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) Secondorder: 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
A blog is that is all about mathematics and calculators, two of my passions in life.
Thursday, August 15, 2013
Differential Equations #4: Homogeneous SecondOrder Differential Equations
Subscribe to:
Post Comments (Atom)
HP 42S and TI60: Dimensions of a Race Track
HP 42S and TI60: Dimensions of a Race Track We have a race track that consists of two rectangular tracks connected by a circular ri...

Casio fx991EX Classwiz Review Casio FX991EX The next incarnation of the fx991 line of Casio calculators is the fx991 EX. ...

HP Prime: Basic CAS Commands for Polynomials and Rational Expressions Define the following variables: poly: a polynomial o...

The Odds of Hitting it Big The number of possible combinations is fairly easy to calculate. You multiply the number symbols each slot has ...
No comments:
Post a Comment