Hello it's a me again drifter1! Today we continue with Mathematics to talk about Differential equations that are second-order, linear and have constant coefficients! Next time we will get into other simple types of 2nd-order ODE's.
So, without further do, let's get straight into it!
Before getting into the main topic of this post, we might first want to talk about 2nd-order ODE's in general.
An 2nd-order ODE is of the form:
y" + P(x, y)y' + Q(x, y)y = R(x)
The right side is a function of x, cause if it had any y then it would be a part of Q(x, y) in the left side.
If y=y(x) is not a part of an 2nd-order ODE, then we can solve it as an 1st-order ODE.
By substituting y' = u we get a 1-st order ODE with variable u.
This "new" ODE will be one of the 6 types we covered in my previous posts.
By solving it and finding u we can then find y by finding the integral of u:
y(x) = integral(u(x)dx) + c
y" + 5xy' = x^2 + 3
By setting y' = u we get:
u' + 5xu = x^2 + 3 [linear 1st-order ODE]
Linear second-order ODE's
A linear ODE of any order contains P(x), Q(x) etc. "coefficient functions" of x.
That way a second-order linear ODE is of the form:
y" + P(x)y' + Q(x)y = R(x)
If R(x) = 0 then we get the homogeneous form:
y" + P(x)y' + Q(x)y = 0
This last one is pretty important cause the following statement is true:
If y1(x) and y2(x) are two solutions of this homogeneous OE then:
y0 = c1*y1(x) + c2*y2(x) is the general solution of the homogeneous ODE,
where c1, c2 are real numbers
If the homogeneous is the corresponding homogeneous of an 2nd-order linear ODE (by setting R(x) = 0) then we can find the solution of the given ODE by using:
y = y0 + yp, where y is the general solution of our 2nd-order linear ODE.
As told before y0 is the general solution of our homogeneous.
The other part (yp) is any solution of the given (non-homogeneous) ODE.
After this introduction let's now get into the main topic of this post.
Linear second-order ODE's with const coeffs
When P(x) and Q(x) don't contain x as an variable and are real numbers then we have an linear second-order ODE with constant coefficients.
Note that R(x) is still a function of x (it can be 0 or any real number too tho).
So, the basic form of such an ODE is:
ay" + by' + cy = R(x)
5y" + 3y' -2y = x^2 + e^x
To solve such a ODE we first get the corresponding homogeneous:
ay" + by" + cy = 0
We suppose that y = e^px is a solution of the homogeneous (can be proven).
Doing that we get the characteristic equation:
ap^2 + bp + c = 0
This is a 2nd-order polynomial equation that can be solved very easily.
There are 3 possible outcomes:
- 2 real solutions: p1, p2 => y1 = e^(p1*x) and y2 = e^(p2*x) solutions of the homogeneous.
- 1 double-solution: p => y1 = e^px and y2 = x*e^px (can be proven) solutions of the homogeneous
- 2 complex solutions: p1, p2 => We use the Euler equation (e^iφ = cosφ + i*sinφ) and get y1= e^(i*x) = cosx, y2 = e^(-i*x) = sinx.
y" - 2y' - 3y = 2cos(x) [given ODE]
y" - 2y' - 3y = 0 [corresponding homogeneous ODE]
p^2 - 2p - 3 = 0 [characteristic equation]
p1 = 3 and p2 = -1 are the 2 real solutions and so:
y1 = e^3x and y2 = e^-x are the solutions of the homogeneous.
The general solutions will be: y0 = c1e^3x + c2e^-x
After that we have to find one solution of the given ODE.
There are some Cases that depend on the form of R(x):
1. If R(x) = 0 then yp = 0 and so:
y = y0 = c1*y1 + c2*y2, where c1, c2 reals
2. If R(x) = e^(k*x) [Exponential], where k a real number
Then we have some cases that depend on the solutions p1, p2:
- k != p1, p2 => yp = λ*e^(k*x) and we find λ with substituting in the given ODE.
- k = p1 or k = p2 => yp = λ*x*e^(k*x) and we again find λ with substitution.
- p is double-solution => yp = λ*x^2*e^(k*x) and we again find λ.
3. If R(x) = an*x^n + ... + a1*x + a0 [Polynomial of x], where ai reals
Then we have cases that depend on c:
- c!=0 => yp = bn*x^n + ... + b1*x + b0, bi reals and we substitute to find those coefficients.
- c==0 => yp = x*(bn*x^n + ... + b1*x + b0), bi reals and we again substitute.
4. If R(x) = k*cos(n*x) + λ*sin(n*x) [Trigonometric], where k, l, n are reals
Then we have cases that depend on the solutions:
- cos(nx) and sin(nx) ARE NOT solutions of the homogeneous => yp = s*cos(n*x) + t*sin(n*x), where s, t are reals and we have to find s, t with substitution.
- cos(nx) and sin(nx) ARE solutions of the homogeneous => yp = x*[s*cos(n*x) + t*sin(n*x)], where we find s, t with substitution.
5. If R(x): Combination of the previous cases
R(x) is of the form:
R(x) = R1(x) + R2(x) + ... + Rn(x), where Ri(x) one of the cases 2-4.
We find the solution for each independently.
ay" + by' + cy = R1(x) => yp1
ay" + by' + cy = R2(x) => yp2
ay" + by' + cy = Rn(x) => ypn
Then we sum all of these together:
yp = yp1 + yp2 + ... + ypn
R(x) = e^2x - 3x^2 + 4x - 5 + 7cos(3x) - 8sin(3x) [Combination]
We can split it into:
R1(x) = e^2x [Exponential]
R2(x) = -3x^2 + 4x - 5 [Polynomial]
R3(x) = 7cos(3x) - 8sin(3x) [Trigonometric]
For all those cases the general solution of our given ODE is then:
y = y0 + yp
I will get into practical examples of 2nd-order linear ODE's when we cover the other special forms too (next post).
Previous posts of the series:
Introduction -> Definition and Applications
First-order part(1) -> Separable, homogeneous and exact 1st-order ODE's
First-order part(2) -> Linear, Bernoulli and Riccati first-order ODE's
First-order exercises -> Exercises for all the 1st-order ODE types
And this is actually it!
Next time we will get into linear 2nd-order ODE's that are of other special forms!