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}?
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 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 equationﬁnd 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
Nonhomogeneous equations; method of undetermined coeﬃcients
Want to solve the nonhomogeneous 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 nonhomogeneous 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 righthand 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 nonhomogeneous equation ay′′ + by′ + cy = g(t).
Then, one can multiply it by 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.
Answer.Since r_{1 }= −1, r_{2} = 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 r_{2} + 16 = 0, with roots r_{1,2} = ±4^{i}, and yH = c_{1 }sin 4t + c_{2} 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} + t^{2} − 3.
Answer.The char equation is r_{2} − 2r + 2 = 0 with roots r_{1,2} = 1 ± i. Then, for the term e^{t} cos t we must multiply by t.
Hermite and Laguerre Special Functions  1 Doc  12 pages 
Hermite and Laguerre Special Functions  2 Doc  6 pages 
1. What is a linear ordinary differential equation of first order? 
2. What is a linear ordinary differential equation of second order? 
3. How can linear ordinary differential equations of first order be solved? 
4. What are the applications of linear ordinary differential equations in physics? 
5. Can linear ordinary differential equations of second order be solved analytically? 
Hermite and Laguerre Special Functions  1 Doc  12 pages 
Hermite and Laguerre Special Functions  2 Doc  6 pages 

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