Linear Ordinary Differential Equations of First and Second Order (Part - 1), UGC - NET Physics Physics Notes | EduRev

Physics for IIT JAM, UGC - NET, CSIR NET

Created by: Akhilesh Thakur

Physics : Linear Ordinary Differential Equations of First and Second Order (Part - 1), UGC - NET Physics Physics Notes | EduRev

The document Linear Ordinary Differential Equations of First and Second Order (Part - 1), UGC - NET Physics Physics Notes | EduRev is a part of the Physics Course Physics for IIT JAM, UGC - NET, CSIR NET.
All you need of Physics at this link: Physics

First Order Differential Equations

We consider the equation

Linear Ordinary Differential Equations of First and Second Order (Part - 1), UGC - NET Physics Physics Notes | EduRev

Linear equations; Method of integrating factors 

The function f (t, y) is a linear function in y, i.e, we can write

f (t, y) = −p(t)y + g(t).

So we will study the equation

y′ + p(t)y = g(t).                                              (A)

We introduce the method of integrating factors (due to Leibniz): We multiply equation (A) by a function µ(t) on both sides

µ(t)y′ + µ(t)p(t)y = µ(t)g(t)

The function µ is chosen such that the equation is integrable, meaning the LHS (Left Hand Side) is the derivative of something. In particular, we require:

µ(t)y′ + µ(t)p(t)y = (µ(t)y)′ , ⇒ µ(t)y′ + µ(t)p(t)y = µ(t)y′ + µ′(t)y

which requires

Linear Ordinary Differential Equations of First and Second Order (Part - 1), UGC - NET Physics Physics Notes | EduRev

Integrating both sides

Linear Ordinary Differential Equations of First and Second Order (Part - 1), UGC - NET Physics Physics Notes | EduRev

which gives a formula to compute µ

Linear Ordinary Differential Equations of First and Second Order (Part - 1), UGC - NET Physics Physics Notes | EduRev

Therefore, this µ is called the integrating factor.
Note that µ is not unique. In fact, adding an integration constant, we will get a different µ. But we don’t need to be bothered, since any such µ will work. We can simply choose one that is convenient.

Putting back into equation (A), we get

Linear Ordinary Differential Equations of First and Second Order (Part - 1), UGC - NET Physics Physics Notes | EduRev

which give the formula for the solution

Linear Ordinary Differential Equations of First and Second Order (Part - 1), UGC - NET Physics Physics Notes | EduRev


Example 1. Solve y′ + ay = b (a = 0).

Answer.We have p(t) = a and g(t) = b. So

Linear Ordinary Differential Equations of First and Second Order (Part - 1), UGC - NET Physics Physics Notes | EduRev

so

Linear Ordinary Differential Equations of First and Second Order (Part - 1), UGC - NET Physics Physics Notes | EduRev

where c is an arbitrary constant. Pay attention to where one adds this integration constant!


Example 2. Solve y′ + y = e2t.
 Answer.
We have p(t) = 1 and g(t) = e2t. So

Linear Ordinary Differential Equations of First and Second Order (Part - 1), UGC - NET Physics Physics Notes | EduRev

and

Linear Ordinary Differential Equations of First and Second Order (Part - 1), UGC - NET Physics Physics Notes | EduRev


Example 3. Solve

Linear Ordinary Differential Equations of First and Second Order (Part - 1), UGC - NET Physics Physics Notes | EduRev

Answer.First, let’s rewrite the equation into the normal form

Linear Ordinary Differential Equations of First and Second Order (Part - 1), UGC - NET Physics Physics Notes | EduRev

so

Linear Ordinary Differential Equations of First and Second Order (Part - 1), UGC - NET Physics Physics Notes | EduRev

Then

Linear Ordinary Differential Equations of First and Second Order (Part - 1), UGC - NET Physics Physics Notes | EduRev

Then

Linear Ordinary Differential Equations of First and Second Order (Part - 1), UGC - NET Physics Physics Notes | EduRev

By the IC y(0) = 1:

Linear Ordinary Differential Equations of First and Second Order (Part - 1), UGC - NET Physics Physics Notes | EduRev

 

Example 4. Solve ty′ − y = t2 e−t , (t > 0).
 Answer
.Rewrite it into normal form

Linear Ordinary Differential Equations of First and Second Order (Part - 1), UGC - NET Physics Physics Notes | EduRev

so

p(t) = −1/t, g(t) = te−t.

We have

Linear Ordinary Differential Equations of First and Second Order (Part - 1), UGC - NET Physics Physics Notes | EduRev

and

Linear Ordinary Differential Equations of First and Second Order (Part - 1), UGC - NET Physics Physics Notes | EduRev


Example 5. Solve Linear Ordinary Differential Equations of First and Second Order (Part - 1), UGC - NET Physics Physics Notes | EduRev with y(0) = a, and discussion the behavior of y as t → ∞, as one chooses different initial value a.
 Answer.
Let’s solve it first. We have

Linear Ordinary Differential Equations of First and Second Order (Part - 1), UGC - NET Physics Physics Notes | EduRev

so

Linear Ordinary Differential Equations of First and Second Order (Part - 1), UGC - NET Physics Physics Notes | EduRev

Plug in the IC to find c

Linear Ordinary Differential Equations of First and Second Order (Part - 1), UGC - NET Physics Physics Notes | EduRev

so

Linear Ordinary Differential Equations of First and Second Order (Part - 1), UGC - NET Physics Physics Notes | EduRev

To see the behavior of the solution, we see that it contains two terms. The first term e−t goes to 0 as t grows. The second term et/3 goes to ∞ as t grows, but the constantLinear Ordinary Differential Equations of First and Second Order (Part - 1), UGC - NET Physics Physics Notes | EduRev

Linear Ordinary Differential Equations of First and Second Order (Part - 1), UGC - NET Physics Physics Notes | EduRev

On the other hand, as t → −∞, the term e−t will blow up to −∞, and will dominate.
Therefore, y → −∞ as t → −∞ for any values of a.
See plot below:

Linear Ordinary Differential Equations of First and Second Order (Part - 1), UGC - NET Physics Physics Notes | EduRev

Example 6. Solve ty′ + 2y = 4t2 , y(1) = 2.
 Answer.
Rewrite the equation first

Linear Ordinary Differential Equations of First and Second Order (Part - 1), UGC - NET Physics Physics Notes | EduRev

So p(t) = 2/t and g(t) = 4t. We have

Linear Ordinary Differential Equations of First and Second Order (Part - 1), UGC - NET Physics Physics Notes | EduRev

and

Linear Ordinary Differential Equations of First and Second Order (Part - 1), UGC - NET Physics Physics Notes | EduRev

By IC y (1) = 2

y(1) = 1 + c = 2, c = 1

we get the solution:

Linear Ordinary Differential Equations of First and Second Order (Part - 1), UGC - NET Physics Physics Notes | EduRev

Note the condition t > 0 comes from the fact that the initial condition is given at t = 1, and we require t ≠ 0.
In the graph below we plot several solutions in the t − y plan, depending on initial data.
The one for our solution is plotted with dashed line where the initial point is marked with a
‘×’.

Linear Ordinary Differential Equations of First and Second Order (Part - 1), UGC - NET Physics Physics Notes | EduRev

 

Separable Equations 

We study first order equations that can be written as

Linear Ordinary Differential Equations of First and Second Order (Part - 1), UGC - NET Physics Physics Notes | EduRev

where M (x) and N (y) are suitable functions of x and y only. Then we have

Linear Ordinary Differential Equations of First and Second Order (Part - 1), UGC - NET Physics Physics Notes | EduRev

and we get implicitly defined solutions of y(x).
Example 1. ConsiderLinear Ordinary Differential Equations of First and Second Order (Part - 1), UGC - NET Physics Physics Notes | EduRev

We can separate the variables:

Linear Ordinary Differential Equations of First and Second Order (Part - 1), UGC - NET Physics Physics Notes | EduRev

If one has IC as y(π) = 2, then

Linear Ordinary Differential Equations of First and Second Order (Part - 1), UGC - NET Physics Physics Notes | EduRev

so the solution y(x) is implicitly given as

Linear Ordinary Differential Equations of First and Second Order (Part - 1), UGC - NET Physics Physics Notes | EduRev

Example 2. Find the solution in explicit form for the equation

Linear Ordinary Differential Equations of First and Second Order (Part - 1), UGC - NET Physics Physics Notes | EduRev

Answer.Separate the variables

Linear Ordinary Differential Equations of First and Second Order (Part - 1), UGC - NET Physics Physics Notes | EduRev

Set in the IC y(0) = −1, i.e., y = −1 when x = 0, we get

Linear Ordinary Differential Equations of First and Second Order (Part - 1), UGC - NET Physics Physics Notes | EduRev

In explicitly form, one has two choices:

Linear Ordinary Differential Equations of First and Second Order (Part - 1), UGC - NET Physics Physics Notes | EduRev

To determine which sign is the correct one, we check again by the initial condition:

Linear Ordinary Differential Equations of First and Second Order (Part - 1), UGC - NET Physics Physics Notes | EduRev must have y(0) = −1.

We see we must choose the ‘-’ sign. The solution in explicitly form is:

Linear Ordinary Differential Equations of First and Second Order (Part - 1), UGC - NET Physics Physics Notes | EduRev

On which interval will this solution be defined ?

Linear Ordinary Differential Equations of First and Second Order (Part - 1), UGC - NET Physics Physics Notes | EduRev

We can also argue that when x = −2, we have y = 1. At this point |dy/dx| → ∞, therefore solution can not be defined at this point.
The plot of the solution is given below, where the initial data is marked with ‘x’. We also include the solution with the ‘+’ sign, using dotted line.

Linear Ordinary Differential Equations of First and Second Order (Part - 1), UGC - NET Physics Physics Notes | EduRev

Example 3. Solve y′ = 3x2 + 3x2 y2, y(0) = 0, and find the interval where the solution is defined.
Answer. Let’s first separate the variables.

Linear Ordinary Differential Equations of First and Second Order (Part - 1), UGC - NET Physics Physics Notes | EduRev

Set in the IC:
arctan 0 = 0 + c, ⇒ c = 0

we get the solution

arctan y = x3 , ⇒ y = tan(x3).

Since the initial data is given at x = 0, i.e., x3 = 0, and tan is defined on the interval  Linear Ordinary Differential Equations of First and Second Order (Part - 1), UGC - NET Physics Physics Notes | EduRev

We have

Linear Ordinary Differential Equations of First and Second Order (Part - 1), UGC - NET Physics Physics Notes | EduRev

Example 4. Solve

Linear Ordinary Differential Equations of First and Second Order (Part - 1), UGC - NET Physics Physics Notes | EduRev

and identify the interval where solution is valid.
Answer.Separate the variables

Linear Ordinary Differential Equations of First and Second Order (Part - 1), UGC - NET Physics Physics Notes | EduRev

Set in the IC: x = 0, y = 1, we get
1 − 3 = c, ⇒ c = −2,

Then,
y− 3y2 = x− x − 2.

Note that solution is given in implicitly form.
To find the valid interval of this solution, we note that y′ is not defined if 3y2 − 6y = 0, i.e., when y = 0 or y = 2. These are the two so-called “bad points” where you can not define the solution. To find the corresponding values of x, we use the solution expression:

y = 0: x3+x−2 = 0,

⇒ (x+ x + 2)(x − 1) = 0, ⇒ x = 1

and 
Linear Ordinary Differential Equations of First and Second Order (Part - 1), UGC - NET Physics Physics Notes | EduRev

(Note that we used the facts x2 + x + 2 = 0 and x2 − x + 2 = 0 for all x.) Draw the real line and work on it as following:

Linear Ordinary Differential Equations of First and Second Order (Part - 1), UGC - NET Physics Physics Notes | EduRev

Therefore the interval is −1 < x < 1.

Differences between linear and nonlinear equations

We will take this chapter before the modeling .
For a linear equation

y′ + p(t)y = g(t), y(t0) = y0,

we have the following existence and uniqueness theorem.
Theorem . If p(t) and g(t) are continuous and bounded on an open interval containing t0, then it has an unique solution on that interval.

Example 1. Find the largest interval where the solution can be defined for the following problems.

(A). ty′ + y = t3, y(−1) = 3.

Linear Ordinary Differential Equations of First and Second Order (Part - 1), UGC - NET Physics Physics Notes | EduRev

Answer. The equation is same as (A), so t = 0. t0 = 1, the interval is t > 0.

(C). (t − 3)y′ + (ln t)y = 2t, y(1) = 2

Answer. Rewrite: Linear Ordinary Differential Equations of First and Second Order (Part - 1), UGC - NET Physics Physics Notes | EduRev and t > 0 for the ln function. Since t0 = 1, the interval is then 0 < t < 3.

(D). y′ + (tan t)y = sin t, y(π) = 100.

Answer. Since t0 = π, and for tan t to be defined we must have  Linear Ordinary Differential Equations of First and Second Order (Part - 1), UGC - NET Physics Physics Notes | EduRev

So the interval is Linear Ordinary Differential Equations of First and Second Order (Part - 1), UGC - NET Physics Physics Notes | EduRev

For non-linear equation

y′ = f (t, y), y(t0) = y0,

we have the following theorem:

Theorem . If  Linear Ordinary Differential Equations of First and Second Order (Part - 1), UGC - NET Physics Physics Notes | EduRev are continuous and bounded on an rectangle (α < t < β , a < y < b) containing (t0, y0 ), then there exists an open interval around t0, contained in (α, β ), where the solution exists and is unique.
We note that the statement of this theorem is not as strong as the one for linear equation.
Below we give two counter examples.

Example 1. Loss of uniqueness. Consider

Linear Ordinary Differential Equations of First and Second Order (Part - 1), UGC - NET Physics Physics Notes | EduRev y(−2) = 0.

We first note that at y = 0, which is the initial value of y, we have y′ = f (t, y) → ∞. So the conditions of the Theorem are not satisfied, and we expect something to go wrong.
Solve the equation as an separable equation, we get

Linear Ordinary Differential Equations of First and Second Order (Part - 1), UGC - NET Physics Physics Notes | EduRev

and by IC we get c = (−2)2 + 0 = 4, so y2 + t2 = 4. In the y − t plan, this is the equation for a circle, centered at the origin, with radius 2. The initial condition is given at t0 = −2, y0 = 0, where the tangent line is vertical (i.e., with infinite slope). We have two solutions: y =   Linear Ordinary Differential Equations of First and Second Order (Part - 1), UGC - NET Physics Physics Notes | EduRev  and  y =  Linear Ordinary Differential Equations of First and Second Order (Part - 1), UGC - NET Physics Physics Notes | EduRev

We lose uniqueness of solutions.

Example 2. Blow-up of solution. Consider a simple non-linear equation: 

y′ = y2, y(0) = 1.

Note that f (t, y) = y2, which is defined for all t and y. But, due to the non-linearity of f , solution can not be defined for all t.
This equation can be easily solved as a separable equation.

Linear Ordinary Differential Equations of First and Second Order (Part - 1), UGC - NET Physics Physics Notes | EduRev

By IC y(0) = 1, we get 1 = −1/(0 + c), and so c = −1, and

Linear Ordinary Differential Equations of First and Second Order (Part - 1), UGC - NET Physics Physics Notes | EduRev

We see that the solution blows up as t → 1, and can not be defined beyond that point.

This kind of blow-up phenomenon is well-known for nonlinear equations.

Dynamic Test

Content Category

Related Searches

Sample Paper

,

Linear Ordinary Differential Equations of First and Second Order (Part - 1)

,

UGC - NET Physics Physics Notes | EduRev

,

UGC - NET Physics Physics Notes | EduRev

,

Summary

,

MCQs

,

Free

,

Objective type Questions

,

past year papers

,

Linear Ordinary Differential Equations of First and Second Order (Part - 1)

,

ppt

,

study material

,

practice quizzes

,

mock tests for examination

,

Important questions

,

shortcuts and tricks

,

Exam

,

Previous Year Questions with Solutions

,

Extra Questions

,

Semester Notes

,

UGC - NET Physics Physics Notes | EduRev

,

Viva Questions

,

video lectures

,

Linear Ordinary Differential Equations of First and Second Order (Part - 1)

,

pdf

;