Differential Equations #4: Homogeneous Second-Order Differential Equations

Solving y'' + A*y' + B*y = 0
(A and B are assumed to be real numbers)

There are two characteristics for this type of equation:

(1) Second-order: The second derivative of y, (y'') is involved, and
(2) Homogeneous: The equation is equal to 0.

We also have constant coefficients A and B. Here is a way to tackle this type of equation:

1. Turn the equation into a characteristic equation. Substitute the following: y'' = λ^2, y' = λ, and y = 1. As a result we have

λ^2 + A * λ + B = 0

2. Solve for λ. Let λ1 and λ2 be the roots of the polynomial.

λ = (-A ± √(A^2 - 4*B))/2 where
λ1 = (-A + √(A^2 - 4*B))/2
λ2 = (-A - √(A^2 - 4*B))/2

3. The nature of λ1 and λ2 determines the nature of solution.

The root is a real double root. λ1 = λ2
The solution takes the form: y = C1 * e^(λ1 * x) + x * C2 * e^(λ1*x)

The roots are distinct and real.
The solution takes the form: y = C1 * e^(λ1*x) + C2 * e^(λ2*x)

The roots are complex, in the form of λ = S ± T*i
The solution takes the form:
y = C1 * e^(S*x) cos(T*x) + C2 * e^(S*x) * sin(T*x)

On initial value problems, you are given y(x0) = y0 and y'(x1) = y1. You can solve for C1 and C2.

There are also boundary value problems, in where you are given y(x0) = y0 and y(x1) = y1. Approach this like solving initial value problems, solve for y first, then solve for C1 and C2.

1. y" - 3*y' - 10*y = 0

The characteristic equation is:

λ^2 - 3*λ - 10 = 0

The roots are λ = 5 and λ = 2. The solution is

y = C1*e^(5*x) + C2*e^(-2*x)

2. y" - 6*y' + 9*y = 0

The characteristic equation is:
λ^2 - 6*λ + 9 = 0

Where the roots are λ = 3 and λ = 3 - a double root. Our solution is:

y = C1*e^(3*x) + C2*x*e^(3*x)

3. y" - 2*y' + 5*y = 0

The characteristic equation is:
λ^2 - 2*λ + 5 = 0

Where the roots are λ = 1 ± 2*i. With complex roots to the characteristic equation, the solution takes the form of:

y = C1*e^x*cos(2*x) + C2*e^x*sin(2*x)

The next time in the series, which will be in a week or so, we will spend several entries regarding the Laplace Transforms and how they can assist in solving differential equations.

Have a great day! Eddie


This blog is property of Edward Shore. 2013