Linear Ordinary Differential Equations of First and Second Order (Part - 5), 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 - 5), UGC - NET Physics Physics Notes | EduRev

The document Linear Ordinary Differential Equations of First and Second Order (Part - 5), UGC - NET Physics Physics Notes | EduRev is a part of the Physics Course Physics for IIT JAM, UGC - NET, CSIR NET.
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Repeated roots; reduction of order 

For the characteristic equation ar2 + br + c = 0, if b2 = 4ac, we will have two repeated roots

Linear Ordinary Differential Equations of First and Second Order (Part - 5), UGC - NET Physics Physics Notes | EduRev

We have one solution Linear Ordinary Differential Equations of First and Second Order (Part - 5), UGC - NET Physics Physics Notes | EduRev. How can we find the second solution which is linearly independent of y1?

From experience in an earlier example, we claim that Linear Ordinary Differential Equations of First and Second Order (Part - 5), UGC - NET Physics Physics Notes | EduRev is a solution. To prove this claim, we plug it back into the equation. If r¯ is the double root, then, the characteristic equation can be written

Linear Ordinary Differential Equations of First and Second Order (Part - 5), UGC - NET Physics Physics Notes | EduRev

which gives the equation

Linear Ordinary Differential Equations of First and Second Order (Part - 5), UGC - NET Physics Physics Notes | EduRev

We can check if y2 satisfies this equation. We have

Linear Ordinary Differential Equations of First and Second Order (Part - 5), UGC - NET Physics Physics Notes | EduRev

Put into the equation, we get

Linear Ordinary Differential Equations of First and Second Order (Part - 5), UGC - NET Physics Physics Notes | EduRev

Finally, we must make sure that y1, y2 are linearly independent. We compute their Wronskian

Linear Ordinary Differential Equations of First and Second Order (Part - 5), UGC - NET Physics Physics Notes | EduRev

We conclude now, the general solution is

Linear Ordinary Differential Equations of First and Second Order (Part - 5), UGC - NET Physics Physics Notes | EduRev

Example 1. (not covered in class) Consider the equation y′′ + 4y′ + 4y = 0. We have r2 + 4r + 4 = 0, and r1 = r2 = r = −2. So one solution is y1 = e−2t . What is y2?
 Method 1. 
Use Wronskian and Abel’s Theorem. By Abel’s Theorem we have

Linear Ordinary Differential Equations of First and Second Order (Part - 5), UGC - NET Physics Physics Notes | EduRev

By the definition of Wronskian we have

Linear Ordinary Differential Equations of First and Second Order (Part - 5), UGC - NET Physics Physics Notes | EduRev

They must equal to each other:

Linear Ordinary Differential Equations of First and Second Order (Part - 5), UGC - NET Physics Physics Notes | EduRev

Solve this for y2,

Linear Ordinary Differential Equations of First and Second Order (Part - 5), UGC - NET Physics Physics Notes | EduRev

Let C = 0, we get y= te−2t , and the general solution is

Linear Ordinary Differential Equations of First and Second Order (Part - 5), UGC - NET Physics Physics Notes | EduRev

Method 2. This is the textbook’s version. We guess a solution of the form y= v(t)y1 = v(t)e−2t , and try to find the function v(t). We have

Linear Ordinary Differential Equations of First and Second Order (Part - 5), UGC - NET Physics Physics Notes | EduRev

Put them in the equation

Linear Ordinary Differential Equations of First and Second Order (Part - 5), UGC - NET Physics Physics Notes | EduRev

Note that the term c2 e−2t is already contained in cy1 .

Therefore we can choose c= 1, c2 = 0, and get y2 = te−2t , which gives the same general solution as Method 1. We observe that this method involves more computation than Method 1.
A typical solution graph is included below:

Linear Ordinary Differential Equations of First and Second Order (Part - 5), UGC - NET Physics Physics Notes | EduRev

We see if c2 > 0, y increases for small t. But as t grows, the exponential (decay) function dominates, and solution will go to 0 as t → ∞.

One can show that in general if one has repeated roots r1 = r2 = r, then y1 = ert and y2 = tert , and the general solution is

Linear Ordinary Differential Equations of First and Second Order (Part - 5), UGC - NET Physics Physics Notes | EduRev

Example 2. Solve the IVP

Linear Ordinary Differential Equations of First and Second Order (Part - 5), UGC - NET Physics Physics Notes | EduRev

Answer.This follows easily now

Linear Ordinary Differential Equations of First and Second Order (Part - 5), UGC - NET Physics Physics Notes | EduRev

The ICs give

y(0) = 2 : c+ 0 = 2, ⇒ c= 2.

y′(t) = (c1 + c2 t)et + ce, y′(0) = c1 + c2 = 1, ⇒ c2 = 1 − c1 = −1.

So the solution is y(t) = (2 − t)et .

Summary: For ay′′ + by′ + cy = 0, and ar2 + br + c = 0 has two roots r1, r2 , we have

Linear Ordinary Differential Equations of First and Second Order (Part - 5), UGC - NET Physics Physics Notes | EduRev

On reduction of order: This method can be used to find a second solution y2 if the first solution yis given for a second order linear equation.

Example 3. For the equation

Linear Ordinary Differential Equations of First and Second Order (Part - 5), UGC - NET Physics Physics Notes | EduRev

given one solution  Linear Ordinary Differential Equations of First and Second Order (Part - 5), UGC - NET Physics Physics Notes | EduRev find a second linearly independent solution.

Answer.Method 1: Use Abel’s Theorem and Wronskian. By Abel’s Theorem, and choose C = 1, we have

Linear Ordinary Differential Equations of First and Second Order (Part - 5), UGC - NET Physics Physics Notes | EduRev

By definition of the Wronskian,

Linear Ordinary Differential Equations of First and Second Order (Part - 5), UGC - NET Physics Physics Notes | EduRev

Solve this for y2:

Linear Ordinary Differential Equations of First and Second Order (Part - 5), UGC - NET Physics Physics Notes | EduRev

Linear Ordinary Differential Equations of First and Second Order (Part - 5), UGC - NET Physics Physics Notes | EduRev

 

Method 2. We will use Abel’s Theorem, and at the same time we will seek a solution of the form y1 = vy1.

By Abel’s Theorem, we have ( worked out in M1) W (y1, y2) Linear Ordinary Differential Equations of First and Second Order (Part - 5), UGC - NET Physics Physics Notes | EduRev . Now, seek y2 = vy1.

By the definition of the Wronskian, we have

Linear Ordinary Differential Equations of First and Second Order (Part - 5), UGC - NET Physics Physics Notes | EduRev

Note that this is a general formula:

Linear Ordinary Differential Equations of First and Second Order (Part - 5), UGC - NET Physics Physics Notes | EduRev

Now putting y1 = 1/t, we get

Linear Ordinary Differential Equations of First and Second Order (Part - 5), UGC - NET Physics Physics Notes | EduRev

Drop the constant 3/2 , we get

Linear Ordinary Differential Equations of First and Second Order (Part - 5), UGC - NET Physics Physics Notes | EduRev

We see that Method 3 is the most efficient one among all three methods. We will focus on this method from now on.

 

Example 4. Consider the equation

Linear Ordinary Differential Equations of First and Second Order (Part - 5), UGC - NET Physics Physics Notes | EduRev

Given y= t, find the general solution.
Answer. We have

Linear Ordinary Differential Equations of First and Second Order (Part - 5), UGC - NET Physics Physics Notes | EduRev

Let y2 be the second solution. By Abel’s Theorem, choosing c = 1, we have

Linear Ordinary Differential Equations of First and Second Order (Part - 5), UGC - NET Physics Physics Notes | EduRev

Linear Ordinary Differential Equations of First and Second Order (Part - 5), UGC - NET Physics Physics Notes | EduRev

(A cheap trick to double check your solution y2 would be: plug it back into the equation and see if it satisfies it.) The general solution is

Linear Ordinary Differential Equations of First and Second Order (Part - 5), UGC - NET Physics Physics Notes | EduRev

We observe here that Method 2 is very efficient.
 

Example 5. Given the equationLinear Ordinary Differential Equations of First and Second Order (Part - 5), UGC - NET Physics Physics Notes | EduRevfind y2

Answer.We will always use method 2. We see that p = 0. By Abel’s Theorem, setting c = 1, we have

Linear Ordinary Differential Equations of First and Second Order (Part - 5), UGC - NET Physics Physics Notes | EduRev

Linear Ordinary Differential Equations of First and Second Order (Part - 5), UGC - NET Physics Physics Notes | EduRev

Linear Ordinary Differential Equations of First and Second Order (Part - 5), UGC - NET Physics Physics Notes | EduRev

So drop the constant Linear Ordinary Differential Equations of First and Second Order (Part - 5), UGC - NET Physics Physics Notes | EduRev , we get

Linear Ordinary Differential Equations of First and Second Order (Part - 5), UGC - NET Physics Physics Notes | EduRev

The general solution is

Linear Ordinary Differential Equations of First and Second Order (Part - 5), UGC - NET Physics Physics Notes | EduRev

Non-homogeneous equations; method of undetermined coefficients

Want to solve the non-homogeneous equation

Linear Ordinary Differential Equations of First and Second Order (Part - 5), UGC - NET Physics Physics Notes | EduRev

Steps:

1. First solve the homogeneous equation

y′′ + p(t)y′ + q(t)y = 0,                         (H )

i.e., find y1, y2, linearly independent of each other, and form the general solution

yH = c1 y1 + c2 y2.

2. Find a particular/specific solution Y for (N), by MUC (method of undetermined coefficients);

3. The general solution for (N) is then

y = yH + Y = c1y1 + c2y2 + Y .

Find c1 , c2 by initial conditions, if given.

Key step: step 2.

Why y = yH + Y ?

A quick proof: If yH solves (H), then

y′′H + p(t)y′H + q(t)yH = 0,                 (A)

and since Y solves (N), we have

Y ′′ + p(t)Y ′ + q(t)Y = g(t),                (B )

Adding up (A) and (B), and write y = yH + Y , we get y′′ + p(t)y′ + q(t)y = g(t).

Main focus: constant coefficient case, i.e.,

ay′′ + by′ + cy = g(t).

Example 1. Find the general solution for y′′ − 3y′ − 4y = 3e2t .

Answer.Step 1: Find yH .

r2 − 3r − 4 = (r + 1)(r − 4) = 0, ⇒ r1 = −1, r2 = 4, so

yH = c1 e−t + c2 e4t.

Step 2: Find Y . We guess/seek solution of the same form as the source term Y = Ae2t , and will determine the coefficient A.

Y ′ = 2Ae2t , Y ′′ = 4Ae2t .

Plug these into the equation:

Linear Ordinary Differential Equations of First and Second Order (Part - 5), UGC - NET Physics Physics Notes | EduRev

Linear Ordinary Differential Equations of First and Second Order (Part - 5), UGC - NET Physics Physics Notes | EduRev

Step 3. The general solution to the non-homogeneous solution is

Linear Ordinary Differential Equations of First and Second Order (Part - 5), UGC - NET Physics Physics Notes | EduRev

Observation: The particular solution Y take the same form as the source term g(t).

But this is not always true.

Example 2. Find general solution for y′′ − 3y′ − 4y = 2e−t .

Answer.The homogeneous solution is the same as Example 1:  Linear Ordinary Differential Equations of First and Second Order (Part - 5), UGC - NET Physics Physics Notes | EduRev For the particular solution Y , let’s first try the same form as g, i.e., Y = Ae−t . So Y ′ = −Ae−t , Y ′′ = Ae−t . Plug them back in to the equation, we get

Linear Ordinary Differential Equations of First and Second Order (Part - 5), UGC - NET Physics Physics Notes | EduRev

So it doesn’t work. Why?

We see r1 = −1 and y1 = e−t, which means our guess Y = Ae−t is a solution to the homogeneous equation. It will never work.

Second try: Y = Ate−t . So

Linear Ordinary Differential Equations of First and Second Order (Part - 5), UGC - NET Physics Physics Notes | EduRev

Plug them in the equation

Linear Ordinary Differential Equations of First and Second Order (Part - 5), UGC - NET Physics Physics Notes | EduRev

we get

Linear Ordinary Differential Equations of First and Second Order (Part - 5), UGC - NET Physics Physics Notes | EduRev

so we have Y  =Linear Ordinary Differential Equations of First and Second Order (Part - 5), UGC - NET Physics Physics Notes | EduRev

Summary 1. If g(t) = aeαt , then the form of the particular solution Y depends on r1 , r2 (the roots of the characteristic equation).

Linear Ordinary Differential Equations of First and Second Order (Part - 5), UGC - NET Physics Physics Notes | EduRev

Example 3. Find the general solution for

Linear Ordinary Differential Equations of First and Second Order (Part - 5), UGC - NET Physics Physics Notes | EduRev

Answer.The yH is the same yH = Linear Ordinary Differential Equations of First and Second Order (Part - 5), UGC - NET Physics Physics Notes | EduRev

Note that g(t) is a polynomial of degree 2. We will try to guess/seek a particular solution of the same form:

Y = At2 + Bt + C,       Y ′ = 2At + B,              Y ′′ = 2A

Plug back into the equation

Linear Ordinary Differential Equations of First and Second Order (Part - 5), UGC - NET Physics Physics Notes | EduRev

Compare the coefficient, we get three equations for the three coefficients A, B , C :

Linear Ordinary Differential Equations of First and Second Order (Part - 5), UGC - NET Physics Physics Notes | EduRev

So we get

Linear Ordinary Differential Equations of First and Second Order (Part - 5), UGC - NET Physics Physics Notes | EduRev

But sometimes this guess won’t work.

Example 4. Find the particular solution for y′′ − 3y′ = 3t2 + 2.

Answer.We see that the form we used in the previous example Y = At2 + Bt + C won’t work because Y′′ − 3Y′ will not have the term t2 .

New try: multiply by a t. So we guess Y = t(At2 + B t + C ) = At3 + B t+ Ct. Then

Y ′ = 3At2 + 2B t + C, Y ′′ = 6At + 2B .

Plug them into the equation
(6At + 2B) − 3(3At2 + 2Bt + C ) = −9At2 + (6A − 6B)t + (2B − 3C ) = 3t2 + 2.

Compare the coefficient, we get three equations for the three coefficients A, B , C :

Linear Ordinary Differential Equations of First and Second Order (Part - 5), UGC - NET Physics Physics Notes | EduRev

Linear Ordinary Differential Equations of First and Second Order (Part - 5), UGC - NET Physics Physics Notes | EduRev

Summary 2. If g(t) is a polynomial of degree n, i.e.,

Linear Ordinary Differential Equations of First and Second Order (Part - 5), UGC - NET Physics Physics Notes | EduRev

the particular solution for

ay′′ + by′ + cy = g(t)

(where a = 0) depends on b, c:

Linear Ordinary Differential Equations of First and Second Order (Part - 5), UGC - NET Physics Physics Notes | EduRev

Example 5. Find a particular solution for

y′′ − 3y′ − 4y = sin t.

Answer.Since g(t) = sin t, we will try the same form. Note that (sin t)′ = cos t, so we must have the cos t term as well. So the form of the particular solution is

Y = A sin t + B cos t.

Then Y ′ = A cos t − B sin t, Y ′′ = −A sin t − B cos t.

Plug back into the equation, we get (−A sin t − B cos t) − 3(A cos t − B sin t) − 4(A sin t + b cos t) = (−5A + 3B ) sin t + (−3A − 5B ) cos t = sin t.

So we must have

Linear Ordinary Differential Equations of First and Second Order (Part - 5), UGC - NET Physics Physics Notes | EduRev

So we get

Linear Ordinary Differential Equations of First and Second Order (Part - 5), UGC - NET Physics Physics Notes | EduRev

We observe that: (1). If the right-hand side is g(t) = a cos t, then the same form would work; (2). More generally, if g(t) = a sin t + b sin t for some a, b, then the same form still work.

However, this form won’t work if it is a solution to the homogeneous equation.

Example 6. Find a general solution for y′′ + y = sin t.

Answer. Let’s first find yH . We have r2 + 1 = 0, so r1,2 = ±i, and yH = c1 cos t + c2 sin t.

For the particular solution Y : We see that the form Y = A sin t + B cos t won’t work because it solves the homogeneous equation.

Our new guess: multiply it by t, so

Y (t) = t(A sin t + B cos t).

Then

Y ′ = (A sin t + B cos t) + t(A cos t + B sin t), Y ′′ = (−2B − At) sin t + (2A − B t) cos t.

Plug into the equation

Linear Ordinary Differential Equations of First and Second Order (Part - 5), UGC - NET Physics Physics Notes | EduRev

So

Linear Ordinary Differential Equations of First and Second Order (Part - 5), UGC - NET Physics Physics Notes | EduRev

The general solution is

Linear Ordinary Differential Equations of First and Second Order (Part - 5), UGC - NET Physics Physics Notes | EduRev

Summary 3. If g(t) = a sin αt + b cos αt, the form of the particular solution depends on the roots r1 , r2.

Linear Ordinary Differential Equations of First and Second Order (Part - 5), UGC - NET Physics Physics Notes | EduRev

Note that case (2) occurs when the equation is y′′ + α2 y = a sin αt + b cos αt.

We now have discovered some general rules to obtain the form of the particular solution for the non-homogeneous equation ay′′ + by′ + cy = g(t).

  • Rule (1). Usually, Y take the same form as g(t);
  • Rule (2). Except, if the form of g(t) provides a solution to the homogeneous equation.

Then, one can multiply it by t.

  • Rule (3). If the resulting form in Rule (2) is still a solution to the homogeneous equation, then, multiply it by another t.

Next we study a couple of more complicated forms of g.


Example 7. Find a particular solution for

y′′ − 3y′ − 4y = te.

Answer. We see that g = P1 (t)eat , where P1 is a polynomial of degree

1. Also we see r= −1, r= 4, so r1 = a and r2 = a. For a particular solution we will try the same form as g, i.e., Y = (At + B )et. So

Linear Ordinary Differential Equations of First and Second Order (Part - 5), UGC - NET Physics Physics Notes | EduRev

Plug them into the equation,

Linear Ordinary Differential Equations of First and Second Order (Part - 5), UGC - NET Physics Physics Notes | EduRev

We must have −6At − A − 6B = t, i.e.,

Linear Ordinary Differential Equations of First and Second Order (Part - 5), UGC - NET Physics Physics Notes | EduRev

However, if the form of g is a solution to the homogeneous equation, it won’t work for a particular solution. We must multiply it by t in that case.

Example 8. Find a particular solution of

y′′ − 3y′ − 4y = te−t .

Answer.Since a = −1 = r1, so the form we used in Example 7 won’t work here. (Can you intuitively explain why?)

Try a new form now Y = t(At + B )e−t = (At+ B t)e−t.
Then

Linear Ordinary Differential Equations of First and Second Order (Part - 5), UGC - NET Physics Physics Notes | EduRev

Plug into the equation

Linear Ordinary Differential Equations of First and Second Order (Part - 5), UGC - NET Physics Physics Notes | EduRev

So we must have −10At + 2A − 5B = t, which means

Linear Ordinary Differential Equations of First and Second Order (Part - 5), UGC - NET Physics Physics Notes | EduRev

Then

Linear Ordinary Differential Equations of First and Second Order (Part - 5), UGC - NET Physics Physics Notes | EduRev

Summary 4. If g(t) = Pn (t)eat where Pn(t) = αntn + · · · + α1 t + α0 is a polynomial of degree n, then the form of a particular solution depends on the roots r1, r2 .

Linear Ordinary Differential Equations of First and Second Order (Part - 5), UGC - NET Physics Physics Notes | EduRev

Other cases of g are treated in a similar way: Check if the form of g is a solution to the homogeneous equation. If not, then use it as the form of a particular solution. If yes, then multiply it by t or t2 .

We summarize a few cases below.

Summary 5. If g(t) = eαt (a cos β t + b sin β t), and r1, r2 are the roots of the characteristic equation. Then

Linear Ordinary Differential Equations of First and Second Order (Part - 5), UGC - NET Physics Physics Notes | EduRev

 

Summary 6. If g(t) = Pn(t)eαt (a cos β t + b sin β t) where Pn (t) is a polynomial of degree n, and r1, r2 are the roots of the characteristic equation. Then

Linear Ordinary Differential Equations of First and Second Order (Part - 5), UGC - NET Physics Physics Notes | EduRev

More terms in the source. If the source g(t) has several terms, we treat each separately and add up later. Let g(t) = g1 (t) + g2 (t) + · · · gn (t), then, find a particular solution Yfor each gi(t) term as if it were the only term in g, then Y = Y1 + Y2 + · · · Yn . This claim follows from the principle of superposition. (Can you provide a brief proof ?)

In the examples below, we want to write the form of a particular solution.

Example 9. y′′ − 3y′ − 4y = sin 4t + 2e4t + e5t − t.
 Answer.
Since r= −1, r2 = 2, we treat each term in g separately and the add up: Y (t) = A sin 4t + B cos 4t + C te4t + De5t + (E t + F ).

Example 10. y′′ + 16y = sin 4t + cos t − 4 cos 4t + 4.
 Answer
.The char equation is r2 + 16 = 0, with roots r1,2 = ±4i, and yH = csin 4t + c2 cos 4t.

We also note that the terms sin 4t and −4 cos 4t are of the same type, and we must multiply it by t. So
Y = t(A sin 4t + B cos 4t) + (C cos t + D sin t) + E .

Example 11. y′′ − 2y′ + 2y = et cos t + 8et sin 2t + te−t + 4e−t + t2 − 3.
 Answer.
The char equation is r2 − 2r + 2 = 0 with roots r1,2 = 1 ± i. Then, for the term et cos t we must multiply by t.
Linear Ordinary Differential Equations of First and Second Order (Part - 5), UGC - NET Physics Physics Notes | EduRev

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