Second-Order Homogeneous Differential Equations — Theory
1. Standard form
A homogeneous second-order linear differential equation with constant coefficients has the form:
\[
ay''+by'+cy=0,
\qquad
a\ne 0.
\]
The coefficients \(a\), \(b\), and \(c\) are constants. The word homogeneous means the right side is zero.
2. Trial solution
The main idea is to try an exponential solution:
\[
y=e^{rx}.
\]
Then:
\[
y'=re^{rx},
\qquad
y''=r^2e^{rx}.
\]
Substitute these into \(ay''+by'+cy=0\):
\[
ar^2e^{rx}+bre^{rx}+ce^{rx}=0.
\]
Factor out \(e^{rx}\):
\[
e^{rx}\left(ar^2+br+c\right)=0.
\]
Since \(e^{rx}\ne 0\), the important part is:
\[
ar^2+br+c=0.
\]
3. Characteristic equation
The equation
\[
ar^2+br+c=0
\]
is called the characteristic equation. Its roots determine the form of the solution.
The roots are found using the quadratic formula:
\[
r=\frac{-b\pm\sqrt{b^2-4ac}}{2a}.
\]
4. Discriminant and root classification
The discriminant is:
\[
\Delta=b^2-4ac.
\]
5. Case 1: Two real distinct roots
If the characteristic equation has two different real roots \(r_1\) and \(r_2\), then the general solution is:
\[
y=C_1e^{r_1x}+C_2e^{r_2x}.
\]
Each root creates one exponential mode. The full solution is a linear combination of the two modes.
6. Case 2: Repeated real root
If the characteristic equation has one repeated root \(r\), the solution is not just \(C_1e^{rx}+C_2e^{rx}\),
because those two terms are the same. A second independent solution is needed.
\[
y=(C_1+C_2x)e^{rx}.
\]
This is the case for the sample equation:
\[
y''+4y'+4y=0.
\]
The characteristic equation is:
\[
r^2+4r+4=0.
\]
Factor:
\[
(r+2)^2=0.
\]
So \(r=-2\) is repeated, and the solution is:
\[
y=(C_1+C_2x)e^{-2x}.
\]
7. Case 3: Complex conjugate roots
If the roots are:
\[
r=\alpha\pm i\beta,
\]
then the real-valued general solution is:
\[
y=e^{\alpha x}\left(C_1\cos(\beta x)+C_2\sin(\beta x)\right).
\]
This form describes oscillation. If \(\alpha<0\), the oscillation decays. If \(\alpha>0\), it grows.
8. Applying initial conditions
A general solution contains arbitrary constants \(C_1\) and \(C_2\). To find a particular solution, use two initial conditions:
\[
y(x_0)=y_0,
\qquad
y'(x_0)=v_0.
\]
Substitute these into the general solution and its derivative. This gives two equations for \(C_1\) and \(C_2\).
9. Physical interpretation
Many second-order homogeneous equations model motion, vibration, and damping.
A common form is:
\[
my''+cy'+ky=0.
\]
Here \(m\) behaves like mass, \(c\) like damping, and \(k\) like stiffness.
The root type describes the motion:
- Real distinct roots: overdamped behavior.
- Repeated root: critical damping.
- Complex roots: oscillatory behavior.
10. Worked example: repeated root
Solve:
\[
y''+4y'+4y=0.
\]
The characteristic equation is:
\[
r^2+4r+4=0.
\]
Factor:
\[
(r+2)^2=0.
\]
Therefore:
\[
r=-2.
\]
Because the root is repeated:
\[
y=(C_1+C_2x)e^{-2x}.
\]
12. Common mistakes
- Forgetting that \(a\ne 0\): if \(a=0\), the equation is not second-order.
- Using \(x\) instead of \(r\) in the characteristic equation: the characteristic variable is \(r\).
- Using the wrong repeated-root form: for a repeated root, use \((C_1+C_2x)e^{rx}\).
- Forgetting sine and cosine for complex roots: complex roots produce real sine-cosine solutions.
- Using only one initial condition: a second-order equation needs two conditions to find \(C_1\) and \(C_2\).
- Mixing up \(y\) and \(y'\): \(y(x_0)\) is position/value, while \(y'(x_0)\) is velocity/slope.
- Ignoring graph units: axis tick labels should include numerical values and units when interpreting a model.
