Linear Ordinary Differential Equations of First and Second Order - 5 | Physics for IIT JAM, UGC - NET, CSIR NET PDF Download

Repeated roots; reduction of order 

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

We have one solution . How can we find the second solution which is linearly independent of y₁?

From experience in an earlier example, we claim that  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

which gives the equation

We can check if y₂ satisfies this equation. We have

Put into the equation, we get

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

We conclude now, the general solution is

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

By the definition of Wronskian we have

They must equal to each other:

Solve this for y₂,

Let C = 0, we get y₂ = te^−2t , and the general solution is

Method 2. This is the textbook’s version. We guess a solution of the form y₂ = v(t)y₁ = v(t)e^−2t , and try to find the function v(t). We have

Put them in the equation

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

Therefore we can choose c₁ = 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:

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 = e^rt and y₂ = te^rt , and the general solution is

Example 2. Solve the IVP

Answer.This follows easily now

The ICs give

y(0) = 2 : c₁ + 0 = 2, ⇒ c₁ = 2.

y′(t) = (c₁ + c₂ t)e^t + c₂ e^t , y′(0) = c₁ + c₂ = 1, ⇒ c₂ = 1 − c₁ = −1.

So the solution is y(t) = (2 − t)e^t .

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

On reduction of order: This method can be used to find a second solution y₂ if the first solution y₁ is given for a second order linear equation.

Example 3. For the equation

given one solution   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

By definition of the Wronskian,

Solve this for y₂:

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)  . Now, seek y₂ = vy₁.

By the definition of the Wronskian, we have

Note that this is a general formula:

Now putting y₁ = 1/t, we get

Drop the constant 3/2 , we get

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

Given y₁ = t, find the general solution.
Answer. We have

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

(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

We observe here that Method 2 is very efficient.
 

Example 5. Given the equationfind y₂

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

So drop the constant  , we get

The general solution is

Non-homogeneous equations; method of undetermined coefficients

Want to solve the non-homogeneous equation

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 = 3e^2t .

Answer.Step 1: Find yH .

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

yH = c₁ e^−t + c₂ e^4t.

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

Y ′ = 2Ae^2t , Y ′′ = 4Ae^2t .

Plug these into the equation:

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

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:   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

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

Plug them in the equation

we get

so we have Y  =

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

Example 3. Find the general solution for

Answer.The yH is the same yH = 

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

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

So we get

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 t² + 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 :

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

the particular solution for

ay′′ + by′ + cy = g(t)

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

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

So we get

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 r_1,2 = ±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

So

The general solution is

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

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^t .

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

1. Also we see r₁ = −1, r₂ = 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

Plug them into the equation,

We must have −6At − A − 6B = t, i.e.,

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 = (At₂ + B t)e^−t.
Then

Plug into the equation

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

Then

Summary 4. If g(t) = P_n (t)e^at where P_n(t) = α_ntⁿ + · · · + α₁ t + α₀ is a polynomial of degree n, then the form of a particular solution depends on the roots r₁, r₂ .

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) = e^αt (a cos β t + b sin β t), and r₁, r₂ are the roots of the characteristic equation. Then

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

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) + · · · g_n (t), then, find a particular solution Y_i for each g_i(t) term as if it were the only term in g, then Y = Y₁ + Y₂ + · · · Y_n . 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 + 2e^4t + e^5t − t.
 Answer.Since r₁ = −1, r₂ = 2, we treat each term in g separately and the add up: Y (t) = A sin 4t + B cos 4t + C te^4t + De^5t + (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 r_1,2 = ±4ⁱ, and yH = c₁ sin 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 = e^t cos t + 8e^t sin 2t + te^−t + 4e^−t + t² − 3.
 Answer.The char equation is r₂ − 2r + 2 = 0 with roots r_1,2 = 1 ± i. Then, for the term e^t cos t we must multiply by t.