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 + Δ1/2)/2a
r2 = (- b - Δ1/2)/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 + Δ1/2)/2a
r2 = (- b - Δ1/2)/2a
r = - b/2a
α = - b/2a 
β = (- Δ)1/2/2a



Example:
Simple harmonix 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:
yc(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:
yp(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 yp of the 
equation (4.3). Then add it to the soltion of the complementary 
equation ycf. The sum y = yc(x) + yp(x) 
is the searched solution for the general equation (4.1)