Diff. Eq. Notes
Chapter 1
First-Order Differential Equation
A first order differential equation is an equation which can be written in the form:
dy
= f (x, y)
dx
where, f : (x, y) ↦ R
Here,
x = independent variable
y = dependent variable
The solution to the differential equation is a function φ−which may be implicitly defined−such that:
dφ
= f (x, y)
dx
First-Order Linear ODEs
A first-order ODE of the form:
dy
+ p(x)y = g(x)
dx
Integrating Factors for First-Order Linear ODEs
A function μ(x) is called an integrating factor for,
dy
+ p(x)y = g(x)
dx
iff:
dy
μ(x) [
+ p(x)y] ≡ (μ(x)y)
′
dx
Theorem 1
An integrating factor of:
dy
+ p(x)y = g(x)
dx
is:
μ = exp ∫
p(x) dx
Proof By the definition of the integrating factor:
dy
μ(x) [
+ p(x)y] ≡ (μ(x)y)
′
dx
dy
μ(x)
Comparing:
′
′
+ μ(x)p(x)y ≡ μ(x)y + μ (x)y
dx
′
μ(x)p(x) = μ (x)
′
μ (x)
= p(x)
μ(x)
′
μ (x)
∫
dx = ∫
p(x) dx
ln μ(x) = ∫
p(x) dx
μ(x)
μ(x) = exp ∫
p(x) dx
Note that since μ can be any function, we have defined μ : x ↦ R
+
Theorem 2: General Solution
A general solution to:
dy
+ p(x)y = g(x)
dx
is:
1
y =
[∫
μ(x)g(x) dx + C]
μ(x)
where, μ is an integrating factor for the ODE.
Proof By def,
dy
μ(x) [
+ p(x)y] = (μ(x)y)
′
= μ(x)g(x)
dx
Integrating both sides wrt to x:
′
∫ (μ(x)y ) dx = ∫
μ(x)g(x) dx + C
μ(x)y = ∫
μ(x)g(x) dx + C
1
y =
[∫
μ(x)g(x) dx + C]
μ(x)
The function above will satisfy the ODE for all values of C . And since a first order ODE only has a single degree of
freedom, one C is enough to determine all particular solutions.
Some Useful Identities
For a, b ∈ R ,
+
∫
e
±ax
e
±ax
b
cos bx dx ≡ ±
cos (bx ∓ arctan
)
a
√a 2 + b 2
∫
e
±ax
e
±ax
b
sin bx dx ≡ ±
sin (bx ∓ arctan
)
a
√a 2 + b 2
a sin ωx ± b cos ωx ≡ √ a
2
a cos ωx ± b cos ωx ≡ √ a
2
+ b
2
+ b
2
b
sin (ωx ± arctan
a
)
b
cos (ωx ∓ arctan
)
a
For any a, b ∈ R, we've:
∫
e
ax
bx
n
dx = e
ax
b
(
x
a
n
bn
−
a
2
x
n−1
bn(n − 1)
+
a
3
x
n−2
− ⋯ + (−1)
n−1
Separable First-Order Differential Equation
n
bn!
a
n+1
) = be
ax
(−1)
∑
i=0
a
i+1
n+1
n!
x
(n − i)!
n−i
A Different Notation
We can rewrite any first-order ODE in the form:
dy
M(x, y) + N(x, y)
dx
= 0
Using differentials, we may rewrite the above expression in its differential form:
M(x, y)dx + N(x, y)dy = 0
Separable First-Order Differential Equation
A first-order differential equation is called separable if it is in the form:
M(x)dx + N(y)dy = 0
Theorem 3: General Solution
The general solution of the equation:
M(x)dx + N(y)dy = 0
is:
∫
M(x) dx + ∫
N(y) dy = C
Proof (The proof will not be rigorous) Rewriting the ODE:
dy
M(x) + N(y)
= 0
dx
Integrating both sides wrt to x:
dy
∫
M(x) + N(y)
dx = C
dx
The above expression satisfies the differential equation as well as solves any initial value problem (IVP) related to
the ODE. Now, we shall use the fact:
d
dy
[∫
N(y) dy] = N(y)
dx
dy
⟺
∫
N(y)
dx
dx = ∫
N(y) dy
dx
to get:
∫
M(x) dx + ∫
N(y) dy = C
Interval of Existence
The interval over which a solution to an IVP is unique and well-defined. (The definition here isn't fully rigorous)
We can find the interval of existence by finding an interval in which the derivatives of our solutions are well defined
and that the domain of the solution contains the initial value of the independent variable.
Constant Solution
Consider the ODE:
dy
dx
= x(y − 2)
This is a separable of differential equation. This equation has three solutions depending where your initial
conditions is. One solution is the constant solution that is y ≡ 2 and the other two can be found by separating
variables,
1
dy = xdx
y − 2
1
∫
dy = ∫
x dx
y − 2
x
2
ln |y − 2| =
+ C
2
2
|y − 2| = Ce
x /2
2
y = 2 ± Ce
x /2
This example extends similarly to differential equations such as:
dy
= x(y − 1)(y − 2)(y − 3)
dx
which have constant solutions:
y ≡ 1
y ≡ 2
y ≡ 3