lecture1 d3

MA 108 - Ordinary Differential Equations

Neela Nataraj

Department of Mathematics,Indian Institute of Technology Bombay,

Powai, Mumbai [email protected]

February 29, 2016

Neela Nataraj Lecture 1: D3

Outline of the lecture

Basic Concepts

Classification based on1 Type2 Order3 Linearity

Solution(s)1 Explicit2 Implicit3 Formal

First Order ODEs, IVPSeparable ODEs


Differential equations

Definition

An equation involving derivatives of one or more dependentvariables with respect to one or more independent variables iscalled a differential equation (DE).

DE’s occur naturally in physics, engineering, biology, economics and soon.

A few examples of physical systems which lead to DE 1:

Radio active decay

Population dynamics

Newton’s law of cooling

Spread of a contagious disease

Chemical reactions

Falling bodies

Suspended cables

1Dennis G. Zill, Differential EquationsNeela Nataraj Lecture 1: D3

Classification based on TYPE: ODE/PDE

Definition

Let y(x) denote a function in the variable x . An ordinarydifferential equation (ODE) is an equation containing one or morederivatives of an unknown function y .

In general, a differential equation involving one or more dependentvariables with respect to a single independent variable is called anODE.

Definition

A differential equation involving partial derivatives of one or moredependent variables with respect to more than one independentvariable is called a partial differential equation (PDE).


Classification based on TYPE: ODE/PDE

Definition

Let y(x) denote a function in the variable x . An ordinarydifferential equation (ODE) is an equation containing one or morederivatives of an unknown function y .In general, a differential equation involving one or more dependentvariables with respect to a single independent variable is called anODE.

Definition

A differential equation involving partial derivatives of one or moredependent variables with respect to more than one independentvariable is called a partial differential equation (PDE).


Examples

dy

dx+ 5y = ex ,

d2y

dx2− dy

dx+ 6y = 0,︸︷︷︸

ODEs with one dependent variable

dx

dt+

dy

dt= 2x + y .︸︷︷︸

an ODE with two dependent variables

∂2u

∂x2+∂2u

∂y2= 0.︸︷︷︸

PDE with two independent variables- Laplace equation

∂2u

∂t2− (

∂2u

∂x2+∂2u

∂y2) = 0.︸︷︷︸

PDE with three independent variables- Wave equation


ODE- general form

Along with derivatives of y (the dependent variable) with respectto x (the independent variable), note that, an an ODE maycontain

y itself (the 0th derivative), and

known functions of x (including constants).

In other words, an ODE is a relation between the derivatives y , y ′

ordy

dx, . . . , y (n) or

dny

dxnand functions of x :

F (x , y , y ′, . . . , y (n)) = 0.


ODE- general form





ordy

dx, . . . , y (n) or

dny

dxn

and functions of x :

F (x , y , y ′, . . . , y (n)) = 0.


ODE- general form





ordy

dx, . . . , y (n) or

dny

dxnand functions of x :

F (x , y , y ′, . . . , y (n)) = 0.


Classification based on ORDER

Further classification according to the appearance of the highestderivative appearing in the equation is done now.

Definition

The order of a differential equation is the order of the highestderivative in the equation.

Examples :

1d2y

dx2+ xy

(dy

dx

)2

= 0

(ODE, 2nd order)

2d4x

dt4+ 5

d2x

dt2+ 3x = sin t (ODE, 4th order)

3∂v

∂t+∂v

∂s= v (PDE, 1st order)

4∂2u

∂x2+∂2u

∂y2+∂2u

∂z2= 0 (PDE, 2nd order)

5dx

dt= f (x , y),

dy

dt= g(x , y), x = x(t), y = y(t). (System of

ODEs, 1st order)




Definition


Examples :

1d2y

dx2+ xy

(dy

dx

)2

= 0 (ODE,

2nd order)

2d4x

dt4+ 5

d2x


3∂v

∂t+∂v


4∂2u

∂x2+∂2u

∂y2+∂2u


5dx

dt= f (x , y),

dy


ODEs, 1st order)




Definition


Examples :

1d2y

dx2+ xy

(dy

dx

)2

= 0 (ODE, 2nd order)

2d4x

dt4+ 5

d2x


3∂v

∂t+∂v


4∂2u

∂x2+∂2u

∂y2+∂2u


5dx

dt= f (x , y),

dy


ODEs, 1st order)




Definition


Examples :

1d2y

dx2+ xy

(dy

dx

)2


2d4x

dt4+ 5

d2x

dt2+ 3x = sin t

(ODE, 4th order)

3∂v

∂t+∂v


4∂2u

∂x2+∂2u

∂y2+∂2u


5dx

dt= f (x , y),

dy


ODEs, 1st order)




Definition


Examples :

1d2y

dx2+ xy

(dy

dx

)2


2d4x

dt4+ 5

d2x

dt2+ 3x = sin t (ODE,

4th order)

3∂v

∂t+∂v


4∂2u

∂x2+∂2u

∂y2+∂2u


5dx

dt= f (x , y),

dy


ODEs, 1st order)




Definition


Examples :

1d2y

dx2+ xy

(dy

dx

)2


2d4x

dt4+ 5

d2x


3∂v

∂t+∂v


4∂2u

∂x2+∂2u

∂y2+∂2u


5dx

dt= f (x , y),

dy


ODEs, 1st order)




Definition


Examples :

1d2y

dx2+ xy

(dy

dx

)2


2d4x

dt4+ 5

d2x


3∂v

∂t+∂v

∂s= v

(PDE, 1st order)

4∂2u

∂x2+∂2u

∂y2+∂2u


5dx

dt= f (x , y),

dy


ODEs, 1st order)




Definition


Examples :

1d2y

dx2+ xy

(dy

dx

)2


2d4x

dt4+ 5

d2x


3∂v

∂t+∂v

∂s= v (PDE,

1st order)

4∂2u

∂x2+∂2u

∂y2+∂2u


5dx

dt= f (x , y),

dy


ODEs, 1st order)




Definition


Examples :

1d2y

dx2+ xy

(dy

dx

)2


2d4x

dt4+ 5

d2x


3∂v

∂t+∂v


4∂2u

∂x2+∂2u

∂y2+∂2u


5dx

dt= f (x , y),

dy


ODEs, 1st order)




Definition


Examples :

1d2y

dx2+ xy

(dy

dx

)2


2d4x

dt4+ 5

d2x


3∂v

∂t+∂v


4∂2u

∂x2+∂2u

∂y2+∂2u

∂z2= 0

(PDE, 2nd order)

5dx

dt= f (x , y),

dy


ODEs, 1st order)




Definition


Examples :

1d2y

dx2+ xy

(dy

dx

)2


2d4x

dt4+ 5

d2x


3∂v

∂t+∂v


4∂2u

∂x2+∂2u

∂y2+∂2u

∂z2= 0 (PDE,

2nd order)

5dx

dt= f (x , y),

dy


ODEs, 1st order)




Definition


Examples :

1d2y

dx2+ xy

(dy

dx

)2


2d4x

dt4+ 5

d2x


3∂v

∂t+∂v


4∂2u

∂x2+∂2u

∂y2+∂2u


5dx

dt= f (x , y),

dy


ODEs, 1st order)




Definition


Examples :

1d2y

dx2+ xy

(dy

dx

)2


2d4x

dt4+ 5

d2x


3∂v

∂t+∂v


4∂2u

∂x2+∂2u

∂y2+∂2u


5dx

dt= f (x , y),

dy

dt= g(x , y), x = x(t), y = y(t).

(System of

ODEs, 1st order)




Definition


Examples :

1d2y

dx2+ xy

(dy

dx

)2


2d4x

dt4+ 5

d2x


3∂v

∂t+∂v


4∂2u

∂x2+∂2u

∂y2+∂2u


5dx

dt= f (x , y),

dy


ODEs, 1st order)Neela Nataraj Lecture 1: D3

Linear equations

Linear equations

- F (x , y , y ′, . . . , y (n)) = 0 is linear if F is a linearfunction of the variables y , y ′, . . . , y (n).Thus, a linear ODE of order n is of the form

a0(x)y (n) + a1(x)y (n−1) + . . .+ an(x)y = b(x)

where a0, a1, . . . , an, b are functions of x and a0(x) 6= 0.

Check list : If the dependent variable is y , derivatives occur uptofirst degree only, no products of y and/or its derivatives are there.


Linear equations

Linear equations - F (x , y , y ′, . . . , y (n)) = 0 is linear if F is a linearfunction of the variables y , y ′, . . . , y (n).

Thus, a linear ODE of order n is of the form

a0(x)y (n) + a1(x)y (n−1) + . . .+ an(x)y = b(x)




Linear equations

Linear equations - F (x , y , y ′, . . . , y (n)) = 0 is linear if F is a linearfunction of the variables y , y ′, . . . , y (n).Thus, a linear ODE of order n is of the form

a0(x)y (n) + a1(x)y (n−1) + . . .+ an(x)y = b(x)




Example : Radioactive decay

A radioactive substance decomposes at a rate proportional to theamount present.

Let y(t) be the amount present at time t. Then

dy

dt= −k · y

where k is a physical constant whose value is found by experiments(−k is called the decay constant).Linear ODE of first order.



A radioactive substance decomposes at a rate proportional to theamount present. Let y(t) be the amount present at time t.

Then

dy

dt= −k · y




A radioactive substance decomposes at a rate proportional to theamount present. Let y(t) be the amount present at time t. Then

dy

dt= −k · y





dy

dt= −k · y

where k is a physical constant whose value is found by experiments

(−k is called the decay constant).Linear ODE of first order.




dy

dt= −k · y

where k is a physical constant whose value is found by experiments(−k is called the decay constant).

Linear ODE of first order.




dy

dt= −k · y



Examples - The motion of an oscillating pendulum

Consider an oscillating pendulum of length L.

Let θ be the angle itmakes with the vertical direction.

d2θ

dt2+

g

Lsin θ = 0.

ODE of second order. not linear - Non-linear DE.



Consider an oscillating pendulum of length L. Let θ be the angle itmakes with the vertical direction.

d2θ

dt2+

g

Lsin θ = 0.





d2θ

dt2+

g

Lsin θ = 0.

ODE of

second order. not linear - Non-linear DE.




d2θ

dt2+

g

Lsin θ = 0.

ODE of second order.

not linear - Non-linear DE.




d2θ

dt2+

g

Lsin θ = 0.



Example : A falling object

A body of mass m falls under the force of gravity.

The drag forcedue to air resistance is c · v2 where v is the velocity and c is aconstant Then,

mdv

dt= mg − c · v2.

An ODE of first order. Linear or non-linear? (NL)Examples :

1 y ′′ + 5y ′ + 6y = 0 - 2nd order, linear

2 y (4) + x2y (3) + x3y ′ = xex - 4th order, linear

3 y ′′ + 5(y ′)3 + 6y = 0 - 2nd order, non-linear.



A body of mass m falls under the force of gravity. The drag forcedue to air resistance is

c · v2 where v is the velocity and c is aconstant Then,

mdv

dt= mg − c · v2.







A body of mass m falls under the force of gravity. The drag forcedue to air resistance is c · v2

where v is the velocity and c is aconstant Then,

mdv

dt= mg − c · v2.







A body of mass m falls under the force of gravity. The drag forcedue to air resistance is c · v2 where v is the velocity and c is aconstant

Then,

mdv

dt= mg − c · v2.







A body of mass m falls under the force of gravity. The drag forcedue to air resistance is c · v2 where v is the velocity and c is aconstant Then,

mdv

dt= mg − c · v2.








mdv

dt= mg − c · v2.

An ODE of first order.

Linear or non-linear? (NL)Examples :







mdv

dt= mg − c · v2.

An ODE of first order. Linear or non-linear?

(NL)Examples :







mdv

dt= mg − c · v2.

An ODE of first order. Linear or non-linear? (NL)

Examples :







mdv

dt= mg − c · v2.


1 y ′′ + 5y ′ + 6y = 0

- 2nd order, linear






mdv

dt= mg − c · v2.








mdv

dt= mg − c · v2.



2 y (4) + x2y (3) + x3y ′ = xex

- 4th order, linear





mdv

dt= mg − c · v2.








mdv

dt= mg − c · v2.




3 y ′′ + 5(y ′)3 + 6y = 0

- 2nd order, non-linear.




mdv

dt= mg − c · v2.






Can we solve it?

Given an equation, you would like to

solve it. At least, try to solveit.Questions:

1 What is a solution?

2 Does an equation always have a solution? If so, how many?

3 Can the solutions be expressed in a nice form? If not, how toget a feel for it?

4 How much can we proceed in a systematic manner?

order - first, second, ..., nth, ...linear or non-linear?


Can we solve it?

Given an equation, you would like to solve it.

At least, try to solveit.Questions:







Can we solve it?

Given an equation, you would like to solve it. At least, try to solveit.

Questions:







Can we solve it?

Given an equation, you would like to solve it. At least, try to solveit.Questions:







Can we solve it?



2 Does an equation always have a solution?

If so, how many?





Can we solve it?








Can we solve it?




3 Can the solutions be expressed in a nice form?

If not, how toget a feel for it?




Can we solve it?








Can we solve it?






order

- first, second, ..., nth, ...linear or non-linear?


Can we solve it?






order - first, second, ..., nth, ...

linear or non-linear?


Can we solve it?






order - first, second, ..., nth, ...linear

or non-linear?


Can we solve it?








What is a solution?

Consider F (x , y , y ′, . . . , y (n)) = 0.

We assume that it is always possibleto solve a differential equation for the highest derivative, obtaining

y (n) = f (x , y , y ′, · · · , y (n−1))

and study equations of this form. This is to avoid the ambiguity whichmay arise because a single equation F (x , y , y ′, . . . , y (n)) = 0 maycorrespond to several equations of the form y (n) = f (x , y , y ′, · · · , y (n−1)).For example, the equation y ′2 + xy ′ + 4y = 0 leads to the two equations

y ′ =−x +

√x2 − 16y

2or y ′ =

−x −√x2 − 16y

2.

Definition

A explicit solution of the ODE y (n) = f (x , y , y ′, · · · , y (n−1)) on theinterval α < x < β is a function φ(x) such that φ′, φ′′, · · · , φ(n) exist andsatisfy

φ(n)(x) = f (x , φ, φ′, · · · , φ(n−1)),

for every x in α < x < β.


What is a solution?

Consider F (x , y , y ′, . . . , y (n)) = 0. We assume that it is always possibleto solve a differential equation for the highest derivative, obtaining

y (n) = f (x , y , y ′, · · · , y (n−1))

and study equations of this form.

This is to avoid the ambiguity whichmay arise because a single equation F (x , y , y ′, . . . , y (n)) = 0 maycorrespond to several equations of the form y (n) = f (x , y , y ′, · · · , y (n−1)).For example, the equation y ′2 + xy ′ + 4y = 0 leads to the two equations

y ′ =−x +

√x2 − 16y

2or y ′ =

−x −√x2 − 16y

2.

Definition


φ(n)(x) = f (x , φ, φ′, · · · , φ(n−1)),



What is a solution?


y (n) = f (x , y , y ′, · · · , y (n−1))

and study equations of this form. This is to avoid the ambiguity whichmay arise because a single equation F (x , y , y ′, . . . , y (n)) = 0 maycorrespond to several equations of the form y (n) = f (x , y , y ′, · · · , y (n−1)).

For example, the equation y ′2 + xy ′ + 4y = 0 leads to the two equations

y ′ =−x +

√x2 − 16y

2or y ′ =

−x −√x2 − 16y

2.

Definition


φ(n)(x) = f (x , φ, φ′, · · · , φ(n−1)),



What is a solution?


y (n) = f (x , y , y ′, · · · , y (n−1))

and study equations of this form. This is to avoid the ambiguity whichmay arise because a single equation F (x , y , y ′, . . . , y (n)) = 0 maycorrespond to several equations of the form y (n) = f (x , y , y ′, · · · , y (n−1)).For example, the equation y ′2 + xy ′ + 4y = 0 leads to the two equations

y ′ =−x +

√x2 − 16y

2or y ′ =

−x −√x2 − 16y

2.

Definition


φ(n)(x) = f (x , φ, φ′, · · · , φ(n−1)),



Implicit solution & Formal solution

Definition

A relation g(x , y) = 0 is called an implicit solution ofy (n) = f (x , y , y ′, · · · , y (n−1)) if this relation defines at least one functionφ(x) on an interval α < x < β, such that, this function is an explicitsolution of y (n) = f (x , y , y ′, · · · , y (n−1)) in this interval.

Examples :

1 x2 + y2 − 25 = 0 is an implicit solution of x + yy ′ = 0 in−5 < x < 5, because it defines two functions

φ1(x) =√

25− x2, φ2(x) = −√

25− x2

which are solutions of the DE in the given interval. Verify!

2 Consider x2 + y2 + 25 = 0 =⇒ x + yy ′ = 0 =⇒ y ′ = −x

y. We say

x2 + y2 + 25 = 0 formally satisfies x + yy ′ = 0. But it is NOT animplicit solution of DE as this relation doesn’t yield φ which is anexplicit solution of the DE on any real interval I .



Definition


Examples :


φ1(x) =√

25− x2, φ2(x) = −√

25− x2

which are solutions of the DE in the given interval.

Verify!


y. We say




Definition


Examples :


φ1(x) =√

25− x2, φ2(x) = −√

25− x2



y.

We say




Definition


Examples :


φ1(x) =√

25− x2, φ2(x) = −√

25− x2



y. We say



First order ODE & Initial Value Problem for first orderODE

We now consider first order ODE of the form F (x , y , y ′) = 0 or

y ′ = f (x , y) .

Consider a linear first order ODE of the form y ′ + a(x)y = b(x) .

If b(x) = 0, then we say that the equation is homogeneous.

MA106 nostalgia : Do the solutions of a homogeneous differentialequation form a vector space under usual addition and scalarmultiplication?

Definition

Initial value problem (IVP) : A DE along with an initial condition isan IVP.

y ′ = f (x , y), y(x0) = y0.


Examples

Given an amount of a radioactive substance, say 1 gm, find theamount present at any later time.

The relevant ODE isdy

dt= −k · y .

Initial amount given is 1 gm at time t = 0. i.e.,

y(0) = 1.

By inspection, y = ce−kt , for an arbitrary constant c , is a solutionof the above ODE. The initial condition determines c = 1. Hence

y = e−kt

is a particular solution to the above ODE with the given initialcondition.


Examples

Given an amount of a radioactive substance, say 1 gm, find theamount present at any later time.The relevant ODE is

dy

dt= −k · y .


y(0) = 1.


y = e−kt



Examples


dy

dt= −k · y .

Initial amount given is 1 gm at time t = 0.

i.e.,

y(0) = 1.


y = e−kt



Examples


dy

dt= −k · y .


y(0) = 1.


y = e−kt



Examples


dy

dt= −k · y .


y(0) = 1.

By inspection, y = ce−kt , for an arbitrary constant c , is a solutionof the above ODE.

The initial condition determines c = 1. Hence

y = e−kt



Examples


dy

dt= −k · y .


y(0) = 1.

By inspection, y = ce−kt , for an arbitrary constant c , is a solutionof the above ODE. The initial condition determines c =

1. Hence

y = e−kt



Examples


dy

dt= −k · y .


y(0) = 1.


y = e−kt



Examples

Find the curve through the point (1, 1) in the xy -plane having at

each of its points, the slope −y

x.

The relevant ODE isy ′ = −y

x.

By inspection,

y =c

xis its general solution for an arbitrary constant c ; that is, a familyof hyperbolas.The initial condition given is

y(1) = 1,

which implies c = 1. Hence the particular solution for the aboveproblem is

y =1

x.


Examples



x.


x.

By inspection,

y =c

xis its general solution for an arbitrary constant c ; that is, a familyof hyperbolas.

The initial condition given is

y(1) = 1,


y =1

x.


Examples



x.


x.

By inspection,

y =c


y(1) = 1,

which implies c = 1.

Hence the particular solution for the aboveproblem is

y =1

x.


Examples



x.


x.

By inspection,

y =c


y(1) = 1,


y =1

x.


Examples

A first order IVP can have

1 NO solution :

|y ′|+ |y | = 0, y(0) = 3.

2 Precisely one solution : y ′ = x , y(0) = 1. What is thesolution?

3 Infinitely many solutions: xy ′ = y − 1, y(0) = 1 The solutionsare y = 1 + cx .

Motivation to study conditions under which the solution wouldexist and the conditions under which it will be unique!

We first start with a few methods for finding out the solution offirst order ODEs, discuss the geometric meaning of solutions andthen proceed to study existence-uniqueness results.


Examples


1 NO solution : |y ′|+ |y | = 0, y(0) = 3.






Examples


1 NO solution : |y ′|+ |y | = 0, y(0) = 3.

2 Precisely one solution :

y ′ = x , y(0) = 1. What is thesolution?





Examples


1 NO solution : |y ′|+ |y | = 0, y(0) = 3.

2 Precisely one solution : y ′ = x , y(0) = 1.

What is thesolution?





Examples


1 NO solution : |y ′|+ |y | = 0, y(0) = 3.


3 Infinitely many solutions:

xy ′ = y − 1, y(0) = 1 The solutionsare y = 1 + cx .




Examples


1 NO solution : |y ′|+ |y | = 0, y(0) = 3.


3 Infinitely many solutions: xy ′ = y − 1, y(0) = 1

The solutionsare y = 1 + cx .




Examples


1 NO solution : |y ′|+ |y | = 0, y(0) = 3.






Separable ODE’s

An ODE of the form

M(x) + N(y)y ′ = 0

is called a separable ODE.

Let H1(x) and H2(y) be any functions such that H ′1(x) = M(x)and H ′2(y) = N(y).Substituting in the DE, we obtain

H ′1(x) + H ′2(y)y ′ = 0.

Using chain rule,d

dxH2(y) = H ′2(y)

dy

dx.

Hence,d

dx(H1(x) + H2(y)) = 0.

Integrating, H1(x) + H2(y) = c, where c is an arbirtaryconstant.Note: This method many times gives us an implicit solution andnot necessarily an explicit one!


Separable ODE’s

An ODE of the form

M(x) + N(y)y ′ = 0

is called a separable ODE.Let H1(x) and H2(y) be any functions such that H ′1(x) = M(x)and H ′2(y) = N(y).

Substituting in the DE, we obtain

H ′1(x) + H ′2(y)y ′ = 0.

Using chain rule,d

dxH2(y) = H ′2(y)

dy

dx.

Hence,d

dx(H1(x) + H2(y)) = 0.



Separable ODE’s

An ODE of the form

M(x) + N(y)y ′ = 0

is called a separable ODE.Let H1(x) and H2(y) be any functions such that H ′1(x) = M(x)and H ′2(y) = N(y).Substituting in the DE, we obtain

H ′1(x) + H ′2(y)y ′ = 0.

Using chain rule,d

dxH2(y) = H ′2(y)

dy

dx.

Hence,d

dx(H1(x) + H2(y)) = 0.



Separable ODE’s

An ODE of the form

M(x) + N(y)y ′ = 0


H ′1(x) + H ′2(y)y ′ = 0.

Using chain rule,d

dxH2(y) = H ′2(y)

dy

dx.

Hence,d

dx(H1(x) + H2(y)) = 0.

Integrating, H1(x) + H2(y) = c, where c is an arbirtaryconstant.Note:

This method many times gives us an implicit solution andnot necessarily an explicit one!


Separable ODE’s

An ODE of the form

M(x) + N(y)y ′ = 0


H ′1(x) + H ′2(y)y ′ = 0.

Using chain rule,d

dxH2(y) = H ′2(y)

dy

dx.

Hence,d

dx(H1(x) + H2(y)) = 0.



Separable ODE - Example 1

Solve the DE :y ′ = −2xy .

Separating the variables, we get :

dy

y= −2xdx .

Integrating both sides, we obtain :

ln |y | = −x2 + c1.

Thus, the solutions arey = ce−x

2.

How do they look?





dy

y= −2xdx .


ln |y | = −x2 + c1.

Thus, the solutions are

y = ce−x2.

How do they look?





dy

y= −2xdx .


ln |y | = −x2 + c1.

Thus, the solutions arey = ce−x

2.

How do they look?


If we are given an initial condition

y(x0) = y0,

then we get:c = y0e

x20

and y = y0ex20−x2 .



Find the solution to the initial value problem:

dy

dx=

y cos x

1 + 2y2; y(0) = 1.

Assume y 6= 0. Then,

1 + 2y2

ydy = cos x dx .

Integrating,ln |y |+ y2 = sin x + c .

As y(0) = 1, we get c = 1. Hence a particular solution to the IVPis

ln |y |+ y2 = sin x + 1.

Note: y ≡ 0 is a solution to the DE but it is not a solution to thegiven IVP.




dy

dx=

y cos x

1 + 2y2; y(0) = 1.

Assume y 6= 0.

Then,

1 + 2y2

ydy = cos x dx .



ln |y |+ y2 = sin x + 1.





dy

dx=

y cos x

1 + 2y2; y(0) = 1.


1 + 2y2

ydy = cos x dx .



ln |y |+ y2 = sin x + 1.





dy

dx=

y cos x

1 + 2y2; y(0) = 1.


1 + 2y2

ydy = cos x dx .

Integrating,

ln |y |+ y2 = sin x + c .


ln |y |+ y2 = sin x + 1.





dy

dx=

y cos x

1 + 2y2; y(0) = 1.


1 + 2y2

ydy = cos x dx .



ln |y |+ y2 = sin x + 1.





dy

dx=

y cos x

1 + 2y2; y(0) = 1.


1 + 2y2

ydy = cos x dx .


As y(0) = 1, we get c = 1.

Hence a particular solution to the IVPis

ln |y |+ y2 = sin x + 1.





dy

dx=

y cos x

1 + 2y2; y(0) = 1.


1 + 2y2

ydy = cos x dx .



ln |y |+ y2 = sin x + 1.





dy

dx=

y cos x

1 + 2y2; y(0) = 1.


1 + 2y2

ydy = cos x dx .



ln |y |+ y2 = sin x + 1.

Note: y ≡ 0 is a solution to the DE

but it is not a solution to thegiven IVP.




dy

dx=

y cos x

1 + 2y2; y(0) = 1.


1 + 2y2

ydy = cos x dx .



ln |y |+ y2 = sin x + 1.




Escape velocity.

A projectile of mass m moves in a direction perpendicular to thesurface of the earth. Suppose v0 is its initial velocity. We want tocalculate the height the projectile reaches.Its weight at height x (from the surface of the earth) is given by,

w(x) = − mgR2

(R + x)2,

where R is the radius of the earth.Neglect force due to air resistance and other celestial bodies.Therefore, the equation of motion is

mdv

dt= − mgR2

(R + x)2; v(0) = v0.



Escape velocity.A projectile of mass m moves in a direction perpendicular to thesurface of the earth.

Suppose v0 is its initial velocity. We want tocalculate the height the projectile reaches.Its weight at height x (from the surface of the earth) is given by,

w(x) = − mgR2

(R + x)2,


mdv

dt= − mgR2

(R + x)2; v(0) = v0.



Escape velocity.A projectile of mass m moves in a direction perpendicular to thesurface of the earth. Suppose v0 is its initial velocity.

We want tocalculate the height the projectile reaches.Its weight at height x (from the surface of the earth) is given by,

w(x) = − mgR2

(R + x)2,


mdv

dt= − mgR2

(R + x)2; v(0) = v0.



Escape velocity.A projectile of mass m moves in a direction perpendicular to thesurface of the earth. Suppose v0 is its initial velocity. We want tocalculate the height the projectile reaches.

Its weight at height x (from the surface of the earth) is given by,

w(x) = − mgR2

(R + x)2,


mdv

dt= − mgR2

(R + x)2; v(0) = v0.



Escape velocity.A projectile of mass m moves in a direction perpendicular to thesurface of the earth. Suppose v0 is its initial velocity. We want tocalculate the height the projectile reaches.Its weight at height x

(from the surface of the earth) is given by,

w(x) = − mgR2

(R + x)2,


mdv

dt= − mgR2

(R + x)2; v(0) = v0.



Escape velocity.A projectile of mass m moves in a direction perpendicular to thesurface of the earth. Suppose v0 is its initial velocity. We want tocalculate the height the projectile reaches.Its weight at height x (from the surface of the earth)

is given by,

w(x) = − mgR2

(R + x)2,


mdv

dt= − mgR2

(R + x)2; v(0) = v0.



Escape velocity.A projectile of mass m moves in a direction perpendicular to thesurface of the earth. Suppose v0 is its initial velocity. We want tocalculate the height the projectile reaches.Its weight at height x (from the surface of the earth) is given by,

w(x)

= − mgR2

(R + x)2,


mdv

dt= − mgR2

(R + x)2; v(0) = v0.




w(x) = − mgR2

(R + x)2,


mdv

dt= − mgR2

(R + x)2; v(0) = v0.




w(x) = − mgR2

(R + x)2,

where R is the radius of the earth.

Neglect force due to air resistance and other celestial bodies.Therefore, the equation of motion is

mdv

dt= − mgR2

(R + x)2; v(0) = v0.




w(x) = − mgR2

(R + x)2,

where R is the radius of the earth.Neglect force due to air resistance and other celestial bodies.

Therefore, the equation of motion is

mdv

dt= − mgR2

(R + x)2; v(0) = v0.




w(x) = − mgR2

(R + x)2,


mdv

dt= − mgR2

(R + x)2; v(0) = v0.




w(x) = − mgR2

(R + x)2,


mdv

dt= − mgR2

(R + x)2;

v(0) = v0.




w(x) = − mgR2

(R + x)2,


mdv

dt= − mgR2

(R + x)2; v(0) = v0.


Separable ODE’s

By chain rule,dv

dt

=dv

dx· dxdt

= v · dvdx.

Thus,

v · dvdx

= − gR2

(R + x)2.

This ODE is separable. Linear or non-linear? (NL)Separating the variables and integrating, we get:

v2

2=

gR2

R + x+ c .

For x = 0, we getv202

= gR + c , hence, c =v202− gR, and,

v = ±

√v20 − 2gR +

2gR2

R + x.


Separable ODE’s

By chain rule,dv

dt=

dv

dx· dxdt

= v · dvdx.

Thus,

v · dvdx

= − gR2

(R + x)2.


v2

2=

gR2

R + x+ c .



v = ±

√v20 − 2gR +

2gR2

R + x.


Separable ODE’s

By chain rule,dv

dt=

dv

dx· dxdt

= v · dvdx.

Thus,

v · dvdx

= − gR2

(R + x)2.

This ODE is

separable. Linear or non-linear? (NL)Separating the variables and integrating, we get:

v2

2=

gR2

R + x+ c .



v = ±

√v20 − 2gR +

2gR2

R + x.


Separable ODE’s

By chain rule,dv

dt=

dv

dx· dxdt

= v · dvdx.

Thus,

v · dvdx

= − gR2

(R + x)2.

This ODE is separable.

Linear or non-linear? (NL)Separating the variables and integrating, we get:

v2

2=

gR2

R + x+ c .



v = ±

√v20 − 2gR +

2gR2

R + x.


Separable ODE’s

By chain rule,dv

dt=

dv

dx· dxdt

= v · dvdx.

Thus,

v · dvdx

= − gR2

(R + x)2.

This ODE is separable. Linear or non-linear?

(NL)Separating the variables and integrating, we get:

v2

2=

gR2

R + x+ c .



v = ±

√v20 − 2gR +

2gR2

R + x.


Separable ODE’s

By chain rule,dv

dt=

dv

dx· dxdt

= v · dvdx.

Thus,

v · dvdx

= − gR2

(R + x)2.

This ODE is separable. Linear or non-linear? (NL)

Separating the variables and integrating, we get:

v2

2=

gR2

R + x+ c .



v = ±

√v20 − 2gR +

2gR2

R + x.


Separable ODE’s

By chain rule,dv

dt=

dv

dx· dxdt

= v · dvdx.

Thus,

v · dvdx

= − gR2

(R + x)2.


v2

2=

gR2

R + x+ c .



v = ±

√v20 − 2gR +

2gR2

R + x.


Separable ODE’s

By chain rule,dv

dt=

dv

dx· dxdt

= v · dvdx.

Thus,

v · dvdx

= − gR2

(R + x)2.


v2

2=

gR2

R + x+ c .


= gR + c ,

hence, c =v202− gR, and,

v = ±

√v20 − 2gR +

2gR2

R + x.


Separable ODE’s

By chain rule,dv

dt=

dv

dx· dxdt

= v · dvdx.

Thus,

v · dvdx

= − gR2

(R + x)2.


v2

2=

gR2

R + x+ c .


= gR + c , hence, c =v202− gR,

and,

v = ±

√v20 − 2gR +

2gR2

R + x.


Separable ODE’s

By chain rule,dv

dt=

dv

dx· dxdt

= v · dvdx.

Thus,

v · dvdx

= − gR2

(R + x)2.


v2

2=

gR2

R + x+ c .



v = ±

√v20 − 2gR +

2gR2

R + x.


Separable ODE’s

Suppose the body reaches the maximum height H.

Then v = 0 atthis height.

v20 − 2gR +2gR2

(R + H)= 0.

Thus,

v20 = 2gR − 2gR2

R + H= 2gR

(H

R + H

).

The escape velocity is found by taking limit as H →∞. Thus,

ve =√

2gR ∼ 11 km/sec.


Separable ODE’s

Suppose the body reaches the maximum height H. Then v =

0 atthis height.

v20 − 2gR +2gR2

(R + H)= 0.

Thus,

v20 = 2gR − 2gR2

R + H= 2gR

(H

R + H

).


ve =√

2gR ∼ 11 km/sec.


Separable ODE’s

Suppose the body reaches the maximum height H. Then v = 0 atthis height.

v20 − 2gR +2gR2

(R + H)= 0.

Thus,

v20 = 2gR − 2gR2

R + H= 2gR

(H

R + H

).


ve =√

2gR ∼ 11 km/sec.


Separable ODE’s


v20 − 2gR +2gR2

(R + H)= 0.

Thus,

v20 = 2gR − 2gR2

R + H= 2gR

(H

R + H

).

The escape velocity is found by taking limit as

H →∞. Thus,

ve =√

2gR ∼ 11 km/sec.


Separable ODE’s


v20 − 2gR +2gR2

(R + H)= 0.

Thus,

v20 = 2gR − 2gR2

R + H= 2gR

(H

R + H

).

The escape velocity is found by taking limit as H →∞.

Thus,

ve =√

2gR ∼ 11 km/sec.


Separable ODE’s


v20 − 2gR +2gR2

(R + H)= 0.

Thus,

v20 = 2gR − 2gR2

R + H= 2gR

(H

R + H

).


ve =√

2gR

∼ 11 km/sec.


Separable ODE’s


v20 − 2gR +2gR2

(R + H)= 0.

Thus,

v20 = 2gR − 2gR2

R + H= 2gR

(H

R + H

).


ve =√

2gR ∼ 11 km/sec.


lecture1 d3

Engineering