Second-Order Non-Homogeneous Equations — Theory
1. Standard form
A second-order linear non-homogeneous differential equation with constant coefficients can be written as:
\[
y''+ay'+by=G(x).
\]
The left side contains the unknown function \(y\), its derivatives, and constant coefficients.
The right side \(G(x)\) is the forcing term.
2. Homogeneous equation
First ignore the forcing term and solve the associated homogeneous equation:
\[
y''+ay'+by=0.
\]
Try a solution of the form:
\[
y=e^{rx}.
\]
Substitution gives the characteristic equation:
\[
r^2+ar+b=0.
\]
3. Complementary solution
The complementary solution \(y_h\) depends on the roots of the characteristic equation.
4. General solution
A non-homogeneous equation has solution:
\[
y=y_h+y_p.
\]
Here \(y_h\) solves the homogeneous equation, and \(y_p\) is one particular solution of:
\[
y''+ay'+by=G(x).
\]
5. Method of undetermined coefficients
The method of undetermined coefficients works when \(G(x)\) belongs to a family whose derivatives repeat in a predictable way:
- polynomials, such as \(x^2+1\),
- exponentials, such as \(e^{kx}\),
- sines and cosines, such as \(\sin(kx)\) and \(\cos(kx)\),
- simple products of these families.
The idea is to guess the form of \(y_p\), substitute it into the differential equation, and solve for the unknown coefficients.
7. Resonance
Resonance occurs when the usual particular-solution guess is already part of the complementary solution.
In that case, the guess is not independent and must be multiplied by \(x\).
\[
y_{p,\text{new}}=x^s y_{p,\text{usual}}.
\]
The exponent \(s\) is chosen just large enough to make the guess independent of \(y_h\).
Usually:
\[
s=1
\quad\text{for a simple overlap},
\qquad
s=2
\quad\text{for a repeated-root overlap}.
\]
8. Worked example: resonance
Solve:
\[
y''+4y=\sin(2x).
\]
The associated homogeneous equation is:
\[
y''+4y=0.
\]
Its characteristic equation is:
\[
r^2+4=0.
\]
Therefore:
\[
r=\pm 2i.
\]
So the complementary solution is:
\[
y_h=C_1\cos(2x)+C_2\sin(2x).
\]
The usual guess for \(\sin(2x)\) would be:
\[
y_p=A\cos(2x)+B\sin(2x).
\]
But this overlaps with \(y_h\), so resonance occurs. Multiply by \(x\):
\[
y_p=x\left(A\cos(2x)+B\sin(2x)\right).
\]
Substitution gives:
\[
y_p=-\frac{x}{4}\cos(2x).
\]
Therefore the general solution is:
\[
y=C_1\cos(2x)+C_2\sin(2x)-\frac{x}{4}\cos(2x).
\]
9. Non-resonant example
Consider:
\[
y''+9y=\cos(2x).
\]
The homogeneous roots are:
\[
r=\pm 3i.
\]
Since \(\cos(2x)\) does not match \(\cos(3x)\) or \(\sin(3x)\), there is no resonance.
Use the usual guess:
\[
y_p=A\cos(2x)+B\sin(2x).
\]
10. Verification
After finding \(y_p\), verify it by applying the differential operator:
\[
L[y_p]=y_p''+ay_p'+by_p.
\]
A correct particular solution satisfies:
\[
L[y_p]=G(x).
\]
Then the full solution is:
\[
y=y_h+y_p.
\]
11. Applying initial conditions
If the problem gives:
\[
y(x_0)=y_0,
\qquad
y'(x_0)=v_0,
\]
then substitute \(x=x_0\) into \(y\) and \(y'\) to solve for \(C_1\) and \(C_2\).
13. Common mistakes
- Forgetting to normalize: the coefficient of \(y''\) must be \(1\).
- Solving only the homogeneous equation: the final answer must include \(y_p\).
- Missing resonance: if the guess overlaps with \(y_h\), multiply by \(x^s\).
- Using only sine or only cosine: for trig forcing, the guess usually needs both sine and cosine.
- Dropping polynomial terms: if \(G(x)\) is a polynomial of degree \(n\), guess a full polynomial of degree \(n\).
- Not verifying: always check that \(y_p''+ay_p'+by_p=G(x)\).
- Forgetting constants: \(C_1\) and \(C_2\) stay in the general solution unless initial conditions are applied.
- Ignoring axis units: graph axes should show both numbered values and units.
