lecture3 ode (1) lamia

Chapter 1

First order differential equations

Lecture3

ODE (Lecture3) Introduction to ODE-Math214 Dr Lamia Jaafar 1 / 22

Linear equations

Follow-up

ExampleLet the ODE (E) defined by

(x2 − 9)y ′ + xy = x .

1 Give the general solution of (E).

2 Precise a domain over which the solution is well defined.3 Discuss existence and uniqueness of the solution if we add the

initial value condition y(5) =14.

SolutionStep1:The general solution of (E) is given by y = yh + yp, whereyh is the solution of the homogeneous equation associated to (E)and (yp) is a particular solution of (E).

ExampleLet the ODE (E) defined by

(x2 − 9)y ′ + xy = x .

1 Give the general solution of (E).

2 Precise a domain over which the solution is well defined.3 Discuss existence and uniqueness of the solution if we add the

initial value condition y(5) =14.

SolutionStep1:The general solution of (E) is given by y = yh + yp, whereyh is the solution of the homogeneous equation associated to (E)and (yp) is a particular solution of (E).

Step2: the homogeneous solution yh is determined by solving theseparate variables equation

(x2 − 9)y ′ + xy = 0⇔ 1y

dy = − xx2 − 9

dx , x 6= ±3

thenyh(x) =

λ√|x2 − 9|

, x 6= ±3, λ ∈ R.

Step3: a particular solution of (E) can be determined by takingyp = 1.We deduce that the general solution of (E) is given by

y(x) =λ√|x2 − 9|

+ 1, x 6= ±3, λ ∈ R.

The points 3 and −3 are said to be singular points.

(x2 − 9)y ′ + xy = 0⇔ 1y

dy = − xx2 − 9

dx , x 6= ±3

thenyh(x) =

λ√|x2 − 9|

, x 6= ±3, λ ∈ R.

y(x) =λ√|x2 − 9|

+ 1, x 6= ±3, λ ∈ R.

(x2 − 9)y ′ + xy = 0⇔ 1y

dy = − xx2 − 9

dx , x 6= ±3

thenyh(x) =

λ√|x2 − 9|

, x 6= ±3, λ ∈ R.

y(x) =λ√|x2 − 9|

+ 1, x 6= ±3, λ ∈ R.

If x ∈]−∞,−3[, the general solution is given by

y1(x) =λ√

x2 − 9+ 1.

If x ∈]− 3,3[, the general solution is given by

y2(x) =λ√

9− x2+ 1.

If x ∈]3,+∞[, the general solution is given by

y3(x) =λ√

x2 − 9+ 1.

By taking y(5) =14, we solve the equation on (3,+∞), on which the

theorem of existence and uniqueness are assured:

∃!y(x) =4√

x2 − 9+ 1, x ∈ (3,+∞).

y1(x) =λ√

x2 − 9+ 1.

y2(x) =λ√

9− x2+ 1.

y3(x) =λ√

x2 − 9+ 1.

∃!y(x) =4√

x2 − 9+ 1, x ∈ (3,+∞).

y1(x) =λ√

x2 − 9+ 1.

y2(x) =λ√

9− x2+ 1.

y3(x) =λ√

x2 − 9+ 1.

∃!y(x) =4√

x2 − 9+ 1, x ∈ (3,+∞).

ExampleLet (E) defined by xy ′ − 4y = x5ex .

1 Verify that yp(x) = x4ex verifies the equation (E).

2 Find the general solution y of (E) and precise a domain overwhich the solution is well defined.

3 Discuss the existence and uniqueness of the IVP if the followinginitial conditions are given

y(0) = 0,y(0) = y0, y0 > 0,y(x0) = y0, x0 > 0, y0 > 0.

1 It is clear that y(x) = x4ex verifies (E).

2 The general solution of (E) is given by y = yh + yp, where yh isthe solution of the homogeneous equation and yp is a particularsolution of (E). For x 6= 0, we havexy ′ − 4y = 0⇔ y = λx4, λ ∈ R, then for x 6= 0, y(x) = λx4 + x4ex .We conclude that I = (0,∞) is a domain over which y is welldefined.

3 We havey(0) = 0⇒ a non uniqueness of the solution.y(0) = y0, y0 > 0⇒ a non existence of the solution.y(x0) = y0, x0 > 0, y0 > 0⇒ existence and uniqueness of thesolution.

Method of variation of parameters

Variation of parameters (or variation of constants) is a general methodfor determining the general solution of a linear ODE

(E) : a(x)y ′(x) + b(x)y(x) = φ(x).

Let(Eh) : a(x)y ′(x) + b(x)y(x) = 0,

thenyh(x) = λef (x), f (x) =

∫−b(x)

a(x)dx , λ ∈ R.

Let us suppose that λ is a function depending of the variable x , i.ey(x) = λ(x) exp(f (x)), then

(E)⇔ λ(x)′ =c(x)

a(x)exp(−f (x)).

We obtainλ(x) =

∫c(x)

a(x)exp(−f (x))dx + C,C ∈ R.

(E) : a(x)y ′(x) + b(x)y(x) = φ(x).

Let(Eh) : a(x)y ′(x) + b(x)y(x) = 0,

∫−b(x)

a(x)dx , λ ∈ R.

(E)⇔ λ(x)′ =c(x)

a(x)exp(−f (x)).

We obtainλ(x) =

∫c(x)

(E) : a(x)y ′(x) + b(x)y(x) = φ(x).

Let(Eh) : a(x)y ′(x) + b(x)y(x) = 0,

∫−b(x)

a(x)dx , λ ∈ R.

(E)⇔ λ(x)′ =c(x)

a(x)exp(−f (x)).

We obtainλ(x) =

∫c(x)

(E) : a(x)y ′(x) + b(x)y(x) = φ(x).

Let(Eh) : a(x)y ′(x) + b(x)y(x) = 0,

∫−b(x)

a(x)dx , λ ∈ R.

(E)⇔ λ(x)′ =c(x)

a(x)exp(−f (x)).

We obtainλ(x) =

∫c(x)

ConclusionWe conclude that the general solution of a linear ordinary equation isgiven by

y(x) = C exp(f (x)) + exp(f (x))∫ c(x)

a(x)exp(−f (x))dx

↓ ↓= yh(x) + yp(x)

ExamplesSolve the IVP {

y ′ + y = xy(0) = 4

and precise the domain of existence and uniqueness of the solution.

ConclusionWe conclude that the general solution of a linear ordinary equation isgiven by

y(x) = C exp(f (x)) + exp(f (x))∫ c(x)

a(x)exp(−f (x))dx

↓ ↓= yh(x) + yp(x)

ExamplesSolve the IVP {

y ′ + y = xy(0) = 4

and precise the domain of existence and uniqueness of the solution.

Exact equations

Definitionan exact differential equation is an ODE of first order which can beexpressed as follows

M(x , y)dx + N(x , y)dy = 0, (1)

where M and N are functions in both the variables x and y , such that∂M∂y

=∂N∂x

RemarkLet us suppose that f is a given real function defined in both variables

x , y , then its differential is given by df =∂f∂x

dx +∂f∂y

If f is constant we obtain the equation∂f∂x

dx +∂f∂y

dy = 0, then the

equation (1) corresponds to a differential of f (x , y) = Cst .

Definitionan exact differential equation is an ODE of first order which can beexpressed as follows

M(x , y)dx + N(x , y)dy = 0, (1)

where M and N are functions in both the variables x and y , such that∂M∂y

=∂N∂x

RemarkLet us suppose that f is a given real function defined in both variables

x , y , then its differential is given by df =∂f∂x

dx +∂f∂y

If f is constant we obtain the equation∂f∂x

dx +∂f∂y

dy = 0, then the

equation (1) corresponds to a differential of f (x , y) = Cst .

Example

x2y3dx + x3y2dy = 0, is an exact equation such that f (x , y) =13

Then the solution is given by an explicit one defined as13

x3y3 = C,C ∈ R.

Examples

2xydx + (x2 − 1)dy = 0,

y ′y(1− x2)− xy2 +12

sin 2x ,

are exact equations too, but f (x , y) cannot be easily deduced.

Example

x2y3dx + x3y2dy = 0, is an exact equation such that f (x , y) =13

Then the solution is given by an explicit one defined as13

x3y3 = C,C ∈ R.

Examples

2xydx + (x2 − 1)dy = 0,

y ′y(1− x2)− xy2 +12

sin 2x ,

are exact equations too, but f (x , y) cannot be easily deduced.

Remark

As M(x , y) =∂f∂x

and N(x , y) =∂f∂y, we deduce that to solve (1), we

have to find M (or N) such that∂f∂x

= M and to integrate with respect

to x (such that∂f∂y

= N and to integrate with respect to y .)

ExamplesSolve the previous exact equations.

Remark

As M(x , y) =∂f∂x

and N(x , y) =∂f∂y, we deduce that to solve (1), we

have to find M (or N) such that∂f∂x

= M and to integrate with respect

to x (such that∂f∂y

= N and to integrate with respect to y .)

ExamplesSolve the previous exact equations.

Solution of (E) : 2xydx + (x2 − 1)dy = 0

Let M(x , y) = 2xy and N(x , y) = x2 − 1.

As∂M∂y

=∂N∂x

then (E) is an exact equation. We deduce that it exists f

defined on a domain I such that∂f∂x = 2xy ,

∂f∂y = x2 − 1

f (x , y) = x2y + F (y),

∂f∂y = x2 − 1 = x2 + F (y)

we deduce thatF ′(y) = −1⇒ F (y) = −y + Cst ⇒ f (x , y) = x2y − y + Cst , then animplicit solution is giving by x2y − y − C,C ∈ R and an explicit solution

is y =C

x2 − 1, x 6= ±1.

The solution of the initial ODE is defined on any interval not containingeither 1 or −1, i.e a domain of the solution y is ]−∞,−1[ or ]− 1,1[ or]1,+∞[.

Solution of (E) : 2xydx + (x2 − 1)dy = 0

Let M(x , y) = 2xy and N(x , y) = x2 − 1.

As∂M∂y

=∂N∂x

then (E) is an exact equation. We deduce that it exists f

defined on a domain I such that∂f∂x = 2xy ,

∂f∂y = x2 − 1

f (x , y) = x2y + F (y),

∂f∂y = x2 − 1 = x2 + F (y)

we deduce thatF ′(y) = −1⇒ F (y) = −y + Cst ⇒ f (x , y) = x2y − y + Cst , then animplicit solution is giving by x2y − y − C,C ∈ R and an explicit solution

is y =C

x2 − 1, x 6= ±1.

The solution of the initial ODE is defined on any interval not containingeither 1 or −1, i.e a domain of the solution y is ]−∞,−1[ or ]− 1,1[ or]1,+∞[.

Integrating factor

Remark

Let (E1) : M(x , y)dx + N(x , y)dy = 0 such that∂M∂y6= ∂N∂x

, then (E1) is

not an exact equation.

QuestionHow we can transform it to an exact one?

The method proposedFind µ(x , y) such that (E2) : µ(x , y)M(x , y)dx + µ(x , y)N(x , y)dy = 0is an exact solution. We obtain∂

∂y(µM) =

∂x(µN)⇒ ∂µ

∂xN − ∂µ

∂yM = µ(

∂M∂y− ∂N∂x

Two cases are then possible to deduce an integrating factor µ.

Integrating factor

Remark

, then (E1) is

∂y(µM) =

∂x(µN)⇒ ∂µ

∂xN − ∂µ

∂yM = µ(

Integrating factor

Remark

, then (E1) is

∂y(µM) =

∂x(µN)⇒ ∂µ

∂xN − ∂µ

∂yM = µ(

Integrating factor

Remark

, then (E1) is

∂y(µM) =

∂x(µN)⇒ ∂µ

∂xN − ∂µ

∂yM = µ(

Let µx = ∂µ∂x , µy = ∂µ

∂y ,Mx = ∂M∂x ,My = ∂M

∂y ,Nx = ∂N∂x ,Ny = ∂N

∂y .

IfMy − Nx

Ndepends of x alone then we take

µ(x) = exp(∫ My − Nx

IfNx −My

Mdepends of y alone then we take

µ(y) = exp(∫ Nx −My

RemarkIf we consider a linear equation a(x)y ′(x)+ b(x)y(x) = φ(x) equivalent

to y ′(x) + P(x)y(x) = ψ(x), where P(x) =b(x)

a(x)and ψ(x) =

Then exp(−∫

P(x)dx)

is also called an integrating factor.

Let µx = ∂µ∂x , µy = ∂µ

∂y ,Mx = ∂M∂x ,My = ∂M

∂y ,Nx = ∂N∂x ,Ny = ∂N

∂y .

IfMy − Nx

Ndepends of x alone then we take

µ(x) = exp(∫ My − Nx

IfNx −My

Mdepends of y alone then we take

µ(y) = exp(∫ Nx −My

RemarkIf we consider a linear equation a(x)y ′(x)+ b(x)y(x) = φ(x) equivalent

to y ′(x) + P(x)y(x) = ψ(x), where P(x) =b(x)

a(x)and ψ(x) =

Then exp(−∫

P(x)dx)

is also called an integrating factor.

Example1 Is (E) : xydx + (2x2 + 3y2 − 20)dy = 0 an exact equation?2 Prove that an implicit solution of (E) is giving by

x2y4 +12

y6 − 5y4 = C,C ∈ R.

3 Solve the initial value problem associated to (E) if y(1) = 1.

Solutions by substitutions

DefinitionA function f is said to be homogeneous of degree α if it verifiesf (tx , ty) = tαf (x , y), α ∈ R.

Examples

f defined by f (x , y) = x3 + y3 is homogeneous of degree 3.g given by g(x , y) = x2 + y2 + xy is homogeneous of degree 2.h verifying h(x , y) = x + y + xy is not homogeneous.

RemarkIf both M and N are homogeneous functions of degree α, then the firstorder differential equation (E) : M(x , y)dx + N(x , y)dy = 0 is said tobe homogeneous of degree α.

Example

The ODE (x2 + y2)dx + (x2 − xy)dy = 0 is an homogeneous firstorder differential equation of degree 2.

Examples

Example

Examples

Example

Examples

Example

Method of solving homogeneous ODE of degree α.Let y = ux then the equation (E) gives M(1,u)dx + N(1,u)dy = 0.Or dy = udx + xdu, then we obtain a separable variables equation in xand u given by

= − N(1,u)duM(1,u) + uN(1,u)

Example

Prove that an implicit solution of (x2 + y2)dx + (x2 − xy)dy = 0, isgiving by (x + y)2 = λx exp(

), λ ∈ R.

Method of solving homogeneous ODE of degree α.Let y = ux then the equation (E) gives M(1,u)dx + N(1,u)dy = 0.Or dy = udx + xdu, then we obtain a separable variables equation in xand u given by

= − N(1,u)duM(1,u) + uN(1,u)

Example

Prove that an implicit solution of (x2 + y2)dx + (x2 − xy)dy = 0, isgiving by (x + y)2 = λx exp(

), λ ∈ R.

Some particular equations

Bernouilli equationThe Bernouilli equation is a first order differential equation of the formy ′(x) + a(x)y(x) = b(x)yα, α ∈ R.

Remarks1 If α = 0 or α = 1 then we obtain a linear equation.2 If α 6= 0 and α 6= 1, the solution can be deduced by taking

u = y1−α.

Example

Solve the ODE xy ′ + y = y2 ln x for x > 0.

u = y1−α.

Example

u = y1−α.

Example

Ricatti equationThe Ricatti equation is a first order differential equation of the formy ′(x) + a(x)y(x) = b(x)y2 + c(x).

Remarks1 If c = 0 then we obtain a Bernouilli equation, with α = 2.2 If c 6= 0, the solution of the ricatti equation can be deduced by

taking z(x) = y(x)− yp(x), where yp is a particular solution of theinitial Ricatti equation.

Example

Let (E) : x3y ′ + y2 + yx2 + 2x4 = 0.1 Verify that yp = −x2 is a particular solution of (E).

2 Find the general solution of (E).

Example

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