Second-Order Nonhomogeneous Linear Differential Equations

A second-order nonhomogeneous linear differential equation with constant coefficients is written as $$ ay'' + by' + cy = r(x) $$ where \( a \), \( b \), and \( c \) are real constants.

This is referred to as the complete form of the differential equation.

One common approach to solving it is the method of variation of parameters.

1. Start by finding the general solution of the associated homogeneous equation: $$ y_A = c_1 y_1 + c_2 y_2 \qquad \text{where } ay'' + by' + cy = 0 $$
2. Next, determine a particular solution of the nonhomogeneous equation in the form \( y_P = c_1(x) y_1 + c_2(x) y_2 \), by solving the following system: $$ \begin{cases} c'_1(x) y_1 + c'_2(x) y_2 = 0 \\ \\ c'_1(x) y'_1 + c'_2(x) y'_2 = r(x) \end{cases} $$

   Note. There are alternative techniques for finding a particular integral, such as the method of undetermined coefficients.

3. The general solution to the original equation is then the sum: $$ y = y_A + y_P $$

A Worked Example

Let’s solve the following second-order nonhomogeneous differential equation:

$$ y'' - 2y' + y = \frac{e^x}{x^4} $$

The associated homogeneous equation is:

$$ y'' - 2y' + y = 0 $$

We solve its characteristic equation by introducing an auxiliary variable \( t \) and using the coefficients \( a = 1 \), \( b = -2 \), \( c = 1 \):

$$ at^2 + bt + c = 0 $$

$$ t^2 - 2t + 1 = 0 $$

This quadratic equation yields:

$$ t = \frac{2 \pm \sqrt{4 - 4(1)(1)}}{2} $$

$$ t = \frac{2 \pm 0}{2} $$

$$ t = \begin{cases} t_1 = 1 \\ \\ t_2 = 1 \end{cases} $$

Since the roots are repeated, the general solution to the homogeneous equation is:

$$ y_A = c_1 e^x + c_2 x e^x $$

We now look for a particular solution in the form:

$$ y_P = c_1(x) e^x + c_2(x) x e^x $$

Here, \( c_1(x) \) and \( c_2(x) \) are functions of \( x \). This is why the method is called variation of parameters.

To determine these functions, we solve the system:

$$ \begin{cases} c'_1(x) e^x + c'_2(x) x e^x = 0 \\ \\ c'_1(x) D[e^x] + c'_2(x) D[xe^x] = \frac{e^x}{x^4} \end{cases} $$

Important. The system involves the derivatives \( c'_1(x) \) and \( c'_2(x) \), not the functions themselves.

Computing the derivatives gives:

$$ \begin{cases} c'_1(x) e^x + c'_2(x) x e^x = 0 \\ \\ c'_1(x) e^x + c'_2(x) (x e^x + e^x) = \frac{e^x}{x^4} \end{cases} $$

Using substitution, we isolate \( c'_1(x) \) from the first equation:

$$ \begin{cases} c'_1(x) = \frac{-c'_2(x) x e^x}{e^x} \\ \\ c'_1(x) e^x + c'_2(x)(x e^x + e^x) = \frac{e^x}{x^4} \end{cases} $$

$$ \begin{cases} c'_1(x) = -c'_2(x) x \\ \\ c'_1(x) e^x + c'_2(x)(x e^x + e^x) = \frac{e^x}{x^4} \end{cases} $$

Substituting into the second equation:

$$ \begin{cases} c'_1(x) = -c'_2(x) x \\ \\ (-c'_2(x) x)e^x + c'_2(x)(x e^x + e^x) = \frac{e^x}{x^4} \end{cases} $$

$$ \begin{cases} c'_1(x) = -c'_2(x) x \\ \\ -c'_2(x) x e^x + c'_2(x) x e^x + c'_2(x) e^x = \frac{e^x}{x^4} \end{cases} $$

$$ \begin{cases} c'_1(x) = -c'_2(x) x \\ \\ c'_2(x) e^x = \frac{e^x}{x^4} \end{cases} $$

$$ \begin{cases} c'_1(x) = -c'_2(x) x \\ \\ c'_2(x) = \frac{1}{x^4} \end{cases} $$

Now substitute \( c'_2(x) = 1/x^4 \) into the first equation:

$$ \begin{cases} c'_1(x) = -\left( \frac{1}{x^4} \right) x \\ \\ c'_2(x) = \frac{1}{x^4} \end{cases} $$

$$ \begin{cases} c'_1(x) = -\frac{1}{x^3} \\ \\ c'_2(x) = \frac{1}{x^4} \end{cases} $$

Now we integrate to find \( c_1(x) \) and \( c_2(x) \):

$$ \begin{cases} c_1(x) = \int -\frac{1}{x^3} \ dx \\ \\ c_2(x) = \int \frac{1}{x^4} \ dx \end{cases} $$

$$ \begin{cases} c_1(x) = \frac{x^{-2}}{2} \\ \\ c_2(x) = \frac{x^{-3}}{-3} \end{cases} $$

Substitute these into the expression for \( y_P \):

$$ y_P = c_1(x) e^x + c_2(x) x e^x $$

$$ y_P = \frac{x^{-2}}{2} e^x + \left( \frac{-x^{-3}}{3} \right) x e^x $$

$$ y_P = \frac{x^{-2}}{2} e^x - \frac{x^{-2}}{3} e^x $$

$$ y_P = \left( \frac{x^{-2}}{2} - \frac{x^{-2}}{3} \right) e^x $$

$$ y_P = \left( \frac{3x^{-2} - 2x^{-2}}{6} \right) e^x $$

$$ y_P = \frac{x^{-2}}{6} e^x $$

$$ y_P = \frac{e^x}{6x^2} $$

Finally, combine the homogeneous and particular solutions:

$$ y = y_A + y_P $$

$$ y = [ c_1 e^x + c_2 x e^x ] + y_P $$

$$ y = c_1 e^x + c_2 x e^x + \frac{e^x}{6x^2} $$

And that completes the solution.