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**Repeated roots; reduction of order **

For the characteristic equation ar^{2} + br + c = 0, if b^{2} = 4ac, we will have two repeated roots

We have one solution . How can we ﬁnd the second solution which is linearly independent of y_{1}?

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 y_{2} satisﬁes this equation. We have

Put into the equation, we get

Finally, we must make sure that y_{1}, y_{2} 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 r ^{2} + 4r + 4 = 0, and r_{1} = r_{2} = r = −2. So one solution is y_{1} = e^{−2t} . What is y_{2}?**Use Wronskian and Abel’s Theorem. By Abel’s Theorem we have

Method 1.

By the deﬁnition of Wronskian we have

They must equal to each other:

Solve this for y_{2},

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

**Method 2.** This is the textbook’s version. We guess a solution of the form y_{2 }= v(t)y_{1} = v(t)e^{−2t} , and try to ﬁnd the function v(t). We have

Put them in the equation

Note that the term c_{2} e^{−2t }is already contained in cy_{1} .

Therefore we can choose c_{1 }= 1, c_{2} = 0, and get y_{2} = 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 c_{2} > 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 r_{1} = r_{2} = r, then y_{1} = e^{rt} and y_{2} = te^{rt} , and the general solution is

**Example 2. Solve the IVP**

**Answer.This follows easily now**

The ICs give

y(0) = 2 : c_{1 }+ 0 = 2, ⇒ c_{1 }= 2.

y′(t) = (c_{1} + c_{2} t)e^{t} + c_{2 }e^{t }, y′(0) = c_{1} + c_{2} = 1, ⇒ c_{2} = 1 − c_{1} = −1.

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

**Summary:** For ay′′ + by′ + cy = 0, and ar^{2} + br + c = 0 has two roots r_{1}, r_{2} , we have

On reduction of order: This method can be used to ﬁnd a second solution y_{2} if the ﬁrst solution y_{1 }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 y_{2}:

**Method 2.** We will use Abel’s Theorem, and at the same time we will seek a solution of the form y_{1} = vy_{1}.

By Abel’s Theorem, we have ( worked out in M_{1}) W (y_{1}, y_{2}) . Now, seek y_{2} = vy_{1}.

By the deﬁnition of the Wronskian, we have

Note that this is a general formula:

Now putting y_{1} = 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 y_{1 }= t, ﬁnd the general solution.**Answer. **We have

Let y_{2} be the second solution. By Abel’s Theorem, choosing c = 1, we have

(A cheap trick to double check your solution y_{2} 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 equation**ﬁnd y_{2}

**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 y_{1}, y_{2}, linearly independent of each other, and form the general solution

yH = c_{1} y_{1} + c_{2} y_{2}.

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 = c_{1}y_{1} + c_{2}y_{2} + Y .

Find c_{1} , c_{2} 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 .

r^{2} − 3r − 4 = (r + 1)(r − 4) = 0, ⇒ r_{1} = −1, r_{2} = 4, so

yH = c_{1} e^{−t} + c_{2} 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 r_{1} = −1 and y_{1} = 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 r_{1} , r_{2} (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 = At^{2} + 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′ = 3t ^{2} + 2.**

**Answer.**We see that the form we used in the previous example Y = At^{2} + Bt + C won’t work because Y′′ − 3Y′ will not have the term t^{2} .

New try: multiply by a t. So we guess Y = t(At^{2} + B t + C ) = At^{3} + B t^{2 }+ Ct. Then

Y ′ = 3At^{2} + 2B t + C, Y ′′ = 6At + 2B .

Plug them into the equation

(6At + 2B) − 3(3At^{2} + 2Bt + C ) = −9At^{2} + (6A − 6B)t + (2B − 3C ) = 3t^{2} + 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 r^{2} + 1 = 0, so r_{1,2 }= ±i, and yH = c_{1} cos t + c_{2} 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 r_{1} , r_{2}.

Note that case (2) occurs when the equation is y′′ + α^{2} 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 = P_{1} (t)e^{at }, where P_{1} is a polynomial of degree

1. Also we see r_{1 }= −1, r_{2 }= 4, so r_{1} = a and r_{2} = 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 = r_{1}, 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_{2 }+ 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_{n}t^{n} + · · · + α_{1} t + α_{0} is a polynomial of degree n, then the form of a particular solution depends on the roots r_{1}, r_{2} .

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 t_{2} .

We summarize a few cases below.

**Summary 5.** If g(t) = e^{αt} (a cos β t + b sin β t), and r_{1}, r_{2} 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 r_{1}, r_{2} 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) = g_{1} (t) + g_{2} (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 = Y_{1} + Y_{2} + · · · 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.**Since r

Answer.

**Example 10. y′′ + 16y = sin 4t + cos t − 4 cos 4t + 4. Answer**.The char equation is r

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} + t^{2} − 3.**The char equation is r

Answer.

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