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```Review of Chapter 4
Math 2280
The Chapter 4 exam will focus mostly on higher-order (i.e. second-order and higher) differential equations, generally with constant coefficients. It will help to first review some general theory about these DEs.
Let y(k) denote the kth derivative of y. Given an nth order DE
an y(n) + an−1 y(n−1) + ... + a1 y0 + a0 y = g(x),
(1)
the solution y should be written as yc + y p , where yc is the solution if g(x) = 0, and y p is the particular
solution when g(x) 6= 0. Of course, if the function g(x) is zero, then y p is zero as well.
To find the solution yc to Equation 1, we make the starting assumption that y = emx for some value of m.
Taking derivatives gives that y( n) = mn emx . This allows us to find the auxiliary equation
an mn + an − 1mn−1 + ... + a1 m + a0 = 0,
(2)
where the coefficients ai are exactly the same as in Equation 1. So finding the values of m that solve
Equation 2 allows us to find the terms in the yc portion to the solution of Equation 1. In particular:
• If m is a real root of Equation 2 that is not repeated, then Ci emx is a term in yc .
• If m is a real root of Equation 2 that is repeated k times, then Ci x j emx is a term in yc for all values of
j from 0 to k − 1. (In other words, for each repetition, multiply by a higher power of x.)
• If m = α + iβ is a non-repeated complex root of Equtaion 2, then Ci eαx cos(βx) + Ci+1 eαx sin(βx)
are terms in yc . (The number m = α − iβ will also be a root, but these two terms account for both of
these values of m.)
• If complex roots are repeated k times, then Ci x j eαx cos(βx) +Ci+1 x j eαx sin(βx) appear in yc , where
again the values of j start at 0 and run to k − 1.
Remember that yc should contain n unknown constants, where n is the same as in Equations 1 and 2. This
allows for initial value problems to be solved once the first n − 1 derivatives at a point are given.
If one solution y1 is known that solves Equation 1, then another solution can be found by setting y2 = uy1
and substituting into Equation 1. All of the terms that include a u will factor out with an equation that is
precisely y1 plugged into Equation 1. As a result, an order (n − 1) differential equation in u can be solved
to evaluate uy1 after making the substitution w = u0 . In the case where n = 2, the explicit formula for this
is
Z − R P(x) dx
e
dx,
y2 = y1
y21
when Equation 1 is written as y00 + P(x)y0 + Q(x)y = 0. (Note that this also works in the case where the
coefficients of y are non-constant.)
To find y p , there are two main techniques: Undetermined Coefficients and Variation of Parameters. These
techniques are only needed when g(x) 6= 0. In both cases, the complementary solution yc should be found
first, as parts of these techniques depend on knowing the form of yc .
Undetermined Coefficients can only be used when g(x) consists of polynomials, exponential (base e)
functions, and the trig functions sine and cosine. The approach is similar to partial fractions. For each
term appearing in g(x), take derivatives until no additional functions appear. Then y p is made up of a
sum of those functions with coefficients that are solved after plugging y p in to Equation 1 and solving the
resulting system. In the event that a term in y p also appears in yc , it should be multiplied by x just as before.
Variation of Parameters is a bit more robust, but also a bit more complicated. Here we assume that the
particular solution takes the form
y p = u1 y1 + u2 y2 + ... + un yn
where the yi are all solutions from yc . Then finding all of the individual ui is done by use of the formula
u0i =
Wi
,
W
where W denotes the determinant of the Wronskian matrix of y1 , y2 , ...yn , and Wi is the determinant of the
matrix W with column i replaced by a column of zeros, with g(x) as the last entry. (Remember that the
idea here comes from Cramer’s Rule.) In particular, with n = 2 we have
0
g(x)
u01 =
y1
y01
y2
y02
−y2 g(x)
,
=
W (y1 , y2 )
y2
y02
y1 0
y01 g(x)
y1 g(x)
0
u1 =
.
=
W (y1 , y2 )
y1 y2
y01 y02
With n = 3,
0
0
g(x)
u01 =
y1
y01
y001
y2
y02
y002
y2
y02
y002
y3
y03
y003
,
y3
y03
y003
y1
y01
y00
u02 = 1
y1
y01
y001
0
0
g(x)
y2
y02
y002
y3
y03
y003
,
y3
y03
y003
y1
y01
y00
u03 = 1
y1
y01
y001
y2
y02
y002
y2
y02
y002
0
0
g(x)
.
y3
y03
y003
Note that these are all solved for u0i , so an integral still needs to be taken to find the actual ui . The +C at
the end of each of these can be taken to be anything, so setting C = 0 is often the choice made.
Lastly, these methods can all be applied to systems of differential equations, wherein we assume we have
functions x = x(t), y = y(t), and sometimes even z = z(t). In these cases, it helps to think of derivatives
as being a “coefficient” D that can be factored out. (So (D + 3)y = Dy + 3y = y0 + 3y.) This allows the
elimination technique from Algebra to be used to reduce to a differential equation of just one function that
can be solved using the above techniques. Then back-substitution can be used to find the other functions.
In each of these cases, both the complementary and particular solutions should be considered.
Other Things:
• Just like on the Chapter 2 exam, integration techniques are vital. Make sure you’re comfortable with
the ones that have been on the homework.
• Some problems have multiple techniques that can work. Any valid technique will receive credit.
Suggested review problems:
1. Solve: y00 + 4y0 − y = 0.
2. Solve: y000 − 4y00 − 5y0 = 0.
3. Solve: y(4) − 7y00 − 18y = 0.
4. Solve: y00 − 2y0 + y = 0, y(0) = 5, y0 (0) = 10.
5. Solve: y000 + 12y00 + 36y0 = 0, y(0) = 0, y0 (0) = 1, y00 (0) = −7.
6. Given that y1 = ln(x) is a solution to xy00 + y0 = 0, find a second solution y2 .
7. Given that y1 = 1 is a solution to (1 − x2 )y00 + 2xy0 = 0, find a second solution y2 .
8. Given that y1 = x sin(ln x) is a solution to x2 y00 − xy0 + 2y = 0, find a second solution y2 .
9. Solve: y00 − y0 = −3
10. Solve: y00 − 16y = 2e4x
11. Solve: y00 − 2y0 + 5y = ex cos(2x)
12. Solve: y(4) − y00 = 4x + 2xe−x
13. Solve: y00 + 4y = cos(2x), y(0) = y0 (0) = 0
14. Solve: y00 + y = cos2 (x)
15. Solve: y00 − 9y =
9x
e3x
16. Solve: y00 − 2y0 + y =
17. Solve:
18. Solve:
19. Solve:
ex
1+x2
dx
dt
dy
dt
= −y + t
= x − t
(D + 1)x + (D − 1)y = 2
3x
+ (D + 2)y = −1
(D − 1)x + (D2 + 1)y = 1
(D2 − 1)x + (D + 1)y = 2
```