Differential equations

1. Terminology

Order:

The order of a differential equation is the highest power of derivative in the equation.

Example y" + 1 = 0

Linearity:

A differential equation is linear if each term in this equation has only one order of derivative.

Examples:

y" + 2y = 3 is linear
yy" + 2x = 0 or
(y')² +3 = 0 or
y² + y' = 0
are nonlinear, differntial equations.

Homogeneity:

A differential equation is homogeneous if it doesn't contain terms other than derivatives.

Examples:

y" + 2y = 0 is homogeneous
y" + y + 2x = 0 is nonhomogeneous
(y')² + y = 0 is nonlinear and homogeneous
y² + y' + 2x = -5 is nonlinear and nonhomogeneous.

2. First-order linear differential equation

y' + P(x) y = Q(x)     (2.1)

is the general form of a first-order differential equation.

This equation is linear and nonhomogeneous. With Q(x) = 0, it will become homogeneous.

To solve this kind of equation (2.1), we use the technique of multiplying both sides of the equation by a function called the integrating factor.

Here is how it works:

I(x)(y' + P(x) y) = I(x)Q(x)     (2.2)

But:

I(x)(y' + P(x) y) = I(x)y' + I(x)P(x) y = (I(x)y)'

with I'(x) = I(x) P(x)

Therefore:

I'(x) / I(x) = P(x)

Integrating both sides yields log I(x) = ∫ P(x) dx + C, that is:

I(x) = A exp{∫ P(x)}

(C and A are constants. A is taken equal to 1 because we need here a particular integrating factor)

We have from (2.1) and (2.2)

(I(x)y)' = I(x)Q(x) , then:

I(x)y = ∫ I(x)Q(x) dx + B = ∫ exp{∫ P(x)} Q(x) dx + B

Where B is a constant. Therefore:

y = [∫ exp{∫ P(x)} Q(x) dx + B ]/I(x)

y = [∫ exp{∫ P(x) dx} Q(x) dx + B ]/I(x)

The results:

The first-order linear equation y' + P(x) y = Q(x) has the solutions:

y = [∫ exp{∫ P(x) dx} Q(x) dx + B ]/exp{∫ P(x)}

Example:

y' + 2x²y = x²

We have:

P(x) = 2x²y
Q(x) = x²
I(x) = exp{∫ P(x)} = exp{2x³/3}
∫ exp{∫ P(x)} Q(x) dx + B = ∫ x² exp{2x³/3} dx + B =
(1/2) exp {2x³/3} + B

Then:

y = (1/2) exp {{2x³/3} + B / exp{2x³/3} =
1/2 + Const. exp{- 2x³/3}

3. Second-Order Linear Differential Equations

A second-order linear differential equation has the form:

P(x) y" + Q(x) y' + R(x) y = S(x)     (3.1)

The related homogeneous linear equation is:

P(x) y" + Q(x) y' + R(x) y = 0     (3.2)

We will use the following theorem:

If y₁and y₂ are linearly independent solutions of the homogeneous 2quation (3.2), and P(x) is not null, then the general solution is given by y(x) = c1y₁(x) + c2 y₂(x), where c1 and c2 are arbitrary constants.

Case of P(x), Q(x), and R(x) are constant. (3.2) becomes:

a y" + b y' + c y = 0     (3.3)

We use the fact that the function exp{rx} remains within its derivatives, and we search solutions of the form: y = C exp{rx}. So (3.3) becomes:

a r² + b r + c = 0     (3.4) that is called the related auxiliary equation (or characteristic equation)

The roots of this equation depend on the discriminant: Δ = b² - 4ac.

1. b² - 4ac > 0

The soltions of (3.4) are the following unequal real:

r₁ = (- b + Δ^1/2)/2a
r₂ = (- b - Δ^1/2)/2a

Therefore the soltutions of (3.3) are:
y = C1 exp{r₁x} + C2 exp{r₂x}

2. b² - 4ac = 0

The soltions of (3.4) are the following equal real:

r = r₁ = r₂ = - b/2a

Therefore the soltution of (3.3) is:

y₁ = C1 exp{rx}

We verify that y₂ = x exp{rx} is also solution. y₁ and y₂ are linearly independant (y₂ is not constant multiple of y₂.
Using the theorem, the solution of (3.3) is then:

y = C1 exp{rx} + C2 x exp{rx}

3. b² - 4ac < 0

In this case the roots and of the auxiliary equation are complex numbers:

r₁ = α + i β
r₂ = α - i β

Where

α = - b/2a
β = (- Δ)^1/2/2a

The soltion of the equation (3.3) is:

y = C1 exp{r₁x} + C2 exp{r₂x} = exp{αx} (c1 cos βx + c2 sin βx)

The results:

Equation:
a y" + b y' + c y = 0

Solutions:

b² - 4ac > 0
y = C1 exp{r₁x} + C2 exp{r₂x}

b² - 4ac = 0
y = C1 exp{rx} + C2 x exp{rx}

b² - 4ac < 0
y = C1 exp{r₁x} + C2 exp{r₂x} =
exp{αx} (c1 cos βx + c2 sin βx)

Δ = b² - 4ac
r₁ = (- b + Δ^1/2)/2a
r₂ = (- b - Δ^1/2)/2a
r = - b/2a
α = - b/2a
β = (- Δ)^1/2/2a

Example:

Simple harmonic oscillator:

x"(t) + ω² x(t) = 0
In this case b = 0 then α = 0 and Δ < 0,
the solution is x = c1 cos βt + c2 sin βt


Where β = (4ac)^1/2/2a = 2ω/2 = ω
a = 1 and c = ω²
x = c1 cos ωt + c2 sin ωt

4. Nonhomogeneous Linear Equations

With constant coefficients, the second-order nonhomogeneous linear differential equation takes the form:

ay" + by' + cy = S(x)     (4.1)

where a, b. and c are constants. The related homogeneous equation is:

ay" + by' + cy = 0     (4.2)

It is called the complementary equation

We use the theorem:

The general solution of the nonhomogeneous differential equation is the sum of a particular solution of the nonhomogeneous and the general solution of of the complementary equation.

We have already above solved the complementary equation (sum of the two linearly independent solutions).

There are two methods for solve the nonhomogeneous equation: The method of undetermined coefficients that works only for a restricted class of functions and the method of variation of parameters that works for every function.

1. For the undetermined coefficients method, we idenfy the coefficients of S(x) to those of the left hand.

2. The Method of Variation of Parameters: Let's suppose that we have already solved the complementary equation the solution is: y_c(x) = c1 y₁(x) + c2 y₂(x) where y₁ and y₂ are linearly independent solutions.

Now it remains to find a particular solution for the general nonhomogeneous equation.

We replace the constants c1 and c2 by arbitrary functions v1(x) and v2(x). So:

y_p(x) = v1(x) y₁(x) + v2(x) y₂(x)     (4.3)

Then:

y'_p = v1(x) y'₁(x) + v1'(x) y₁(x)
+ v2(x) y'₂(x) + v2'(x) y₂(x)

v1(x) and v2(x) are arbitrary functions. We can then impose any condition that will simplify the calculations.

Let's write:

v1'(x) y₁(x) + v2'(x) y₂(x) = 0  (4.4)

Then:

y'_p = v1'(x) y'₁(x) + v2'(x) y'₂(x) + v1(x) y"₁(x) + v2(x) y"₂(x)

Substituting in the general differential equation, and according to y₁(x) and y₁(x) are solution of the complementary equation, we get:

a(v1'(x) y'₁(x) + v2'(x) y'₂(x)) = S(x)   (4.5)

The two equations (4.4) and (4.5) give the expressions of v1('x) and v2'(x). After integrating them to v1(x) and v2(x), we will have the particular solution y_p of the equation (4.3). Then add it to the soltion of the complementary equation y_cf. The sum y = y_c(x) + y_p(x) is the searched solution for the general equation (4.1).