# Math 217: Differential Equations Final Review ```Math 217: Differential Equations
Final Review
Mark Pedigo
1
First-Order DEs
Background Info
Definition 1.1 (Differential Equation). A differential equation (abbreviated DE) is an equation relating an unknown function and its derivatives.
Note: The solution is not a number, as in algebra, but a function.
Methods of Solution
Separable Equations
• Solution Strategy: Put all y terms on one side of the equation, all x terms on the other
and integrate.
Example 1.2. Solve the DE:
dy
dx
=
y 2 −1
;
x
y(1) = 2.
2
= y x−1 implies
Z
Z
1
1
dx =
dy
x
y2 − 1
Z
Z
1
1
1
1
=−
dy +
dy (by partial fractions).
2
y+1
2
y−1
Cx2 +1
1
Thus, 2 ln x + C = ln y−1
, so Cx2 = y−1
. Consequently, y = 1−Cx
2.
y+1
y+1
Solution.
dy
dx
Now, use the initial conditions to find C. We’re given y(1) = 2, so 2 =
consequently C = 13 . Therefore,
y=
1
1/3 x2 + 1
3 + x2
=
.
1 − 1/3 x2
3 − x2
Don’t worry too much about absolute value – the sign washes out in C.
1
(1)
(2)
C+1
1−C
and
Linear First-Order Equations
y 0 + P (x)y = Q(x)
• Solution Strategy: Reduce to separable equation by multiplying through by an integrating factor.
• Procedure
1. Calculate the integrating factor
R
ρ(x) = e
P (x) dx
2. Multiply both sides of the DE by ρ(x)
dy
+ P (x)y = ρ(x)Q(x)
ρ(x)
dx
d
3. The left-hand side should equal dx
[ρ(x)y]. (Mentally check this to make sure you
haven’t erred.)
Then equation is now
d
[ρ(x)y] = ρ(x)Q(x)
dx
4. Integrate both sides with respect to x.
Z
ρ(x)y = ρ(x)Q(x) dx.
Now, solve for y.
Example 1.3. Solve the DE: y 0 + 2xy = x, y(0) = 2.
Solution. The steps to find a solution are as follows.
1. Find the integrating factor: ρ(x) = e
R
2x dx
= ex
2
2
2
2. Multiply both sides by the integrating factor: ex (y 0 + 2xy) = ex &middot; x
h
i
h
i
2
2
2
2
d
d
3. By the product rule, dx
ex y = ex (y 0 + 2xy). Therefore, dx
ex y = xex
4. Integrate both sides:
x2
e y=
2
Therefore, y = e−x
1 x2
e
2
Z
2
xex dx (Substitute u = x2 )
1 2
= ex + C.
2
2
+ C = 12 + Ce−x .
2
Substitutions
• Solution Strategy: Use an appropriate substitution to reduce the DE to one we know
how to solve.
1. Homogeneous First-Order Linear DE
dy
= F xy
dx
dy
vx, dx
=v+
• A first-order linear DE that can take the form
y
x
• Solution Strategy: Substitute v =
(so y =
.
dv
x dx
).
dy
Example 1.4. Solve: 2xy dx
= 4x2 + 3y 2 .
Solution. Dividing through by 2xy yields
dy
x 3 y
=2&middot; + &middot; .
dx
y 2 x
Let v = xy . Then y = vx and
yields
dy
dx
= v+x
v+x
Solving for x
dv
dx
dv
.
dx
(3)
Plugging the substitutions into (3)
1 3
dv
=2&middot; + v
dx
v 2
gives us
x
dv
2 3
= + v−v
dx
v 2
2 1
= + v
v 2
4 + v2
=
.
2v
This equation is separable. (That was our goal.) We separate variables and
integrate both sides:
Z
Z
1
2v
dx =
dv.
2
x
v +4
This implies ln x + C = ln(v 2 + 4), so Cx = v 2 + 4. Finally, we substitute
in for v:
y2
Cx = 2 + 4.
x
Thus, a solution (given in implicit form) is
y 2 + 4x2 = Cx3 .
2. Bernoulli Equation
3
y
x
back
•
dy
dx
+ P (x)y = Q(x)y n .
• Solution Strategy: Use a substitution to reduce the Bernoulli DE to a first-order
linear equation in v. Solve this, back-substitute y.
• Procedure: Substitute v = y 1−n (so y = v 1/(1−n) ,
Example 1.5. Solve:
dy
dx
−
3
y
2x
=
dy
dx
=
dy
dv
&middot;
dv
).
dx
2x
.
y
dy
3
− 2x
y = 2xy −1 . Then n = −1, so we let
Solution. Rewrite the given DE as dx
dy
dy
dv
dv
v = y 1−(−1) = y 2 . Then y = v 1/2 and dx = dv &middot; dx
= 12 v −1/2 dx
. Substituting these
expressions into our DE yields
1 −1/2 dv
3
v
− v 1/2 = 2xv −1/2 .
2
dx 2x
Multiplying through by 2v 1/2 yields
3
dv
− v = 4x,
dx x
a first-order linear DE in v. Solving this equation using an appropriate integrating
factor yields
v = −4x2 + Cx3 .
Finally, substituting v = y 2 back into the equation gives us the solution (in
implicit form)
y 2 = −4x2 + Cx3 .
3. Reducible Second-Order DEs
• Reduce to first-order linear DE
There are two special cases of second-order DEs which can be reduced to first-order
DEs.
(a) y missing from DE
• Substitute p = y 0 =
dy
dx
(so y 00 =
dp
).
dx
Example 1.6. Solve: xy 00 + 2y 0 = 6x using reduction of order.
Solution. Employing our substitutions yields
dp
+ 2p = 6x
dx
dp 2
⇒
+ p = 6.
dx x
xy 00 + 2y 0 = 6x ⇒ x
4
This is a first-order linear DE in p, which is readily solved using an appropriate
dy
dy
, so dx
= y0 =
integrating factor to give us p = 2 + C1 x−2 . But p = dx
2x + C1 x−2 . Integrating both sides with respect to x gives the solution y =
x2 + Cx1 + C2 .
(b) x missing from DE
• Procedure: Substitute p = y 0 (so y 00 =
dp
dx
=
dp
dy
&middot;
dy
dx
=p&middot;
dp
).
dy
Example 1.7. Solve: yy 00 = (y 0 )2 using reduction of order.
dp
. Substituting these expressions into our
Solution. Let p = y 0 . Then y 00 = p dy
dp
DE yields yp dy = p2 , which is a separable equation in p with the solution
dy
dy
p = C1 y. But p = dx
, so dx
= C1 y, which is also separable, and has the
Bx
solution y = Ae .
Exact Equations
• M dx + N dy = 0, where My = Nx .
• This means there exists a function F such that F =
R
M dx =
R
N dy = C.
Example 1.8. Solve: (4x − y) dx + (6y − x) dy = 0.
Solution. Applying the exactness test, we see that My = Nx = −1, so the equation is exact.
Thus,
Z
Z
F = M dx = (4x − y) dx = 2x2 − xy + g(y) = C
and
Z
F =
Z
N dy =
(6y − x) dy = 3y 2 − xy + f (x) = C.
Comparing these two expressions for F , we get
2x2 − xy + 3y 2 = C.
2
Linear Equations of Higher Order
Background Info
Definition 2.1 (Complementary function). Let y1 , y2 , . . . , yn be n solutions of the nth-order
homogeneous equation. The complementary solution is yc = c1 y1 + c2 y2 + &middot; &middot; &middot; + cn yn .
• The solution to a homogenous DE is its complementary solution.
5
• The solution to a nonhomogenous DE (its “general solution”) is the sum of complementary solution and any particular solution (i.e., solution that satisfies the nonhomogeneous DE).
Definition 2.2 (Linear Independence/Linear Dependence). The functions f1 , f2 , . . . , fn are
said to be linearly dependent on the interval I provided there exist constants c1 , c2 , . . . , cn
not all zero, such that
c1 f 1 + c2 f 2 + &middot; &middot; &middot; cn f n = 0
on I. Otherwise, the functions f1 , f2 , . . . , fn are said to be linearly independent.
Definition 2.3 (Wronskian). The Wronskian of f1 , f2 , . . . , fn is
W =
f1
f10
..
.
(n−1)
f1
f2
f20
..
.
(n−1)
f2
&middot;&middot;&middot;
&middot;&middot;&middot;
&middot;&middot;&middot;
fn
fn0
..
.
.
(n−1)
fn
Theorem 2.4 (Wronskian test for linear dependence/independence).
If W (f1 , f2 , . . . , fn ) = 0, then f1 , f2 , . . . , fn are linearly dependent.
If W (f1 , f2 , . . . , fn ) 6= 0, then f1 , f2 , . . . , fn are linearly independent.
Example 2.5. Show that y1 = e−3 , y2 = cos(2x) and y3 = sin(2x) are linearly independent
using the Wronskian.
Solution.
e−3x
cos(2x)
sin(2x)
W = −3e−3x −2 sin(2x) 2 cos(2x)
9e−3x −4 cos(2x) −4 sin(2x)
= 26e−3x
6= 0.
Methods of Solutions
Homogeneous DE: Characteristic Equation
• Assume the solution of the DE is of the form y = erx . Now, find r.
There are three types of roots we need to consider.
1. The roots are real and distinct. If the roots of the characteristic equation, r1 , r2 , . . . , rn ,
are real and distinct, then the general solution of the DE is
y(x) = c1 er1 x + c2 er2 x + &middot; &middot; &middot; + cn ern x .
6
Example 2.6. Solve the IVP: y (3) + 3y 00 − 10y 0 = 0; y(0) = 7, y 0 (0) = 0, y 00 (0) = 70.
Solution. The characteristic equation is 0 = r3 + 3r2 − 10r = r(r + 5)(r − 2). Thus,
the roots are r = 0, −5, 2, and so the general solution of the DE is
y(x) = c1 + c2 e−5x + c3 e2x .
Solve for c1 , c2 and c3 using the ICs: c1 = 0, c2 = 2, c3 = 5. Thus,
y(x) = 2e−5x + 5e2x .
2. The roots are complex and distinct. If the characteristic equation has an unrepeated
pair of complex conjugate roots a &plusmn; bi (b 6= 0), then the corresponding part of the
general solution is
eax (c1 cos(bx) + c2 sin(bx)).
Example 2.7. Solve: y 00 − 4y + 5y = 0.
Solution. The roots of the characteristic equation 0 = r2 − 4r + 5 are r = 2 &plusmn; i. Thus,
y(x) = e2x (c1 cos x + c2 sin x).
3. The roots are repeated. If the characteristic equation has a repeated root r of multiplicity k, then the part of the general solution corresponding to r is of the form
(c1 + c2 x + &middot; &middot; &middot; + ck xk−1 )erx .
Example 2.8. Solve: 9y (5) − 6y (4) + y (3) = 0.
Solution. The characteristic equation is 0 = 9r5 − 6r4 + r3 = r3 (3r − 1)2 . The roots
are 0 (with multiplicity 3) and 31 (with multiplicity 2). Thus, the solution to the DE is
y(x) = (c1 + c2 x + c3 x2 ) + (c4 + c5 x)e1/3x .
Example 2.9. The roots of a characteristic equation of a certain homogeneous DE are
0, 0, 0, 0, 3, −5, −5, 2 &plusmn; 3i, 2 &plusmn; 3i. What is a general solution of this DE?
Solution. The solution is
y(x) = c1 + c2 x + c3 x2 + c4 x3 + c5 e3x + (c6 + c7 x)e−5x +
e2x [c8 cos(3x) + c9 sin(3x)] + xe2x [c10 cos(3x) + c11 sin(3x)]
7
Method of Undetermined Coefficients
• Method to find the particular solution yp of a nonhomogeneous equation with constant
coefficients.
• For the method of undetermined coefficients to work, the right-hand side must have
finitely many linearly independent derivatives.
• Procedure:
1. Find yc , the complementary solution (i.e., solution of the associated homogeneous
equation) by the characteristic equation, as discussed above.
2. Find a trial solution yp by examining the right-hand side of the DE. Adjust yp to
avoid linear dependence with yc , if necessary.
3. Plug yp and its derivatives into the DE. Solve for the coefficients.
Example 2.10. Find a particular solution of y (3) + y 00 = 3ex + 4x2 .
Solution. The steps of the solution are as follows.
1. Find yc . The characteristic equation is 0 = r3 +r2 = r2 (r +1), which has roots 0, 0, −1.
Therefore, yc = c1 + c2 x + c3 e−x .
2. Find the form of the trial solution yp . By examining the right-hand side of the DE,
our first try would be yp = Aex + (B + Cx + Dx2 ), where Aex is included to cover
the 3ex term and B + Cx + Dx2 is included to cover 4x2 and its derivatives. However,
B + Cx + Dx2 “overlaps” with terms of yc , specifically c1 and c2 x. To ensure linear
independence of the terms of yc and yp , we need to multiply the term B + Cx + Dx2
of yp by x2 . Consequently, the appropriate form of yp is
yp = Aex + (B + Cx + Dx2 )x2
= Aex + Bx2 + Cx3 + Dx4 .
3. Plug yp and its derivatives into the DE and solve for the unknown coefficients:
3ex + 4x2 = yp(3) + yp00
= (Aex + 6C + 24D) + (Aex + 2B + 6Cx + 12Dx2 )
= 2Aex + (2B + 6C) + (6C + 24D)x + 12Dx2 .
Comparing coefficients, we see that 2A = 3, 2B +6C = 0, 6C +24D = 0, and 12D = 4.
Thus, A = 23 , B = 4, C = − 34 and D = 13 . Consequently,
4
1
3
yp = ex + 4x2 − x3 + x4 .
2
3
3
8
Variation of Parameters
• Method to find the particular solution yp of a nonhomogeneous equation with constant
coefficients.
• The right-hand side does not have to have only finitely many linearly independent
derivatives for the method of variation of parameters to work.
• We studied variation of Parameters formula for the nonhomogenous equation y 00 +
P (x)y 0 + Q(x)y = f (x)
• Procedure
1. Find solutions to the associated homogeneous equation: y1 and y2
2. Calculate W (y1 , y2 ), the Wronskian of y1 and y2 .
3. Use the following formula:
Z
Z
y1 f (x)
y2 f (x)
dx y1 +
dx y2 ,
yp = −
W (y1 , y2 )
W (y1 , y2 )
Example 2.11. Find a particular solution of the DE y 00 + 3y 0 + 2y = 4ex using variation of
parameters.
Solution. The steps of the solution are as follows.
1. Find y1 and y2 . The characteristic equation of the associated homogeneous equation
is 0 = r2 + 3r + 2 = (r + 1)(r + 2), which has roots -1 and -2. Thus, y1 = e−x and
y2 = e−2x .
2. The Wronskian is given by
W =
e−x
e−2x
−e−x −2e−2x
= −e−3x
3. The solution of yp is
Z
y1 f (x)
y2 f (x)
dx y1 +
dx y2
yp = −
W (y1 , y2 )
W (y1 , y2 )
Z −2x
Z −x
e
&middot; 4ex
e &middot; 4ex
−x
=−
dx e +
dx e−2x
−3x
−3x
−e
−e
4
= 2ex − ex
3
2 x
= e .
3
Z
9
3
Power Series Methods
• Power series methods work for variable coefficients. (The method of undetermined
coefficients and the method of variation of parameters only work for constant coefficients.)
• Procedure
1. Assume a power series solution of the form y =
P∞
n=0 cn x
n
2. Plug the power series for y and its derivatives into the DE
3. Solve for the coefficients
(a) Find a recurrence relation for the coefficients
(b) Use this recurrence relation to find a closed form for cn
P
n
4. Plug cn back into y = ∞
n=0 cn x . This is your power series solution
5. If possible, write the solution in terms of elementary functions
6. Find the radius of convergence.
P
n
– Given y = ∞
n=0 cn (x − a) , the radius of convergence is is at least the distance
from a to the nearest singularity.
– ρ := limn→∞
cn
cn+1
∗ ρ = 0 ⇒ series diverges for all nonzero x
∗ ρ = ∞ ⇒ series converges for all x
∗ 0 &lt; ρ &lt; ∞ ⇒ series converges for |x| &lt; ρ; diverges for |x| &gt; ρ
Example 3.1. Solve the DE: (x − 3)y 0 + 2y = 0 and find its radius of convergence.
P∞
P
n
0
x−1
. Substituting these into the DE
Solution. Let y = ∞
n=0 cn x . Then y =
n=1 cn nx
10
gives:
0 = (x − 3)y 0 + 2y
= xy 0 − 3y 0 + 2y
∞
∞
∞
X
X
X
=x
cn nxn−1 − 3
cn nxx−1 + 2
cn x n
n=1
=
=
∞
X
n=1
∞
X
n=1
cn nxn − 3
n
cn nx − 3
n=0
∞
X
n=1
∞
X
cn nxn−1 + 2
n=0
∞
X
cn x n
n=0
n
cn+1 (n + 1)x + 2
n=0
∞
X
cn x n
n=0
∞
X
=
[cn n − 3cn+1 (n + 1) + 2cn ]xn
n=0
∞
X
=
[(n + 2)cn − 3cn+1 (n + 1)]xn
n=0
By the identity principle, (n + 2)cn − 3cn+1 (n + 1) = 0, from which we get
cn+1 =
1 n+2
&middot;
&middot; cn , n ≥ 0.
3 n+1
Reindexing for n ≥ 1, we see that
cn =
1 n+1
&middot;
&middot; cn−1 , n ≥ 1.
3
n
c1 = 32 c0
c2 =
3
c
3&middot;2 1
=
3
3&middot;2
2
c
3 0
c3 =
4
c
3&middot;3 2
=
4
3&middot;3
3
c
32 0
..
.
cn =
n+1
c,
3n 0
=
=
3
c
32 0
4
c.
33 0
n≥1
11
Therefore,
y=
∞
X
cn x n
n=0
∞ X
n+1
=
c0 x n
n
3
n=0
!
∞
X
n+1
xn .
= c0
n
3
n=0
From the recurrence relation,
cn
cn+1
=3
n+1
n+2
. Therefore, the radius of convergence is
1+
ρ = 3 lim
n→∞ 1 +
1
n
2
n
= 3.
Consequently, the series converges if |x| &lt; 3 and diverges if |x| &gt; 3.
12
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