September 12, 2018
APM 346
Justin Ko
Week 1: ODE Review
Introduction
We summarize the main ODEs we will see in this course
1. Separable Equations: An ODE of the form
dy
= f (x)(g(y))−1
dx
is separable. The ODE is solved by integrating
Z
Z
f (x) dx = g(y) dy.
1*. Homogeneous Equations (Almost Separable): An ODE of the form
dy
= f (x, y)
dx
is homogeneous of order n if
f (tx, ty) = tn f (x, y).
The ODE is solved by using the change of variable u = y/x or u = x/y, which reduces the ODE
to a separable equation.
2. First Order Linear Equations: An ODE of the form
dy
+ P (x)y = Q(x)
dx
is a first Rorder linear ODE. The ODE is solved by multiplying both sides by an integrating factor
I(x) = e P (x) dx and integrating both sides.
2*. Bernoulli Equations (Almost Linear): An ODE of the form
dy
+ P (x)y = Q(x)y n
dx
is called a Bernoulli equation. The ODE is solved by using the change of variable z = y 1−n ,
which reduces the ODE to a first order linear equation.
3. Homogeneous Second Order Constant Coefficients: An ODE of the form
ay 00 + by 0 + cy = 0
is a homogeneous second order constant coefficient ODE. The ODE is solved by finding the roots
r1 and r2 of the characteristic polynomial
C(r) = ar2 + br + c = 0
and the general form of the solution is given by


r1 , r2 ∈ R, r1 6= r2
C1 cosh(r1 x) + C2 sinh(r2 x)
.
y(x) = C1 erx + C2 xerx
r1 , r2 ∈ R, r1 = r2 = r


αx
αx
C1 e cos(βx) + C2 e sin(βx) r1 = α + iβ, r2 = α − iβ, β 6= 0
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September 12, 2018
APM 346
Justin Ko
3*. Euler Equations (Almost Constant Coefficient): An ODE of the form
ax2 y 00 + bxy 0 + cy = 0
are called Euler ODEs. The ODE is solved by finding the roots r1 and r2 of the characteristic
polynomial
C(x) = ax(x − 1) + bx + c = 0
and the general form of the solution is given by

r1
r2

r1 , r2 ∈ R, r1 6= r2
C1 x + C2 x
r
r
y(x) = C1 x + C2 log(x)x
.
r1 , r2 ∈ R, r1 = r2 = r


α
α
C1 x cos(β log x) + C2 x sin(β log x) r1 = α + iβ, r2 = α − iβ, β 6= 0
Problems:
Problem 1. Find a general solution of
p
dy
− 1 + y2
=
.
dx
x log x
Solution 1. We first separate the variables
1
1
p
dx = 0, x 6= 0, 1.
dy +
2
x log x
1+y
Integrating, we have
log |y +
p
y 2 + 1| + log | log |x|| = C
or equivalently (we can remove the absolute value by absorbing the signs into C)
p
log |x|(y + y 2 + 1) = C.
Note: The integral with respect to y is a bit tricky. First using trigonometric substitution with
y = tan θ,
Z
Z
p
1
sec2 θ
p
dy =
dθ = ln | sec θ + tan θ| = log |y + y 2 + 1|.
2
sec θ
1+y
Problem 2. Solve the ODE
xey/x + y
dy
=
,
dx
x
y(1) = 0.
Solution 2. We have
xey/x + y
dy
=
dx
x
is a homogeneous equation of order n = 0. Using the substitution y = ux, we have the following
separable equation
du
du
eu
x
+ u = eu + u ⇒
=
.
dx
dx
x
Integrating, we see for x 6= 0
−e−u = log |x| + C ⇒ −e−y/x = log |x| + C.
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September 12, 2018
APM 346
Justin Ko
Plugging in the initial data y(1) = 0, we have
−e0 = 0 + C ⇒ C = −1.
The final particular solution is
e−y/x + log x = 1, e > x > 0.
The upper bound on x can be found by seeing e−y/x = 1 − log x and the right hand side must be
positive.
Problem 3. Find a general solution of
y 0 + 2y =
3 −2x
e
.
4
Solution 3. This is a first order linear equation. We first find the integrating factor
R
I(x) = e
2 dx
= e2x .
Using the integrating factor, we see
Z
3
3
3
e2x y =
dx ⇒ e2x y = x + C ⇒ y = xe−2x + Ce−2x .
4
4
4
Problem 4. Find a general solution of
xy 0 + y = y 2 log x.
Solution 4. We have a Bernoulli equation. Rearranging terms, we see
1 dy
1
+
= log x.
y 2 dx xy
Letting z = y1 , we have
dz
1
log x
− z=−
.
dx x
x
This is a first order linear equation, so first solving for the integrating factor
I(x) = e−
R
1
x
dx
=
1
.
x
Using the integrating factor, we see
Z
1
log x
1
log(x) 1
z=−
dx ⇒
=
+ + C ⇒ 1 = y log x + y + Cxy.
2
x
x
xy
x
x
Problem 5. Find a general solution of
r2 R00 (r) + rR0 (r) −
nπ 2
β
where n > 0 and β > 0.
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R(r) = 0.
September 12, 2018
APM 346
Justin Ko
2
Solution 5. This is an Euler ODE with characteristic equation C(r) = r(r − 1) + r − ( nπ
β ) with roots
nπ
r = ± β . The general solution is of the form
R(r) = Ar
nπ
β
+ Br−
nπ
β
for some coefficients A and B.
Problem 6. Find a general solution of
X 00 (x) − β 2 X(x) = 0
where β > 0.
Solution 6. This is an second order constant coefficient ODE with characteristic equation C(x) =
x2 − β 2 with roots r = ±β, which corresponds to the solution
X(x) = A cosh(βx) + B sinh(βx)
for some coefficients A and B.
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