DE.

Why?

One of the most important application of calculus is differential equations, which often arise in describing some phenomenon in engineering, physical science etc.• Population Dynamics• Circuit Design• Astrophysics• Geodesics (Pure Math)

13

Differential Equations

The following are examples of physical phenomena involving rates of change: Motion of fluids Motion of mechanical systems Flow of current in electrical circuits Dissipation of heat in solid objects Seismic waves Population dynamics

14

Differential Equations (DE’s)

• Equations that describe rates of change• Equations that involve derivatives:

(Falling object)

(Logistic Equation)

(Wave Equation)

(Current)

𝑦𝑦″ + 𝑦𝑦 = 4𝑥𝑥2

𝑦𝑦″(𝑥𝑥) + 𝑦𝑦(𝑥𝑥) = 4𝑥𝑥2

𝑑𝑑2𝑦𝑦(𝑥𝑥)𝑑𝑑𝑥𝑥2

+ 𝑦𝑦(𝑥𝑥) = 4𝑥𝑥2

y : function of xx : independent variable

e.g.,

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Two Types

Ordinary (ODE)(Only One Independent Variable)

Partial (PDE)(Multiple Independent Variables)

involves only total derivativesinvolves partial derivatives

16

4505.0,2.08.9 −=−= pdtdpv

dtdv

equation) (wave ),(),(

equation)(heat ),(),(

2

2

2

22

2

22

ttxu

xtxua

ttxu

xtxu

∂∂

=∂

∂∂

∂=

∂∂α

𝜕𝜕𝑓𝑓(𝑥𝑥.𝑦𝑦)𝜕𝜕𝑥𝑥

,𝜕𝜕𝑓𝑓(𝑥𝑥.𝑦𝑦)𝜕𝜕𝑦𝑦

𝑑𝑑𝑓𝑓(𝑥𝑥)𝑑𝑑𝑥𝑥

This Class

Ordinary(Only One Independent Variable)

𝑑𝑑𝑦𝑦𝑑𝑑𝑥𝑥

+ 2 = 0,

𝑑𝑑2𝑦𝑦𝑑𝑑𝑥𝑥2

+𝑑𝑑𝑦𝑦𝑑𝑑𝑥𝑥

= 𝑒𝑒𝑥𝑥

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Definition: A differential equation is an equation containing anunknown function and its derivatives.

𝑑𝑑𝑦𝑦𝑑𝑑𝑥𝑥

= 2𝑥𝑥 + 3

𝑑𝑑2𝑦𝑦𝑑𝑑𝑥𝑥2

+ 3𝑑𝑑𝑦𝑦𝑑𝑑𝑥𝑥

+ 𝑎𝑎𝑦𝑦 = 0

𝑑𝑑3𝑦𝑦𝑑𝑑𝑥𝑥3

+𝑑𝑑𝑦𝑦𝑑𝑑𝑥𝑥

4

+ 6𝑦𝑦 = 3

Examples:(ex.’s)

y is dependent variable (dep. var.) and x is independent variable (ind. var.)

1.

2.

3.

18

Ordinary Differential Equations (ODE)

Definition: Partial Differential Equation: Differential equations thatinvolve two or more independent variables are called partialdifferential equations.

ex.’s :

y & u are dependent variables and x and t are independent variables

1.

2.

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Partial Differential Equations (PDE)

𝜕𝜕2𝑦𝑦𝜕𝜕𝑥𝑥2

=𝜕𝜕𝑦𝑦𝜕𝜕𝜕𝜕

3.

Order (O)

The order of a differential equation is the highest derivative thatappears in the differential equation.

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Equation ind. var. dep. var. Order𝑑𝑑𝑦𝑦𝑑𝑑𝑥𝑥

2

+ 2 = 0𝑥𝑥 𝑦𝑦 1

𝑥𝑥𝑦𝑦′ − 𝑦𝑦2 = 3𝑥𝑥 𝑥𝑥 𝑦𝑦 1𝑑𝑑2𝑦𝑦𝑑𝑑𝑥𝑥2 +

𝑑𝑑𝑦𝑦𝑑𝑑𝑥𝑥 = 𝑒𝑒𝑥𝑥 𝑥𝑥 𝑦𝑦 2

𝑢𝑢(𝜕𝜕, 𝑥𝑥)𝑡𝑡𝑡𝑡= 𝑐𝑐𝑢𝑢(𝜕𝜕, 𝑥𝑥)𝑥𝑥𝑥𝑥 𝑥𝑥, 𝜕𝜕 𝑢𝑢 2𝑧𝑧(𝑖𝑖𝑖𝑖) + 𝑧𝑧𝑧 = −y 𝑦𝑦 𝑧𝑧 4𝑑𝑑4𝑤𝑤𝑑𝑑𝜕𝜕4 − 𝑤𝑤2 = 𝑒𝑒−2𝑡𝑡

𝜕𝜕 𝑤𝑤 4

Degree (D)

The degree of a differential equation is the power of the highestderivative term.

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Equation ind. var. dep. var. Order Degree

𝑑𝑑𝑦𝑦𝑑𝑑𝑥𝑥

2

+ 2 = 0𝑥𝑥 𝑦𝑦 1 2

𝑑𝑑2𝑦𝑦𝑑𝑑𝑥𝑥2

+𝑑𝑑𝑦𝑦𝑑𝑑𝑥𝑥

= 𝑒𝑒𝑥𝑥𝑥𝑥 𝑦𝑦 2 1

𝑧𝑧(𝑖𝑖𝑖𝑖) + 𝑧𝑧𝑧 = −y 𝑦𝑦 𝑧𝑧 4 1

Linear Differential Equations (LDE)

If a differential equation can be written as:

𝑎𝑎0 𝜕𝜕 y 𝑛𝑛 + 𝑎𝑎1 𝜕𝜕 y 𝑛𝑛−1 + ⋯+ 𝑎𝑎𝑛𝑛𝑦𝑦 = 𝑔𝑔(𝜕𝜕)

A differential equation is called linear if there are

no multiplications among dependent variables and

their derivatives.

The term g(t) is called a forcing function.

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Linear Differential Equations (LDE)

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Equation ind. var. dep. var. Order Degree Linear

𝑑𝑑𝑦𝑦𝑑𝑑𝑥𝑥

2

+ 2 = 0𝑥𝑥 𝑦𝑦 1 2 -

𝑑𝑑2𝑦𝑦𝑑𝑑𝑥𝑥2

+𝑑𝑑𝑦𝑦𝑑𝑑𝑥𝑥

= 𝑒𝑒𝑥𝑥𝑥𝑥 𝑦𝑦 2 1 OK

𝑧𝑧(𝑖𝑖𝑖𝑖) + 𝑧𝑧𝑧 = −y 𝑦𝑦 𝑧𝑧 4 1 OK

In other words, all coefficients are functions of independent variables.

Non-linear Differential Equations

Differential equations that do not satisfy the definition of linear arenon-linear.

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Equation ind. var. dep. var. Order Degree Linear Non-Linear

𝑑𝑑𝑦𝑦𝑑𝑑𝑥𝑥

2

+ 2 = 0𝑥𝑥 𝑦𝑦 1 2 - OK

𝑑𝑑2𝑦𝑦𝑑𝑑𝑥𝑥2

+𝑑𝑑𝑦𝑦𝑑𝑑𝑥𝑥

= 𝑒𝑒𝑥𝑥𝑥𝑥 𝑦𝑦 2 1 OK -

𝑧𝑧(𝑖𝑖𝑖𝑖) + 𝑧𝑧𝑧 = −y 𝑦𝑦 𝑧𝑧 4 1 OK -

Constant Coefficient Linear ODE

Linear Differential equations with Constant Coefficient

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𝑏𝑏𝑚𝑚𝑑𝑑𝑚𝑚𝑦𝑦𝑑𝑑𝜕𝜕𝑚𝑚

+ 𝑏𝑏𝑚𝑚−1𝑑𝑑𝑚𝑚−1𝑦𝑦𝑑𝑑𝜕𝜕𝑚𝑚−1 +. . . . +𝑏𝑏1

𝑑𝑑𝑦𝑦𝑑𝑑𝜕𝜕

+ 𝑏𝑏0𝑦𝑦 = 𝑓𝑓(𝜕𝜕)

3𝑑𝑑4𝑦𝑦𝑑𝑑𝜕𝜕4

+ 5𝑑𝑑3𝑦𝑦𝑑𝑑𝜕𝜕3

+ 7𝑑𝑑2𝑦𝑦𝑑𝑑𝜕𝜕2

+ 8𝑑𝑑𝑦𝑦𝑑𝑑𝜕𝜕

+ 4𝑦𝑦 = cos 5 𝜕𝜕 + 𝜕𝜕2

Equation ind. var. dep. var. Order Degree Linear Non-Linear

Type

𝑑𝑑𝑦𝑦𝑑𝑑𝑥𝑥

2

+ 2 = 0𝑥𝑥 𝑦𝑦 1 2 - OK ODE

𝑑𝑑2𝑦𝑦𝑑𝑑𝑥𝑥2 +

𝑑𝑑𝑦𝑦𝑑𝑑𝑥𝑥

= 𝑒𝑒𝑥𝑥𝑥𝑥 𝑦𝑦 2 1 OK - ODE

𝑧𝑧(𝑖𝑖𝑖𝑖) + 𝑧𝑧𝑧 = −y 𝑦𝑦 𝑧𝑧 4 1 OK - ODE

𝑦𝑦′ + 𝑎𝑎𝑦𝑦 = 𝑏𝑏𝜕𝜕 𝜕𝜕 𝑦𝑦 1 1 OK - ODE

𝑓𝑓𝑧 + 𝑓𝑓𝒚 + 𝑔𝑔(𝜕𝜕)𝑓𝑓 = ℎ(𝜕𝜕) 𝜕𝜕 𝑓𝑓 2 1 OK - ODE𝑑𝑑𝑦𝑦𝑑𝑑𝑥𝑥 = 2𝑥𝑥 + 3 𝑥𝑥 𝑦𝑦 1 1 OK - ODE

𝑑𝑑3𝑦𝑦𝑑𝑑𝑥𝑥3 +

𝑑𝑑𝑦𝑦𝑑𝑑𝑥𝑥

4

+ 6𝑦𝑦 = 3𝑥𝑥 𝑦𝑦 3 1 - OK ODE

𝜕𝜕2𝑦𝑦𝜕𝜕𝑥𝑥2 =

𝜕𝜕𝑦𝑦𝜕𝜕𝜕𝜕

𝑥𝑥, 𝜕𝜕 𝑦𝑦 2 1 OK - PDE

𝑢𝑢(𝜕𝜕, 𝑥𝑥)𝑡𝑡𝑡𝑡= 𝑐𝑐𝑢𝑢(𝜕𝜕, 𝑥𝑥)𝑥𝑥𝑥𝑥or

𝜕𝜕2𝑢𝑢𝜕𝜕𝜕𝜕2 = 𝑐𝑐

𝜕𝜕2𝑢𝑢𝜕𝜕𝑥𝑥2

𝑥𝑥, 𝜕𝜕 𝑢𝑢 2 1 OK - PDE

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ex.’s :

Solutions to Differential Equations

Three important questions in the study of differential equations:

• Is there a solution? (Existence)

• If there is a solution, is it unique? (Uniqueness)

• If there is a solution, how do we find it?

• (Analytical Solution, Numerical Approximation, etc.)

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General Solution (gen. sol’n)• Solutions obtained from integrating the differential equations are

called general solutions.

• The general solution of an order ODE contains arbitrary constants

resulting from integrating times.

• Geometrically, the general solution of an ODE is a family of infinitely

many solution curves, one for each value of the constant c.e.g., 𝑦𝑦″ + 4𝑦𝑦 = 0 , sol’n: 𝑦𝑦 = sin 2𝑥𝑥 for x∈R

𝑦𝑦′ = 𝑦𝑦𝑥𝑥

+ 1 , sol’n: 𝑦𝑦 = 𝑥𝑥 ln 𝑥𝑥 for x > 028

Particular Solution

• Particular solutions are the solutions obtained by assigning specific

values to the arbitrary constants in the general solutions.

• A particular solution does not contain any arbitrary constants.

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Any solution that is given in the form y = f(x) (independent variable).

ex.’s:𝐴𝐴 𝑟𝑟 = 𝜋𝜋𝑟𝑟2 + 𝑐𝑐𝑦𝑦 𝜕𝜕 = 𝑒𝑒−𝑡𝑡 + 𝑐𝑐

etc.

For most differential equations, it is impossible to find an explicit

formula for the solution.

Explicit Solution

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Any solution that isn’t in explicit form f(x.y)=0.

ex.’s:𝑦𝑦2𝑥𝑥2 = 𝑐𝑐

sin 𝑥𝑥𝑦𝑦 = 𝑐𝑐𝑥𝑥𝑒𝑒𝑦𝑦 = 𝑐𝑐

etc.

Implicit Solution

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An ODE together with an initial condition is called an initial value

problem (IVP) .

𝐹𝐹 𝑥𝑥,𝑦𝑦,𝑦𝑦′ = 0, 𝑦𝑦 𝑥𝑥0 = 𝑦𝑦0 ⇒ initial condition

Graphically, the particular integral curve passes through point 𝑥𝑥0,𝑦𝑦0

32

Initial Value Problems

Problems with specified boundary conditions are called boundary

value problems (BVP’s).

𝐹𝐹 𝑥𝑥,𝑦𝑦,𝑦𝑦′,𝑦𝑦𝒚𝒚 = 0, 𝑦𝑦 𝑥𝑥0 = 𝑦𝑦0, 𝑦𝑦 𝑥𝑥1 = 𝑦𝑦1 ⇒ boundary conditions

The objective is to obtain a unique solution

33

Boundary Value Problems

Initial Condition (IC)

In many physical problems we need to find the particular solution that

satisfies a condition of the form y(x0)=y0. Constrains that are specified at

the initial point, generally time point, are called initial conditions.

ex.’s: 1) y’+y=0 y(0)=12) z’=1 z(1)=−13) y’+y=2 y(1)=−5

General solution to (3): 𝑦𝑦 = 2 + 𝑘𝑘𝑒𝑒−𝑥𝑥

∵ 𝑦𝑦(1) = 2 + 𝑘𝑘𝑒𝑒−1 = −5 ∴ 𝑘𝑘 = −7𝑒𝑒, 𝑦𝑦 = 2 − 7𝑒𝑒1−𝑥𝑥34

Boundary Condition (BC)

Constrains that are specified at the boundary points, generally space

points, are called boundary conditions.

ex.’s:z”+z’+z=1 z(0)=2 z(2)=−1

f”+f=0 f(0)=0 f(𝜋𝜋/2)=1

35

A graph of the solution

of an ODE is called a

solution curve, or an

integral curve of the

equation.

Help to comprehend the

behavior of solution

36

Integral Curves, Solution Curves

Integral curves of y’+ y = 2for k = 0, 3, –3, 6, and –6

xkey −+= 2

A solution containing an arbitrary constant (parameter) represents a set

𝐺𝐺 𝑥𝑥,𝑦𝑦, 𝑐𝑐 = 0 of solutions to an ODE called a one-parameter family of

solutions.

A solution to an n−th order ODE is a n-parameter family of solutions

𝐹𝐹 𝑥𝑥,𝑦𝑦,𝑦𝑦′,⋯ ,𝑦𝑦𝑛𝑛 = 0.

Since the parameter can be assigned an infinite number of values, an ODE

can have an infinite number of solutions.

Families of Solutions

37

Suppose we are asked to sketch the graph of the solution of the initial-

value problem 𝒅𝒅 = 𝒇𝒇 𝒅𝒅,𝒅𝒅 ,𝒅𝒅 𝒅𝒅𝟎𝟎 = 𝒅𝒅𝟎𝟎. The equation tells us that the

slope at any point 𝒅𝒅,𝒅𝒅 on the graph is 𝒇𝒇 𝒅𝒅,𝒅𝒅 .

A plot of short line segments drawn at various points in the 𝒅𝒅,𝒅𝒅 -plane

showing the slope of the solution curve this is called a “direction field”

for the differential equation.

The direction field gives us the “ flow of solutions ”.

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Direction Field

We plot small lines representing slopes at each coordinate (x,y) in the

xy-plane. From this, we can infer solution curves. Short tangent

segments suggest the shape of the curve

Giving , instead of solving for y,

solving for

39

𝐹𝐹(𝑥𝑥,𝑦𝑦,𝑦𝑦′) = 0

𝑦𝑦′ =𝑑𝑑𝑦𝑦𝑑𝑑𝑥𝑥

= 𝑓𝑓(𝑥𝑥,𝑦𝑦) �slopedirection field

Direction Field

ex. 1: Sketch a direction field for 𝑦𝑦′ = 𝑥𝑥 + 𝑦𝑦.

At each (x,y) coordinate, we determine 𝑦𝑦′. Some examples are:

At (1,1) we have 𝑦𝑦′ = 2.

At (2,-3), we have 𝑦𝑦′ = −1. And so on.

We do this for “all” possible points in the plane. We get…

40

Direction Field

Direction field for

𝑦𝑦′ = 𝑥𝑥 + 𝑦𝑦

41

example 1: Continuation

We can infer possible solution

curves.

If you know a point on a particular

curve, the rest of the curve can be

inferred by “following” the

direction field. (We always read left

to right, i.e. x is increasing).

42

ex. 1: Continuation

ex. 2: Sketch the direction field for 𝑦𝑦′ = 𝑦𝑦3 − 4𝑦𝑦.

Hint: note that this only depends on y. Thus, if we find one slope-line for a

particular y-value, we can extend it left and right.

Another hint: Note that 𝑦𝑦′ = 0 represents a horizontal slope-line and this

occurs when 𝑦𝑦3 − 4𝑦𝑦 = 0. Factoring, we have

𝑦𝑦 𝑦𝑦2 − 4 = 𝑦𝑦 𝑦𝑦 + 2 𝑦𝑦 − 2 = 0.

Thus, when 𝑦𝑦 = 2,−2 or 0, then 𝑦𝑦′ = 0.

43

Direction Field

The direction field for

𝑦𝑦′ = 𝑦𝑦3 − 4𝑦𝑦

Note the horizontal slopes when

y = -2, y = 0 or y = 2. These are

equilibrium solutions.

44

ex. 2: Continuation

Note how the flow lines

(representing possible solution

curves) veer towards the

equilibrium solution 𝑦𝑦 = 0.

Thus, 𝑦𝑦 = 0 is a stable

equilibrium.

45

ex. 2: Continuation

The flow lines veer away

from the equilibriums 𝑦𝑦 = 2

and 𝑦𝑦 = −2 . These are

unstable equilibriums.

46

ex. 2: Continuation

A third type of equilibrium is semi-stable, in which the solution curves approach the

equilibrium asymptotically from one side, yet veer away from the other.

47

Direction Field

Direction field for y'= y² and integral curves through(0,1), (0,2), (0,3), - 1, (0, - 2), and (0, - 3).

48

2

2

1

1

dyy ydx

y dy dx

y x c

yx c

−

−

′ = =

=

− = +−

=+

ex. 3:𝑦𝑦′ = 𝑦𝑦2

2y

𝑦𝑦 = −1

𝑥𝑥 + 𝑘𝑘General Solution:

Slope:

Direction Field

Worksheet 1Equation ind. var. dep. var. Order Degree Linearity Type

𝑑𝑑𝑦𝑦𝑑𝑑𝑧𝑧

3

+ 2y = 0

𝑑𝑑2𝑦𝑦𝑑𝑑𝑥𝑥2

+𝑑𝑑𝑦𝑦𝑑𝑑𝑥𝑥

= 𝑒𝑒𝑥𝑥

𝑧𝑧(𝑖𝑖𝑖𝑖) + 𝑧𝑧𝑧 = 2𝑝𝑝′ + 𝑎𝑎𝑝𝑝 = 𝑏𝑏𝜕𝜕𝑓𝑓𝑧 + 𝑓𝑓 + 𝑔𝑔(𝜕𝜕) = ℎ(𝜕𝜕)𝑑𝑑4𝑦𝑦𝑑𝑑t4

− t𝑑𝑑2𝑦𝑦𝑑𝑑t2

+ 1 = t2

𝑢𝑢𝑥𝑥𝑥𝑥 + 𝑢𝑢𝑢𝑢𝑦𝑦𝑦𝑦 = sin 𝜕𝜕𝑢𝑢𝑥𝑥𝑥𝑥 + sin𝑢𝑢 𝑢𝑢𝑦𝑦𝑦𝑦 = cos 𝜕𝜕𝑦𝑦′′ + 3𝑒𝑒y𝑦𝑦′ − 2𝜕𝜕 = 0𝑦𝑦′′ + 3𝑦𝑦′ − 2t2 = 0

49

Worksheet 2

Equation ind. var. dep. var. Order Degree Linearity Type𝑑𝑑𝑦𝑦𝑑𝑑𝑥𝑥

+ 𝑥𝑥2𝑦𝑦 = 𝑥𝑥𝑒𝑒𝑥𝑥

𝑑𝑑3𝑦𝑦𝑑𝑑t3 = 𝑥𝑥 + 𝑦𝑦

𝜕𝜕2u𝜕𝜕𝑥𝑥2

+𝜕𝜕2u𝜕𝜕𝑦𝑦2

= 0

𝑑𝑑2𝑦𝑦𝑑𝑑𝑥𝑥2

+ 𝑥𝑥 sin𝑦𝑦 = 0

𝑑𝑑2𝑦𝑦𝑑𝑑𝑥𝑥2 + 𝑦𝑦 sin 𝑥𝑥 = 0

𝑑𝑑r𝑑𝑑𝑠𝑠

3

=𝑑𝑑2r𝑑𝑑s2 + 1

𝜕𝜕4u𝜕𝜕𝑥𝑥2𝜕𝜕𝑦𝑦2

+𝜕𝜕2u𝜕𝜕𝑥𝑥2

+𝜕𝜕2u𝜕𝜕𝑦𝑦2

+ u = 0

𝑢𝑢𝑥𝑥𝑥𝑥 + sin𝑢𝑢 𝑢𝑢𝑦𝑦𝑦𝑦 = cos 𝜕𝜕

𝑥𝑥′′ + 𝜕𝜕𝑥𝑥2 = 𝜕𝜕

𝑦𝑦′′ + 3𝑦𝑦′ + 5𝑥𝑥 = 0 50

ex. 4: 𝑦𝑦′′ − 2𝑦𝑦′ + 𝑦𝑦 = 0; 𝑦𝑦 = 𝑥𝑥𝑒𝑒𝑥𝑥

𝑦𝑦′ = 𝑥𝑥 + 1 𝑒𝑒𝑥𝑥 , 𝑦𝑦′′ = 𝑥𝑥 + 2 𝑒𝑒𝑥𝑥

left hand side:

𝑦𝑦′′ − 2𝑦𝑦′ + 𝑦𝑦 = (𝑥𝑥𝑒𝑒𝑥𝑥 + 2𝑒𝑒𝑥𝑥) − 2(𝑥𝑥𝑒𝑒𝑥𝑥 + 𝑒𝑒𝑥𝑥) + 𝑥𝑥𝑒𝑒𝑥𝑥 = 0

right-hand side: 0

The DE possesses the constant y=0 trivial solution

Verification of a solution by substitution

Worksheet 2

Equation Solution to Check

𝑦𝑦𝒚𝒚 + 4𝑦𝑦 = 0 𝑦𝑦(𝑥𝑥) = 𝑐𝑐1sin2𝑥𝑥 + 𝑐𝑐2cos2𝑥𝑥(𝑦𝑦𝒚)4 + 𝑦𝑦2 = −1 𝑦𝑦 = 𝑥𝑥2 − 1𝑦𝑦𝒚 = 𝑎𝑎𝑦𝑦, 𝑦𝑦(0) = 1 𝑦𝑦(𝜕𝜕) = 𝑒𝑒𝑎𝑎𝑡𝑡

𝑦𝑦𝒚 + 𝑦𝑦 = 10 𝑦𝑦 𝜕𝜕 = 10 − 𝑐𝑐𝑒𝑒 − 𝜕𝜕𝑤𝑤𝑡𝑡 + 3𝑤𝑤𝑥𝑥 = 0 𝑤𝑤(𝑥𝑥, 𝜕𝜕) = 1/(1 + (𝑥𝑥 − 3𝜕𝜕)2)𝑥𝑥’’ + 4x = 0 𝑥𝑥 = cos(2𝜕𝜕) + sin(2𝜕𝜕) + 𝑐𝑐𝑦𝑦𝒚𝒚 + 𝑦𝑦 = 0 𝑦𝑦1 𝜕𝜕 = sin𝜕𝜕, 𝑦𝑦2 𝜕𝜕 = −cos𝜕𝜕, 𝑦𝑦3 𝜕𝜕 = 2sin𝜕𝜕𝑑𝑑𝑦𝑦𝑑𝑑𝑥𝑥

= −(1 + y𝑒𝑒𝑥𝑥𝑦𝑦)(1 + 𝑥𝑥𝑒𝑒𝑥𝑥𝑦𝑦)

x + y + 𝑒𝑒𝑥𝑥𝑦𝑦 = 0

Check whether the given expression a solution of the corresponding equation?

52

53

Simple Integrable Forms

𝑏𝑏𝑘𝑘𝑑𝑑𝑘𝑘𝑦𝑦𝑑𝑑𝜕𝜕𝑘𝑘

= 𝑓𝑓(𝜕𝜕)

In theory, this equation may be solved by integrating both sides k

times. It may be convenient to introduce new variables so that only

first derivative forms need be integrated at each step.

54

ex. 5:

An object is initially at rest and dropped from a height h at t = 0. Determine

velocity v and displacement y.

𝑑𝑑𝑖𝑖𝑑𝑑𝑡𝑡

= 𝑔𝑔𝑑𝑑𝑑𝑑 = 𝑔𝑔𝑑𝑑𝜕𝜕𝑑𝑑 = 𝑔𝑔𝜕𝜕 + 𝐶𝐶1 𝐶𝐶1 = 0 𝑑𝑑 = 𝑔𝑔𝜕𝜕

𝑑𝑑𝑦𝑦𝑑𝑑𝑡𝑡

= 𝑔𝑔𝜕𝜕𝑑𝑑𝑦𝑦 = 𝑔𝑔𝜕𝜕𝑑𝑑𝜕𝜕

𝑦𝑦 = 12𝑔𝑔𝜕𝜕2 + 𝐶𝐶2 𝐶𝐶2 = 0 𝑦𝑦 = 1

2𝑔𝑔𝜕𝜕2

55

ex. 6

θ

x

y

0v

Consider situation below and solve for velocity and displacement in

both x and y directions.

56

ex. 6: Continuation𝑑𝑑𝑑𝑑𝑦𝑦𝑑𝑑𝜕𝜕

= −𝑔𝑔

𝑑𝑑𝑦𝑦 = −𝑔𝑔𝜕𝜕 + 𝐶𝐶1𝑑𝑑𝑦𝑦(0) = 𝑑𝑑0 sin𝜃𝜃 𝐶𝐶1 = 𝑑𝑑0 sin𝜃𝜃 𝑑𝑑𝑦𝑦 = −𝑔𝑔𝜕𝜕 + 𝑑𝑑0 sin𝜃𝜃

𝑑𝑑𝑦𝑦𝑑𝑑𝜕𝜕

= 𝑑𝑑𝑦𝑦 = −𝑔𝑔𝜕𝜕 + 𝑑𝑑0 sin𝜃𝜃

𝑦𝑦 = −12𝑔𝑔𝜕𝜕2 + (𝑑𝑑0 sin𝜃𝜃)𝜕𝜕 + 𝐶𝐶2

𝑦𝑦(0) = 0 𝐶𝐶2 = 0 𝑦𝑦 = −12𝑔𝑔𝜕𝜕2 + (𝑑𝑑0 sin𝜃𝜃)𝜕𝜕

57

ex. 6: Continuation

𝑑𝑑𝑑𝑑𝑥𝑥𝑑𝑑𝜕𝜕

= 0

𝑑𝑑𝑥𝑥 = 𝐶𝐶3𝑑𝑑𝑥𝑥(0) = 𝑑𝑑0 cos𝜃𝜃 𝐶𝐶3 = 𝑑𝑑0 cos𝜃𝜃 𝑑𝑑𝑥𝑥 = 𝑑𝑑0 cos𝜃𝜃

𝑑𝑑𝑥𝑥𝑑𝑑𝜕𝜕

= 𝑑𝑑𝑥𝑥 = 𝑑𝑑0 cos𝜃𝜃

𝑥𝑥 = (𝑑𝑑0 cos𝜃𝜃)𝜕𝜕 + 𝐶𝐶4𝑥𝑥(0) = 0 𝐶𝐶4 = 0 𝑥𝑥 = (𝑑𝑑0 cos𝜃𝜃)𝜕𝜕

Standard Integrals

Standard Integrals