Forced Second-Order ODEs: Trial Choice and Overlap

Start with the equation type

The method below applies to a linear equation with constant coefficients,

$$y''+ay'+by=f(t),$$

where $a$ and $b$ are real constants. If the coefficient of $y''$ is a nonzero constant other than 1, divide every term by that constant first. The associated homogeneous equation has the same left-hand side and a zero right-hand side.

Write a solution as

$$y=y_h+y_p.$$

Here $y_h$ is the general homogeneous solution, including its two arbitrary constants, and $y_p$ is one particular solution of the forced equation. Linearity makes the sum work: the left-hand side applied to $y_h$ gives zero, while the same operation applied to $y_p$ gives $f(t)$. Two different particular solutions can describe the same full solution family if their difference is homogeneous.

Initial conditions belong to the complete solution. A particular solution does not usually satisfy them by itself.

If your equation is first-order and linear, start with the integrating-factor lesson. If its coefficients depend on $t$, or a term such as $y^2$ or $(y')^2$ makes it nonlinear, this constant-coefficient recipe does not apply unchanged.

Choose the whole trial family

Undetermined coefficients works when the forcing is a finite sum of polynomials, exponentials, sines and cosines, or products of those forms. Differentiating these functions stays within a finite family, so a finite set of unknown coefficients can be enough.

• For a polynomial of degree $n$, include every power from $t^n$ down to a constant, even if the forcing omits a power.
• For $e^{\alpha t}$ times a polynomial, multiply the polynomial trial by $e^{\alpha t}$.
• For a sine or cosine at frequency $\beta$, include both sine and cosine. Differentiation can exchange them.
• For $e^{\alpha t}$ times a degree-$n$ polynomial times sine or cosine, use $$\begin{aligned}e^{\alpha t}\bigl[&P_n(t)\cos(\beta t)\\&+Q_n(t)\sin(\beta t)\bigr]\end{aligned}$$. Here $P_n$ and $Q_n$ each contain all powers from $t^n$ down to a constant, with separate unknown coefficients; some can turn out to be zero. Apply the overlap check below to the whole expression.
• For sums, find a particular solution for each forcing term and add them. This step uses linearity.

Before calculating coefficients, compare the trial with the homogeneous solution. If the trial contains terms already in $y_h$, the differential operator will cancel those terms. Multiplying by a new power of $t$ removes that overlap.

In this notation, $\alpha$ is a real exponential rate, $\beta$ is a real trigonometric frequency, and $i$ is the imaginary unit with $i^2=-1$. For a polynomial or a sine/cosine without an exponential factor, take $\alpha =0$. The characteristic polynomial is $p(r)=r^2+ar+b$; $r$ is an auxiliary variable. A root has multiplicity $s$ when its factor $(r-r_0)$ occurs exactly $s$ times in that polynomial.

For an exponential-polynomial trial, multiply the entire trial by $t^s$, where $s$ is the multiplicity of $\alpha$ in the characteristic polynomial. For an exponential-sinusoidal trial with $\beta \ne 0$, use the multiplicity of the complex root $\alpha +i\beta$. Use $s=0$ when there is no matching root. For a real second-order equation, an exponential root can have multiplicity 0, 1 or 2.

This is an overlap check, not a rule to multiply every guess by $t$. Determine $s$ from the equation you actually have.

Worked example: one overlapping exponential

Solve

$$\begin{gathered}y''+3y'+2y=e^{-t},\\ y(0)=0,\quad y'(0)=0.\end{gathered}$$

Find the homogeneous family

The characteristic equation factors as

$$r^2+3r+2=(r+1)(r+2)=0.$$

Therefore

$$y_h=C_1e^{-t}+C_2e^{-2t}.$$

Test the tempting guess

A first guess $y_p=Ae^{-t}$ fails for every $A$. Substitution gives

$$\begin{aligned}y_p''+3y_p'+2y_p\\=(1-3+2)Ae^{-t}\\=0.\end{aligned}$$

No choice of $A$ can turn zero into $e^{-t}$. The problem is the trial family, not an algebra error.

The root $-1$ occurs once, so try

$$y_p=At e^{-t}.$$

Its derivatives are

$$\begin{gathered}y_p'=A(1-t)e^{-t},\\ y_p''=A(t-2)e^{-t}.\end{gathered}$$

Substitution gives

$$y_p''+3y_p'+2y_p=Ae^{-t}.$$

Hence $A=1$, and one particular solution is $y_p=te^{-t}$.

Apply both conditions to the full solution

Now

$$\begin{aligned}y={}&C_1e^{-t}+C_2e^{-2t}\\&+te^{-t}.\end{aligned}$$

At $t=0$, the two conditions give

$$\begin{gathered}C_1+C_2=0,\\ -C_1-2C_2+1=0.\end{gathered}$$

Thus $C_1=-1$ and $C_2=1$:

$$\boxed{y=(t-1)e^{-t}+e^{-2t}.}$$

The particular term contributes 1 to the initial derivative. Omitting it while solving for the constants imposes the wrong condition.

Check the equation and the conditions separately

The derivatives are

$$y'=(2-t)e^{-t}-2e^{-2t},$$

$$y''=(t-3)e^{-t}+4e^{-2t}.$$

In $y''+3y'+2y$, the $e^{-2t}$ terms cancel and the coefficient of $e^{-t}$ is

$$(t-3)+3(2-t)+2(t-1)=1.$$

The left-hand side is exactly $e^{-t}$, as required. Equivalently, the residual $y''+3y'+2y-e^{-t}$ is zero. Also $y(0)=0$ and $y'(0)=0$. Passing the differential-equation check alone would not establish the initial conditions.

Although the answer contains $te^{-t}$, it tends to zero as $t$ tends to positive infinity. An extra factor of $t$ does not automatically mean a growing physical oscillation.

When one factor of $t$ is still not enough

Consider

$$y''+4y'+4y=e^{-2t}.$$

The associated homogeneous family is $(C_1+C_2t)e^{-2t}$. Both $Ae^{-2t}$ and $Ate^{-2t}$ are already in it. The characteristic root $-2$ occurs twice, so choose

$$y_p=At^2e^{-2t}.$$

A short substitution explains the power. Set $y=e^{-2t}u$. Differentiating and combining the terms gives

$$y''+4y'+4y=e^{-2t}u''.$$

For $u=At^2$, $u''=2A$. Matching the forcing gives $A=1/2$. The full family is therefore

$$\begin{aligned}y={}&(C_1+C_2t)e^{-2t}\\&+\tfrac12t^2e^{-2t}.\end{aligned}$$

This example has no stated initial conditions, so the two constants remain arbitrary. The homogeneous repeated-root reasoning is developed in the ENGR 213 worked example.

Sinusoidal forcing and an ideal resonance

For

$$y''+4y=\cos(2t),$$

the homogeneous family is $C_1\cos (2t)+C_2\sin (2t)$. Both sinusoidal trial terms overlap, so use

$$y_p=t[A\cos(2t)+B\sin(2t)].$$

Substitution gives

$$\begin{aligned}y_p''+4y_p={}&-4A\sin(2t)\\&+4B\cos(2t).\end{aligned}$$

Match both coefficients: $A=0$ and $B=1/4$. One particular solution is

$$y_p=\frac{t}{4}\sin(2t).$$

For $t \ge 0$, the factor $t$ gives an increasing envelope $t/4$. In the ideal undamped oscillator represented by this normalized equation, the driving frequency equals the natural frequency, producing resonance. This is a property of this model; damping or a different driving frequency changes the response. It is distinct from the decaying exponential-overlap example above.

Even when the forcing contains only cosine, start with both sine and cosine. After substitution, one coefficient may be zero. Deciding that in advance can discard the term that is actually needed.

Know when to choose another method

The finite trial recipe does not cover every forcing term. For example, $\ln (t)$ and $e^{t^2}$ do not remain inside the polynomial-exponential-sinusoidal families above under repeated differentiation. Variation of parameters can be an alternative for a linear equation once a homogeneous basis is known. Work on an interval where the normalized coefficients and forcing are defined and continuous.

A variable-coefficient equation also needs a separate method choice. Do not form a constant-coefficient characteristic equation by treating $t$ as a fixed number.

For a course assessment, use the methods and topics assigned by your instructor. This lesson teaches a selected method, not a complete ENGR 213 review or a predicted exam.

Practise the decision before doing the algebra

Try the six guided questions on forced second-order equations. They change the forcing, overlap and initial conditions. Explain the trial before calculating, then check both the equation and the data.

If you want help applying this to your course, bring the equation, your proposed trial and the line where the calculation stopped. A tutor can check whether the obstacle is method choice, overlap, differentiation or the initial conditions. See ENGR 213 tutoring for coverage confirmation and booking.