Second Order Non-Homogeneous Differential Equations

You push a child on a swing. Left to itself, the swing would oscillate and gradually slow down — that is the homogeneous equation, and you already know how to solve it. But you are pushing — applying a periodic external force. The swing now does something different: it settles into a steady rhythm that matches your push, superimposed on its natural oscillation.

The equation governing this is

The left side is the familiar homogeneous part (spring, friction). The right side — F\cos(\Omega t) — is the push. This right side is what makes the equation non-homogeneous.

You already know how to handle the left side. This article is about the right side: how to find a solution that accounts for the external forcing.

The structure of the solution

Here is the key structural fact. Suppose you have

and you somehow find one particular solution y_p(x) — any function at all that, when plugged into the left side, produces f(x).

Now take any solution y_h(x) of the homogeneous equation ay'' + by' + cy = 0. Then y_p + y_h is also a solution of the non-homogeneous equation, because

The linearity of differentiation makes the two parts separate cleanly.

The complementary function y_h handles the natural behaviour of the system (oscillation, growth, decay). The particular integral y_p handles the external forcing. Together, they capture everything.

Finding y_h is the subject of the previous article — use the characteristic equation. The new task here is finding y_p. Two methods do this: undetermined coefficients (fast, when it applies) and variation of parameters (always works, but heavier).

Method 1: Undetermined coefficients

The idea is simple: guess the form of y_p based on the form of f(x), with unknown constants, then determine those constants by substituting back into the ODE.

This works when f(x) belongs to a family of functions that reproduce their own form under differentiation — polynomials, exponentials, sines, cosines, and products of these. The table below shows the standard guesses.

There is one important caveat: if your trial y_p is already a solution of the homogeneous equation (i.e., it appears in y_h), it will produce 0 on the left side, not f(x). In that case, multiply the trial by x (or x^2 if the root is repeated). This is called the modification rule.

The method is best shown by example.

When the modification rule kicks in

Consider y'' - 3y' + 2y = 5e^{2x}. The complementary function is the same: y_h = c_1 e^x + c_2 e^{2x}. But now f(x) = 5e^{2x}, and e^{2x} already appears in y_h — because m = 2 is a root of the characteristic equation.

If you naively try y_p = Ae^{2x}, substituting gives A(4 - 6 + 2)e^{2x} = 0 for all A — you get 0, not 5e^{2x}. The guess fails.

The fix: multiply by x. Try y_p = Axe^{2x}. Compute:

Substitute: A(4+4x)e^{2x} - 3A(1+2x)e^{2x} + 2A \cdot x e^{2x} = Ae^{2x}(4 + 4x - 3 - 6x + 2x) = Ae^{2x}(1) = 5e^{2x}.

So A = 5, and y_p = 5xe^{2x}.

The rule extends: if m = 2 were a double root, you would need Ax^2 e^{2x}.

Method 2: Variation of parameters

Undetermined coefficients is fast but limited — it only works when f(x) is a polynomial, exponential, sine, cosine, or a product of these. What if f(x) = \tan x, or f(x) = 1/x, or some other function that does not neatly reproduce itself under differentiation?

Variation of parameters handles any f(x), with no restriction on its form.

The idea: start with the complementary function y_h = c_1 y_1 + c_2 y_2, then let the "constants" c_1, c_2 become functions of x — hence "variation of parameters." You look for a particular solution of the form

where u_1 and u_2 are unknown functions to be determined. Substituting into the ODE and imposing a simplifying condition leads to the formulas:

where W = y_1 y_2' - y_2 y_1' is the Wronskian of y_1 and y_2.

The simplifying condition that makes this work is u_1' y_1 + u_2' y_2 = 0 — this is imposed deliberately so that y_p' does not contain u_1'' or u_2'', which would make the system intractable. With this condition, the two equations for u_1' and u_2' are a simple 2 \times 2 linear system, solved by the Wronskian.

Choosing between the two methods

The choice is straightforward:

• Undetermined coefficients when f(x) is a polynomial, exponential, sine/cosine, or a product of these. It is faster, less error-prone, and gives cleaner algebra.
• Variation of parameters when f(x) is anything else (\sec x, \tan x, \ln x, 1/x, etc.), or when you want a method that works universally and don't mind the integration.

In practice, most textbook problems use undetermined coefficients. Variation of parameters is the safety net — the method that never refuses a problem.

Handling sums of forcing terms

If f(x) = f_1(x) + f_2(x), you can find particular integrals y_{p_1} and y_{p_2} separately for each piece and add them. This works because the equation is linear:

Adding: a(y_{p_1}+y_{p_2})'' + b(y_{p_1}+y_{p_2})' + c(y_{p_1}+y_{p_2}) = f_1(x) + f_2(x).

So for y'' - 3y' + 2y = 4e^{5x} + 3\sin x, you would find the particular integral for 4e^{5x} (as in Example 1: \frac{1}{3}e^{5x}) and then separately for 3\sin x, and add them.

For 3\sin x, try y_p = A\cos x + B\sin x. Substituting gives

Collecting: (A - 3B)\cos x + (3A + B)\sin x = 3\sin x

So A - 3B = 0 and 3A + B = 3. From the first equation A = 3B; substituting into the second: 9B + B = 3, giving B = 3/10 and A = 9/10.

The combined particular integral is \frac{1}{3}e^{5x} + \frac{9}{10}\cos x + \frac{3}{10}\sin x.

Common confusions

• "The particular integral must be unique." It is unique up to homogeneous solutions. If y_p is one particular integral, then y_p + c_1 y_1 + c_2 y_2 is another. But the general solution y_h + y_p is the same in both cases — the extra homogeneous terms get absorbed into c_1 and c_2.

• "I should include the complementary function when testing my trial for undetermined coefficients." The test is simpler: just check whether the trial function satisfies the homogeneous equation. If y_p = Ae^{2x} and e^{2x} is a solution of the homogeneous equation, the trial needs modification. You do not need to consider coefficients.

• "Variation of parameters requires the equation to be in the form y'' + py' + qy = f(x)." True — the standard formulas assume the coefficient of y'' is 1. If it is not, divide through by a first. This is a common source of errors: if the original equation is 2y'' + y' - y = e^x, divide everything by 2 to get y'' + \frac{1}{2}y' - \frac{1}{2}y = \frac{1}{2}e^x before applying the formulas.

• "I forgot the modification rule and got 0 = f(x)." This is the classic sign. If your substitution produces 0 on the left regardless of A, it means your trial is already a homogeneous solution. Multiply by x and try again.

• "For f(x) = \cos x, I only need y_p = A\cos x." You need both: y_p = A\cos x + B\sin x. Differentiation mixes sines and cosines, so both must appear in the trial.

Going deeper

Deriving the variation of parameters formulas

Start with y_p = u_1 y_1 + u_2 y_2. Differentiate:

Impose the condition u_1' y_1 + u_2' y_2 = 0. This is a choice, not a consequence — you are free to impose one condition because you have two unknown functions but only one equation. With this condition:

Differentiate again:

Now substitute y_p, y_p', y_p'' into y'' + py' + qy = f(x) (assuming a = 1):

Regroup:

The bracketed terms vanish because y_1 and y_2 are solutions of the homogeneous equation. What remains is:

Together with the condition u_1' y_1 + u_2' y_2 = 0, you have a 2 \times 2 system in u_1' and u_2':

Solving by Cramer's rule (or elimination):

where W = y_1 y_2' - y_2 y_1' is the Wronskian. These are exactly the formulas used in Example 2.

The operator method (D-operator)

For undetermined coefficients with exponential and polynomial forcing, there is a slick shortcut using the operator D = d/dx. Write the ODE as p(D)y = f(x), where p(D) = aD^2 + bD + c. Then formally,

The symbol 1/p(D) is the inverse operator. It is not a function you can always evaluate directly, but it obeys useful rules:

• \dfrac{1}{p(D)}e^{\alpha x} = \dfrac{e^{\alpha x}}{p(\alpha)} when p(\alpha) \neq 0
• \dfrac{1}{p(D)}\sin\beta x = \dfrac{\sin\beta x}{p(-\beta^2)} (replacing D^2 by -\beta^2 in the polynomial)

These rules reproduce the undetermined coefficients results instantly, without setting up and solving a system. They are widely used in Indian competitive exam preparation.

Where this leads next

With both homogeneous and non-homogeneous second-order equations in hand, you can tackle the physical and geometric problems that motivated differential equations in the first place.

• Applications of DE — growth, decay, cooling, population models, and curves with prescribed slope properties.
• Orthogonal and Isogonal Trajectories — families of curves that cross at specified angles.
• Second Order Homogeneous — the prerequisite: the characteristic equation and the three cases.
• Higher Order Derivatives — the derivatives that appear in these equations.
• Quadratic Equations — the algebra behind every characteristic equation.