### Diffrential Equations

In this post, I'll introduce about Diffrential Equations.

First-order differential
equations
Reference: Boyce and DiPrima, Chapter 2
The general first-order differential equation for the function 𝑦 = 𝑦(𝑥) is written
as
𝑑𝑦
𝑑𝑥 = 𝑓(𝑥, 𝑦), (2.1)
where 𝑓(𝑥, 𝑦) can be any function of the independent variable 𝑥 and the dependent
variable 𝑦. We first show how to determine a numerical solution of this
equation, and then learn techniques for solving analytically some special forms
of (2.1), namely, separable and linear first-order equations.
2.1 The Euler method
view tutorial
Although it is not always possible to find an analytical solution of (2.1) for
𝑦 = 𝑦(𝑥), it is always possible to determine a unique numerical solution given
an initial value 𝑦(𝑥0) = 𝑦0, and provided 𝑓(𝑥, 𝑦) is a well-behaved function.
The differential equation (2.1) gives us the slope 𝑓(𝑥0, 𝑦0) of the tangent line
to the solution curve 𝑦 = 𝑦(𝑥) at the point (𝑥0, 𝑦0). With a small step size
Δ𝑥, the initial condition (𝑥0, 𝑦0) can be marched forward in the x-coordinate
to 𝑥 = 𝑥0 + Δ𝑥, and along the tangent line using Euler’s method to obtain the
y-coordinate
𝑦(𝑥0 + Δ𝑥) = 𝑦(𝑥0) + Δ𝑥𝑓(𝑥0, 𝑦0).
This solution (𝑥0 + Δ𝑥, 𝑦0 + Δ𝑦) then becomes the new initial condition and is
marched forward in the x-coordinate another Δ𝑥, and along the newly determined
tangent line. For small enough Δ𝑥, the numerical solution converges to
the exact solution.
13
14 CHAPTER 2. FIRST-ORDER ODES
2.2 Separable equations
view tutorial
A first-order ode is separable if it can be written in the form
𝑔(𝑦)
𝑑𝑦
𝑑𝑥 = 𝑓(𝑥), 𝑦(𝑥0) = 𝑦0, (2.2)
where the function 𝑔(𝑦) is independent of 𝑥 and 𝑓(𝑥) is independent of 𝑦. Integration
from 𝑥0 to 𝑥 results in
∫︁ 𝑥
𝑥0
𝑔(𝑦(𝑥))𝑦
′
(𝑥)𝑑𝑥 =
∫︁ 𝑥
𝑥0
𝑓(𝑥)𝑑𝑥.
The integral on the left can be transformed by substituting 𝑢 = 𝑦(𝑥), 𝑑𝑢 =
𝑦
′
(𝑥)𝑑𝑥, and changing the lower and upper limits of integration to 𝑦(𝑥0) = 𝑦0
and 𝑦(𝑥) = 𝑦. Therefore,
∫︁ 𝑦
𝑦0
𝑔(𝑢)𝑑𝑢 =
∫︁ 𝑥
𝑥0
𝑓(𝑥)𝑑𝑥,
and since 𝑢 is a dummy variable of integration, we can write this in the equivalent
form
∫︁ 𝑦
𝑦0
𝑔(𝑦)𝑑𝑦 =
∫︁ 𝑥
𝑥0
𝑓(𝑥)𝑑𝑥. (2.3)
A simpler procedure that also yields (2.3) is to treat 𝑑𝑦/𝑑𝑥 in (2.2) like a fraction.
Multiplying (2.2) by 𝑑𝑥 results in
𝑔(𝑦)𝑑𝑦 = 𝑓(𝑥)𝑑𝑥,
which is a separated equation with all the dependent variables on the left-side,
and all the independent variables on the right-side. Equation (2.3) then results
directly upon integration.
Example: Solve 𝑑𝑦
𝑑𝑥 +
1
2
𝑦 =
3
2
, with 𝑦(0) = 2.
We first manipulate the differential equation to the form
𝑑𝑦
𝑑𝑥 =
1
2
(3 − 𝑦), (2.4)
and then treat 𝑑𝑦/𝑑𝑥 as if it was a fraction to separate variables:
𝑑𝑦
3 − 𝑦
=
1
2
𝑑𝑥.
We integrate the right-side from the initial condition 𝑥 = 0 to 𝑥 and the left-side
from the initial condition 𝑦(0) = 2 to 𝑦. Accordingly,
∫︁ 𝑦
2
𝑑𝑦
3 − 𝑦
=
1
2
∫︁ 𝑥
0
𝑑𝑥. (2.5)
2.2. SEPARABLE EQUATIONS 15
0 1 2 3 4 5 6 7
0
1
2
3
4
5
6
x
y
dy/dx + y/2 = 3/2
Figure 2.1: Solution of the following ode: 𝑑𝑦
𝑑𝑥 +
1
2
𝑦 =
3
2
.
The integrals in (2.5) need to be done. Note that 𝑦(𝑥) < 3 for finite 𝑥 or the
integral on the left-side diverges. Therefore, 3 − 𝑦 > 0 and integration yields
− ln (3 − 𝑦)
]︀𝑦
2
=
1
2
𝑥
]︀𝑥
0
,
ln (3 − 𝑦) = −
1
2
𝑥,
3 − 𝑦 = 𝑒
− 1
2
𝑥
,
𝑦 = 3 − 𝑒
− 1
2
𝑥
.
Since this is our first nontrivial analytical solution, it is prudent to check our
result. We do this by differentiating our solution:
𝑑𝑦
𝑑𝑥 =
1
2
𝑒
− 1
2
𝑥
=
1
2
(3 − 𝑦);
and checking the initial conditions, 𝑦(0) = 3 − 𝑒
0 = 2. Therefore, our solution
satisfies both the original ode and the initial condition.
Example: Solve 𝑑𝑦
𝑑𝑥 +
1
2
𝑦 =
3
2
, with 𝑦(0) = 4.
This is the identical differential equation as before, but with different initial
conditions. We will jump directly to the integration step:
∫︁ 𝑦
4
𝑑𝑦
3 − 𝑦
=
1
2
∫︁ 𝑥
0
𝑑𝑥.
16 CHAPTER 2. FIRST-ORDER ODES
Now 𝑦(𝑥) > 3, so that 𝑦 − 3 > 0 and integration yields
− ln (𝑦 − 3)]︀𝑦
4
=
1
2
𝑥
]︀𝑥
0
,
ln (𝑦 − 3) = −
1
2
𝑥,
𝑦 − 3 = 𝑒
− 1
2
𝑥
,
𝑦 = 3 + 𝑒
− 1
2
𝑥
.
The solution curves for a range of initial conditions is presented in Fig. 2.1.
All solutions have a horizontal asymptote at 𝑦 = 3 at which 𝑑𝑦/𝑑𝑥 = 0. For
𝑦(0) = 𝑦0, the general solution can be shown to be 𝑦(𝑥) = 3+(𝑦0−3) exp(−𝑥/2).
Example: Solve 𝑑𝑦
𝑑𝑥 =
2 cos 2𝑥
3+2𝑦
, with 𝑦(0) = −1. (i) For what values of
𝑥 > 0 does the solution exist? (ii) For what value of 𝑥 > 0 is 𝑦(𝑥)
maximum?
Notice that the solution of the ode may not exist when 𝑦 = −3/2, since 𝑑𝑦/𝑑𝑥 →
∞. We separate variables and integrate from initial conditions:
(3 + 2𝑦)𝑑𝑦 = 2 cos 2𝑥 𝑑𝑥
∫︁ 𝑦
−1
(3 + 2𝑦)𝑑𝑦 = 2 ∫︁ 𝑥
0
cos 2𝑥 𝑑𝑥
3𝑦 + 𝑦
2
]︀𝑦
−1
= sin 2𝑥
]︀𝑥
0
𝑦
2 + 3𝑦 + 2 − sin 2𝑥 = 0
𝑦± =
1
2
[−3 ±
√
1 + 4 sin 2𝑥].
Solving the quadratic equation for 𝑦 has introduced a spurious solution that
does not satisfy the initial conditions. We test:
𝑦±(0) = 1
2
[−3 ± 1] = {︂
-1;
-2.
Only the + root satisfies the initial condition, so that the unique solution to the
ode and initial condition is
𝑦 =
1
2
[−3 + √
1 + 4 sin 2𝑥]. (2.6)
To determine (i) the values of 𝑥 > 0 for which the solution exists, we require
1 + 4 sin 2𝑥 ≥ 0,
or
sin 2𝑥 ≥ −
1
4
. (2.7)
Notice that at 𝑥 = 0, we have sin 2𝑥 = 0; at 𝑥 = 𝜋/4, we have sin 2𝑥 = 1;
at 𝑥 = 𝜋/2, we have sin 2𝑥 = 0; and at 𝑥 = 3𝜋/4, we have sin 2𝑥 = −1 We
therefore need to determine the value of 𝑥 such that sin 2𝑥 = −1/4, with 𝑥 in
the range 𝜋/2 < 𝑥 < 3𝜋/4. The solution to the ode will then exist for all 𝑥
between zero and this value.
2.3. LINEAR EQUATIONS 17
0 0.5 1 1.5
−1.6
−1.4
−1.2
−1
−0.8
−0.6
−0.4
−0.2
0
x
y
(3+2y) dy/dx = 2 cos 2x, y(0) = −1
Figure 2.2: Solution of the following ode: (3 + 2𝑦)𝑦
′ = 2 cos 2𝑥, 𝑦(0) = −1.
To solve sin 2𝑥 = −1/4 for 𝑥 in the interval 𝜋/2 < 𝑥 < 3𝜋/4, one needs to
recall the definition of arcsin, or sin−1
, as found on a typical scientific calculator.
The inverse of the function
𝑓(𝑥) = sin 𝑥, −𝜋/2 ≤ 𝑥 ≤ 𝜋/2
is denoted by arcsin. The first solution with 𝑥 > 0 of the equation sin 2𝑥 = −1/4
places 2𝑥 in the interval (𝜋, 3𝜋/2), so to invert this equation using the arcsine
we need to apply the identity sin (𝜋 − 𝑥) = sin 𝑥, and rewrite sin 2𝑥 = −1/4 as
sin (𝜋 − 2𝑥) = −1/4. The solution of this equation may then be found by taking
the arcsine, and is
𝜋 − 2𝑥 = arcsin (−1/4),
or
𝑥 =
1
2
(︂
𝜋 + arcsin
1
4
)︂
.
Therefore the solution exists for 0 ≤ 𝑥 ≤ (𝜋 + arcsin (1/4)) /2 = 1.6971 . . . ,
where we have used a calculator value (computing in radians) to find arcsin(0.25) =
0.2527 . . . . At the value (𝑥, 𝑦) = (1.6971 . . . , −3/2), the solution curve ends and
𝑑𝑦/𝑑𝑥 becomes infinite.
To determine (ii) the value of 𝑥 at which 𝑦 = 𝑦(𝑥) is maximum, we examine
(2.6) directly. The value of 𝑦 will be maximum when sin 2𝑥 takes its maximum
value over the interval where the solution exists. This will be when 2𝑥 = 𝜋/2,
or 𝑥 = 𝜋/4 = 0.7854 . . . .
The graph of 𝑦 = 𝑦(𝑥) is shown in Fig. 2.2.
2.3 Linear equations
view tutorial
18 CHAPTER 2. FIRST-ORDER ODES
The first-order linear differential equation (linear in 𝑦 and its derivative) can be
written in the form
𝑑𝑦
𝑑𝑥 + 𝑝(𝑥)𝑦 = 𝑔(𝑥), (2.8)
with the initial condition 𝑦(𝑥0) = 𝑦0. Linear first-order equations can be integrated
using an integrating factor 𝜇(𝑥). We multiply (2.8) by 𝜇(𝑥),
𝜇(𝑥)
[︂
𝑑𝑦
𝑑𝑥 + 𝑝(𝑥)𝑦
]︂
= 𝜇(𝑥)𝑔(𝑥), (2.9)
and try to determine 𝜇(𝑥) so that
𝜇(𝑥)
[︂
𝑑𝑦
𝑑𝑥 + 𝑝(𝑥)𝑦
]︂
=
𝑑
𝑑𝑥[𝜇(𝑥)𝑦]. (2.10)
Equation (2.9) then becomes
𝑑
𝑑𝑥[𝜇(𝑥)𝑦] = 𝜇(𝑥)𝑔(𝑥). (2.11)
Equation (2.11) is easily integrated using 𝜇(𝑥0) = 𝜇0 and 𝑦(𝑥0) = 𝑦0:
𝜇(𝑥)𝑦 − 𝜇0𝑦0 =
∫︁ 𝑥
𝑥0
𝜇(𝑥)𝑔(𝑥)𝑑𝑥,
or
𝑦 =
1
𝜇(𝑥)
(︂
𝜇0𝑦0 +
∫︁ 𝑥
𝑥0
𝜇(𝑥)𝑔(𝑥)𝑑𝑥)︂
. (2.12)
It remains to determine 𝜇(𝑥) from (2.10). Differentiating and expanding (2.10)
yields
𝜇
𝑑𝑦
𝑑𝑥 + 𝑝𝜇𝑦 =
𝑑𝜇
𝑑𝑥𝑦 + 𝜇
𝑑𝑦
𝑑𝑥;
and upon simplifying,
𝑑𝜇
𝑑𝑥 = 𝑝𝜇. (2.13)
Equation (2.13) is separable and can be integrated:
∫︁ 𝜇
𝜇0
𝑑𝜇
𝜇
=
∫︁ 𝑥
𝑥0
𝑝(𝑥)𝑑𝑥,
ln 𝜇
𝜇0
=
∫︁ 𝑥
𝑥0
𝑝(𝑥)𝑑𝑥,
𝜇(𝑥) = 𝜇0 exp (︂∫︁ 𝑥
𝑥0
𝑝(𝑥)𝑑𝑥)︂
.
Notice that since 𝜇0 cancels out of (2.12), it is customary to assign 𝜇0 = 1. The
solution to (2.8) satisfying the initial condition 𝑦(𝑥0) = 𝑦0 is then commonly
written as
𝑦 =
1
𝜇(𝑥)
(︂
𝑦0 +
∫︁ 𝑥
𝑥0
𝜇(𝑥)𝑔(𝑥)𝑑𝑥)︂
,
with
𝜇(𝑥) = exp (︂∫︁ 𝑥
𝑥0
𝑝(𝑥)𝑑𝑥)︂
the integrating factor. This important result finds frequent use in applied mathematics.
2.3. LINEAR EQUATIONS 19
Example: Solve 𝑑𝑦
𝑑𝑥 + 2𝑦 = 𝑒
−𝑥
, with 𝑦(0) = 3/4.
Note that this equation is not separable. With 𝑝(𝑥) = 2 and 𝑔(𝑥) = 𝑒
−𝑥
, we
have
𝜇(𝑥) = exp (︂∫︁ 𝑥
0
2𝑑𝑥)︂
= 𝑒
2𝑥
,
and
𝑦 = 𝑒
−2𝑥
(︂
3
4
+
∫︁ 𝑥
0
𝑒
2𝑥
𝑒
−𝑥
𝑑𝑥)︂
= 𝑒
−2𝑥
(︂
3
4
+
∫︁ 𝑥
0
𝑒
𝑥
𝑑𝑥)︂
= 𝑒
−2𝑥
(︂
3
4
+ (𝑒
𝑥 − 1))︂
= 𝑒
−2𝑥
(︂
𝑒
𝑥 −
1
4
)︂
= 𝑒
−𝑥
(︂
1 −
1
4
𝑒
−𝑥
)︂
.
Example: Solve 𝑑𝑦
𝑑𝑥 − 2𝑥𝑦 = 𝑥, with 𝑦(0) = 0.
This equation is separable, and we solve it in two ways. First, using an integrating
factor with 𝑝(𝑥) = −2𝑥 and 𝑔(𝑥) = 𝑥:
𝜇(𝑥) = exp (︂
−2
∫︁ 𝑥
0
𝑥𝑑𝑥)︂
= 𝑒
−𝑥
2
,
and
𝑦 = 𝑒
𝑥
2
∫︁ 𝑥
0
𝑥𝑒−𝑥
2
𝑑𝑥.
The integral can be done by substitution with 𝑢 = 𝑥
2
, 𝑑𝑢 = 2𝑥𝑑𝑥:
∫︁ 𝑥
0
𝑥𝑒−𝑥
2
𝑑𝑥 =
1
2
∫︁ 𝑥
2
0
𝑒
−𝑢
𝑑𝑢
= −
1
2
𝑒
−𝑢
]︀𝑥
2
0
=
1
2
(︁
1 − 𝑒
−𝑥
2
)︁
.
Therefore,
𝑦 =
1
2
𝑒
𝑥
2
(︁
1 − 𝑒
−𝑥
2
)︁
=
1
2
(︁
𝑒
𝑥
2
− 1
)︁
.
20 CHAPTER 2. FIRST-ORDER ODES
Second, we integrate by separating variables:
𝑑𝑦
𝑑𝑥 − 2𝑥𝑦 = 𝑥,
𝑑𝑦
𝑑𝑥 = 𝑥(1 + 2𝑦),
∫︁ 𝑦
0
𝑑𝑦
1 + 2𝑦
=
∫︁ 𝑥
0
𝑥𝑑𝑥,
1
2
ln (1 + 2𝑦) = 1
2
𝑥
2
,
1 + 2𝑦 = 𝑒
𝑥
2
,
𝑦 =
1
2
(︁
𝑒
𝑥
2
− 1
)︁
.
The results from the two different solution methods are the same, and the choice
of method is a personal preference.

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