# MST210/Unit 1/Book A/Inhomogeneous Second-Order Differential Equations

This section is concerned with finding solutions to *inhomogeneous* second-order differential equations. These are equations of the form

\[a \frac{d^2y}{dx^2} + b \frac{dy}{dx} + cy = f(x)\]

where \(a \neq 0\), and \(f(x)\) is a given continuous real-valued function.

# General Method of Solution

Let

\[a \frac{d^2y}{dx^2} + b \frac{dy}{dx} + cy = f(x)\]

be an inhomogeneous linear constast-coefficient second-order differential equation.

Its *associated homogeneous equation* is

\[a \frac{d^2y}{dx^2} + b \frac{dy}{dx} + cy = 0\]

The *general solution* \(y_c\) of the associated homogeneous equation is known as the *complementary function* for the original equation.

Any *particular solution* \(y_p\) of the original inhomogeneous equation is referred to as a *particular integral* for that equation.

- Exercise 23
The Principle of Superposition states that if we have two particular solutions \(y_{p1}\), \(y_{p2}\), then \(y = k_1 y_{p1} + k_2 y_{p2}\) is also a solution. Hence \(y = y_{p1} - y_{p2}\) is a solution, where \(k_1 = 1\), \(k_2 = -1\).

- Exercise 24
The associated homogeneous equation is \[\frac{d^2 y}{dx^2} + 4y = 0\]

The auxiliary equation for this homogeneous equation is \[\lambda^2 + 4 = 0\]

As seen in Exercise 21, the solutions are a pair of complex conjugate roots, \(\lambda_{1,2} = \pm 2i\). Following Procedure 5, the general solution is

\[y_c = C \cos 2 x + D \sin 2 x\] where \(C\) and \(D\) are arbitrary constants.

A particular solution to this equation is \(y_p = 2\). This can be verified by seeing that \(y_p = 2\), \(y_p' = 0\), \(y_p'' = 0\). Substituting into the original differential equation, we have

\[\frac{d^2 y_p}{dx^2} + 4y_p = 0 + 4 \cdot 2 = 8\]

Therefore, the general solution of the inhomogeneous differential equation is

\[y = C \cos 2x + D \sin 2x + 2\] where \(C\) and \(D\) are arbitrary constants.

The associated homogeneous equation is \[\frac{d^2y}{dx^2} - 3 \frac{dy}{dx} + 2y = 0\]

We solved this homogeneous differential equation in Example 8, and found \[y_c = Ce^{x} + De^{2x}\]

A particular solution to the inhomogeneous differential equation is \[y_p = 3\]

This can be verified, as \(y_p = 3\), \(y_p' = 0\), \(y_p'' = 0\), and thus

\[\begin{array}{lcl} \frac{d^2 y}{dx^2} - 3 \frac{dy}{dx} + 2 y &=& 0 - 3 \cdot 0 + 2 \cdot 3 \\ &=& 6 \end{array}\]

Therefore, the general solution to the inhomogeneous differential equation is \[y = Ce^x + De^{2x} + 3\] where \(C\) and \(D\) are arbitrary constants.

# Finding a Particular Integral by the Method of Undetermined Coefficients

We have seen that we can find the general solution of a second-order inhomogeneous different equation of the form

\[a \frac{d^2y}{dx^2} + b \frac{dy}{dx} + cy = f(x)\]

by first finding the general solution of the associated homogeneous differential equation (\(y_c\)), and a particular solution to the inhomogeneous differential equation. In this section, we will look at ways to find the particular solution, when \(f(x)\) is a polynomial, exponential or sinusoidal function - or a sum of such functions.

In general, we try a function of the form as \(f(x)\) as our *trial solution*, and then solve to find values for variables in this function.

## Polynomial Functions

Consider a trial solution of the form \[f(x) = m_n x^n + m_{n-1} x^{n-1} + \dotsb + m_1 x + m_0\]

This is a polynomial function of degree \(m\), and \(m\) should be chosen to have the same degree as the original function in the inhomogeneous equation.

Next, find the first- and second-derivatives \(f'\) and \(f''\), and substitute in to the inhomogeneous differential equation. Compare coefficients to find suitable values of \(m_n, m_{n-1}, \dotsb, m_0\).

- Exercise 25
Consider the trial solution \[y = p_1 x + p_0\]

We have \(y' = p_1\) and \(y'' = 0\). Substituting into the original inhomogeneous differential equation we have

\[-2 p_1 + 2 (p_1 x + p_0) = 2p_1 x + (2p_0 - 2p_1) = 2x + 3\]

Comparing coefficients of \(x\), we have \(2p_1 = 2\), \(2p_0 - 2p_1 = 3\). Thus \(p_1 = -1\) and \(p_0 = \frac{5}{2}\).

Therefore, we have the particular integral

\[y_p = x + \frac{5}{2}\]

Checking, we have \(y_p' = 1\), \(y_p'' = 0\), and so

\[-2 \cdot 1 + 2(x + \frac{5}{2}) = -2 + 2x + 5 = 2x + 3\]

Consider the trial solution \[y = p_1 x + p_0\]

We have \(y'' = p_1\) and \(y'' = 0\). Substituting into the original inhomogeneous differential equation we have

\[2 p_1 + p_1x + p_0 = p_1 x + (p_0 + 2p_1) = 2x\]

Comparing coefficients of \(x\), we have \(p_1 = 2\) and \(p_0 + 2p_1 = 0\). Therefore, \(p_1 = 2\), \(p_0 = -4\).

Therefore, we have the particular integral

\[y_p = 2x-4\]

Checking, we have \(y_p' = 2\), \(y_p'' = 0\), and so

\[2 \cdot 2 + 2x - 4 = 2 + 2x - 4 = 2x\]

- Exercise 26
Consider the trial solution \[y = p_2 t^2 + p_1 t + p_0\]

We have \(y' = 2 p_2 t + p_1\) and \(y'' = 2 p_2\). Substituting into the original inhomogeneous differential equation we have

\[2 p_2 - (p_2 t^2 + p_1 t + p_0) = -p_2 t^2 - p_1 t + (2 p_2 - p_0) = t^2\]

Comparing coefficients of \(x\), we have \(-p_2 = 1\), \(p_1 = 0\) and \(2 p_2 - p_0 = 0\). Therefore, \(p_2 = -1\), \(p_0 = - 2\).

Therefore, the particular integral is

\[y = -t^2 - 2\]

Checking, we have \(y' = - 2 t\), \(y'' = -2\). Thus

\[-2 + t^2 + 2 = t^2\]

## Exponential Functions

For exponential functions, consider a trial solution of the form \[y(x) = me^{kx}\] matching \(k\) to the coefficient of \(x\) in the original differential equation.

- Exercise 27
Consider the trial solution \[y = pe^{-x}\]

We have \(y' = -pe^{-x}\) and \(y'' = pe^{-x}\). Substituting into the original inhomogeneous differential equation, we have

\[2pe^{-x} - 2 (-pe^{-x}) + pe^{-x} = 2e^{-x}\]

That is

\[5pe^{-x} = 2e^{-x}\] and so \(p = \frac{2}{5}\). Therefore, the particular integral is \[y = \frac{2}{5}e^{-x}\]

## Sinusoidal Functions

For sinusoidal functions, consider a trial solution of the form \[f(x) = m \cos \Omega x + n \sin \Omega x\] where \(\Omega\) is chosen to match the multiple of \(x\) in the original differential equation.

- Exercise 28
Consider the trial solution

\[y = m \cos 3t + n \sin 3t\]

We have \(y' = 3n \cos 3t - 3m \sin 3t\) and \(y'' = -9 m \cos 3t - 9n \sin 3t\).

Substituting into the original inhomogeneous differential equation, we have

\[(-9m \cos 3t - 9n \sin 3t) - (3n \cos3t - 3m\sin 3t) = \cos 3t + \sin 3t\]

That is

\[-(9m + 3n) \cos 3t + (3m -9n) \sin3t = \cos 3t + \sin 3t\]

Equating coefficients of \(\cos 3t\) and \(\sin 3t\)$, we have the pair of simultaneous equations

\[\begin{cases} -9m - 3n = 1 \\ 3m - 9n = 1 \end{cases}\]

We can eliminate \(m\) as

\[\begin{cases} -30n = 4 \\ 3m - 9n = 1 \end{cases}\]

Therefore \(n = \frac{-2}{15}\) and \(m = \frac{1 - 9 \frac{2}{15}}{3} = \frac{-1}{15}\). Thus the particular integral is

\[y = -\frac{1}{15} \cos 3 t - \frac{2}{15} \sin 3 t\]

## Method of Undetermined Coefficients

(See handbook)

- Exercise 29
- \[f(x) = pe^{2x}\]
- \[f(x) = p\cos4x + q\sin 4x\]

- Exercise 30
Following the hint, we know the roots of the auxiliary equation are a pair of complex conjugate roots, so the complementary function is

\[y_c = e^{-x}(C \cos x + D \sin x)\] where \(C\) and \(D\) are arbitrary constants.

As the right-hand side of the inhomogeneous differential equation is a constant, the trial solution is \[f(x) = p\]

Therefore

\[2p = 4\]

Thus \(p = 2\). The particular integral is \[y_p = 2\]

and so the general solution is

\[y = e^{-x}(C \cos x + D \sin x) + 2\]

Following the hint, we know the roots of the auxiliary equation are distinct real roots, so the complementary function is

\[\theta_c = C + De^{-3t}\] where \(C\) and \(D\) are arbitrary constants.

To find the particular integral, we consider as a trial solution \[f(t) = p \cos 2t + q \sin 2t\]

The first- and second-derivatives are \[f'(t) = 2q \cos 2t - 2p \sin 2t\] \[f''(t) = -4p \cos 2t - 4q \sin 2t\]

Substituting into the inhomogeneous differential equation, we have

\[(-4p \cos 2t - 4q \sin 2t) + 3 (2q \cos 2t - 2p \sin 2t) = 13 \cos 2t\]

That is

\[\begin{array}{ll} 13 \cos 2t &= -4p \cos 2t - 4q \sin 2t + 6q \cos 2t - 6p \sin 2t = 13 \cos 2t \\ &= (6q - 4p) \cos 2t - (4q + 6p) \sin 2t \end{array}\]

Comparing coefficients, we have the pair of simultaneous equations

\[\begin{cases} -4p + 6q = 13 \\ -6p - 4q = 0 \end{cases}\]

Substituting \(p = -\frac{2q}{3}\), we have

\[\begin{cases} \frac{8q}{3} + 6q = 13 \\ p = -\frac{2q}{3} \end{cases}\]

Thus \(q = \frac{3}{2}\) and \(p = -1\)

Therefore the general solution is

\[\theta = C + De^{-3t} - \cos 2t + \frac{3}{2} \sin 2t\] where \(C\) and \(D\) are arbitrary constants.

- Exercise 31
Setting \(R = S = Q = 1\) yields the differential equation

\[y'' - y + \frac{1}{2}(l - x) x = 0\]

Rearranging this, we have the inhomogeneous differential equation

\[y'' - y = \frac{1}{2}x^2 - \frac{l}{2}x\]

The associated homogeneous equation is \[y'' - y = 0\] The auxiliary equation for this is \[\lambda^2 - 1 = 0\] which has solutions \[\lambda_{1,2} = \pm 1\]

As we have two distinct real roots, following Procedure 5 we have the complementary function

\[y_c = Ce^{-x} + De^{x}\] where \(C\) and \(D\) are arbitrary constants.

The right-hand side of the inhomogeneous equation is a polynomial function of degree 2 in \(x\), so we consider the trial function

\[f(x) = p_2 x^2 + p_1 x + p_0\]

The first- and second-derivatives are

\[f'(x) = 2p_2 x + p_1\]

\[f''(x) = 2p_2\]

Substituting in to the inhomogeneous equation, we have

\[2p_2 - (p_2 x^2 + p_1 x + p_0) = \frac{1}{2}x^2 - \frac{l}{2}x\]

That is

\[- p_2 x^2 - p_1 x + (2p_2 - p_0) = \frac{1}{2}x^2 - \frac{l}{2}x\]

On comparing coefficients of \(x\), we have

\[\begin{cases} -p_2 = \frac{1}{2} \\ -p_1 = -\frac{l}{2} \\ 2p_2 - p_0 = 0 \end{cases}\]

Therefore \(p_2 = -\frac{1}{2}\), \(p_1 = \frac{l}{2}\), and \(p_0 = -1\).

Thus the particular integral is

\[y_p = -\frac{1}{2} x ^ 2 + \frac{l}{2} x - 1\]

Therefore the general solution is

\[y = Ce^{-x} + De^{x} -\frac{1}{2} x ^ 2 + \frac{l}{2} x - 1\] where \(C\) and \(D\) are arbitrary constants.

## Exceptional Cases

It is important to check if *all* or *sub-expressions* of the trial solution is also a solution of the associated homogeneous differential equation. In this scenario, therefore trial solution will not help find a particular integral. However, multiplying the trial solution by \(x\) often solves this problem. It may be required to perform this multiplication more than once.

- Exercise 32
The complementary function is \[y_c = Ce^{x} + De^{2x}\] Thus it can be seen that the traditional choice of trial functions, \(y = pe^x\) is a particular solution to the homogeneous differential equation, where \(C = p\), \(D = 0\).

Instead, we try the trial function \(y = pxe^x\)

We have \[y' = pe^x + pxe^x = p(1 + x)e^x\] and \[y'' = pe^x + p(1+x)e^x = p(2+x)e^x\]

Substituting into the inhomogeneous differential equation, we have

\[p(2+x)e^x - 3(p(1 + x)e^x) + 2 pxe^x = 4e^x\]

Expanding out and collecting terms, we have

\[\begin{array}{ll} 4e^x &= p(2+x)e^x - 3(p(1 + x)e^x) + 2 pxe^x \\ &= 2pe^x + pxe^x - 3(pe^x + pxe^x) + 2pxe^x \\ &= 2pe^x + pxe^x - 3pe^x - 3pxe^x + 2pxe^x \\ &= -pe^x \\ \end{array}\]

Comparing coefficients, we have \(p = -4\). Therefore the particular integral is

\[y_p = -4xe^x\]

In Exercise 18b, we saw the complementary function was \[y_c = C + De^{-\frac{3x}{2}}\]

However, the trial function \(f(x) = p\) is a particular solution to the associated homogeneous equation, with \(C = p\), \(D = 0\). Therefore, we try the trial solution \[f(x) = px\]

We have \[f'(x) = p\] \[f''(x) = 0\]

Substituting in to the original inhomogeneous equation, we have

\[3p = 1\]

So \(p = \frac{1}{3}\), and therefore the particular integral is

\[y_p = \frac{1}{3}x\]

- Exercise 33
We can rewrite in the form of homogeneous differential equation as \[m\ddot{x} + r\dot{x} = mg\]

The associated homogeneous equation has the auxiliary equation \[m \lambda^2 + r\lambda = 0\]

That is, \(\lambda(m\lambda + r) = 0\), so the auxiliary equation has two distinct real solutions: \(\lambda_1 = 0\) and \(\lambda_2 = -\frac{r}{m}\).

Following Procedure 5, the complementary function is thus

\[x_c = C + De^{-\frac{rt}{m}}\] where \(C\) and \(D\) are arbitrary constants.

We now find the particular integral. The trial solution \(f(t) = p\) will not help, as this is also a particular solution to the homogeneous differential equation, with \(C = p\), \(D = 0\). Instead, we consider the trial solution \[f(t) = pt\]

We have \[f'(t) = p\] \[f''(t) = 0\]

Substituting into the inhomogeneous differential equation, we have

\[rp = mg\]

So \(p = \frac{mg}{r}\), and therefore the particular integral is

\[x_p = \frac{mg}{r} t\]

Thus the general solution is

\[x = C + De^{-\frac{rt}{m}} + \frac{mgt}{r}\] where \(C\) and \(D\) are arbitrary constants.

- Exercise 34
In Exercise 21d, we saw that the complementary function was \[y_c = (C + Dx)e^x\]

If we consider the trial solution \[f(x) = pe^x\] this can be seen as a particular solution to the associated homogeneous equation, where \(C = p\), \(D = 0\).

Multipling by the independent variable, we try another trial solution \[f(x) = pxe^x\]

However, this is also a solution to the associated homogeneous equation, where \(C = 0\), \(D = p\). We follow Procedure 8 again, and consider another trial function:

\[f(x) = px^2e^x\]

Differentiating, we have

\[f'(x) = 2pxe^x + px^2e^x = p(x^2 + 2x)e^x\]

\[f''(x) = p(x^2 + 4x + 2)e^x\]

Substituting into the inhomogeneous differential equation, we have

\[p(x^2 + 4x + 2)e^x - 2(p(x^2 + 2x)e^x) + px^2e^x = e^x\]

Expanding out and collecting terms, we have

\[\begin{array}{ll} e^x &= p(x^2 + 4x + 2)e^x - 2(p(x^2 + 2x)e^x) + px^2e^x \\ &= pe^x((x^2 + 4x + 2) - 2(x^2 + 2x) + x^2) \\ &= pe^x(x^2 + 4x + 2 - 2x^2 - 4x + x^2) \\ &= pe^x(2) \\ \end{array}\]

Comparing coefficients of \(e^x\), we have \(p = \frac{1}{2}\). Thus the particular integral is \[y_p = \frac{x^2}{2}e^x\]

Therefore, the general solution is

\[y = (C + Dx)e^x + \frac{x^2}{2}e^x\] where \(C\) and \(D\) are arbitrary constants.

# Combining Cases

In this section, we will see how to find the particular integral when \(f(x)\) is a *combination* of polynomial, exponential or sinusoidal functions. We will use the Principle of Superposition to combine functions. We find the particular integrals involving each part of \(f(x)\) separately, and then combine the solutions using the Principle of Superposition.

- Exercise 35
In Exercise 24, we saw that \(\frac{x^2}{2}e^x\) is a particular integral for \[\frac{d^2y}{dx^2} - 2\frac{dy}{dx} + y = e^x\]

Next, we find a particular integral for \[\frac{d^2y}{dx^2} - 2\frac{dy}{dx} + y = e^{2x}\]

As a trial solution, we take \[f(x) = pe^{2x}\]

Differentiating, we have

\[f'(x) = 2pe^{2x}\] \[f''(x) = 4pe^{2x}\]

Substituting in to the original inhomogeneous equation, we have

\[4pe^{2x} - 4pe^2x + pe^{2x} = e^{2x}\]

Thus \(p = 1\). Therefore the particular integral is \[y_p = e^{2x}\]

Therefore, by the Principle of Superposition, the particular integral of \[\frac{d^2y}{dx^2} - 2\frac{dy}{dx} + y = 4e^x - 3e^{2x}\] is

\[y_p = 2x^2e^x - 3e^{2x}\]

We first consider \[2 \frac{d^2x}{dt^2} + 3 \frac{dx}{dt} + 2x = 12 \cos3t t\]

We take \[f(t) = p \cos 2t + q \sin 2t\] as our trial solution. Differentiating, we have

\[f'(t) = 2q \cos 2t - 2p \sin 2t\] \[f'(t) = - 4p \cos 2t - 4q \sin 2t\]

Substituting in to the inhomogeneous differential equation, we have

\[2 (-4p \cos 2t - 4q \sin 2t) + 3 (2q \cos 2t - 2p \sin 2t) + 2 (p \cos 2t + q \sin 2t) = 12 \cos 2t\]

Collecting terms, we have

\[\begin{array}{ll} 12 \cos 2t &= -8p \cos 2t -8 q \sin 2t + 6q \cos 2t - 6p \sin 2t + 2p \cos 2t + 2q \sin 2t &= (-8p + 6q + 2p) \cos 2t + (-8q - 6p + 2q) \sin 2t &= (-6p + 6q) \cos 2t + (-6q - 6p) \sin 2t &= (-6p + 6q) \cos 2t - (6q + 6p) \sin 2t &= 6(q - p) \cos 2t - 6(p + q) \sin 2t \end{array}\]

Comparing coefficients, we have the simultaneous equations

\[\begin{cases} q - p = 2 \\ p + q = 0 \end{cases}\]

Hence \(p = -1\), \(q = 1\)

Therefore, the particular integral is \[x_p = \sin 2t - \cos 2t\]

Next, we consider \[2 \frac{d^2x}{dt^2} + 3 \frac{dx}{dt} + 2x = 10\]

Now take \[f(t) = p\] Differentiating, \(f'(t) = 0\), \(f''(t) = 0\). Substituting, we have

\[2p = 10\]

Thus \(p = 5\). So the particular integral is \(x_p = 5\)

Therefore, by the Principle of Superposition, the particular integral for \[2 \frac{d^2x}{dt^2} + 3 \frac{dx}{dt} + 2x = 12 \cos3t t\]

We take \[f(t) = p \cos 2t + q \sin 2t\] as our trial solution. Differentiating, we have

\[f'(t) = 2q \cos 2t - 2p \sin 2t\] \[f'(t) = - 4p \cos 2t - 4q \sin 2t\]

Substituting in to the inhomogeneous differential equation, we have

\[2 (-4p \cos 2t - 4q \sin 2t) + 3 (2q \cos 2t - 2p \sin 2t) + 2 (p \cos 2t + q \sin 2t) = 12 \cos 2t\]

Collecting terms, we have

\[\begin{array}{ll} 12 \cos 2t &= -8p \cos 2t -8 q \sin 2t + 6q \cos 2t - 6p \sin 2t + 2p \cos 2t + 2q \sin 2t &= (-8p + 6q + 2p) \cos 2t + (-8q - 6p + 2q) \sin 2t &= (-6p + 6q) \cos 2t + (-6q - 6p) \sin 2t &= (-6p + 6q) \cos 2t - (6q + 6p) \sin 2t &= 6(q - p) \cos 2t - 6(p + q) \sin 2t \end{array}\]

Comparing coefficients, we have the simultaneous equations

\[\begin{cases} q - p = 2 \\ p + q = 0 \end{cases}\]

Hence \(p = -1\), \(q = 1\)

Therefore, the particular integral is \[x_p = \sin 2t - \cos 2t\]

Next, we consider \[2 \frac{d^2x}{dt^2} + 3 \frac{dx}{dt} + 2x = 10\]

Now take \[f(t) = p\] Differentiating, \(f'(t) = 0\), \(f''(t) = 0\). Substituting, we have

\[2p = 10\]

Thus \(p = 5\). So the particular integral is \(x_p = 5\)

Therefore, by the Principle of Superposition, the particular integral for \[2 \frac{d^2x}{dt^2} + 3 \frac{dx}{dt} + 2x = 12 \cos3t t + 10\] is \[y_p = \sin 2t - \cos 2t + 5\]