Special solution of inhomogeneous linear differential equations
The general solution of an inhomogeneous linear differential equation can be obtained as the sum of a special solution and the general solution of the corresponding homogeneous differential equation. We study the method of variation of parameters in particular.
Consider the linear differential equation of the form
\[F(D)y = q(x)\]
where \(F(t)\) is a polynomial and \(q(x)\) is a function. When \(q(x) = 0\), this is a homogeneous linear differential equation. Here, we assume \(q(x) \neq 0\).
Suppose we can factorize \(F(t)\) as \(F(t) = G(t)H(t)\). If \(y = y(x)\) is a solution of \(F(D)y = q\), then \(z = H(D)y\) is a solution of \(G(D)z = q\). Thus, the given differential equation \(F(D)y = q\) is decomposed into two parts:
- \(G(D)z = q\) (a linear differential equation of \(z\), given \(q\)),
- \(H(D)y = z\) (a linear differential equation of \(y\), given \(z\)),
and we can process one after the other. Thus, by factorizing the polynomial \(F(D)\), we only need to consider the case where \(F(t) = t - \alpha\) (\(\alpha \in \mathbb{C}\)). That is, \(y' - \alpha y = q.\) This ODE can be readily solved by using the method of variation of parameters. The general solution of this ODE is obtained as
\[y = e^{\alpha x}\left\{\int e^{-\alpha x}q(x)dx + C\right\}\]
where \(C\) is a constant.
Example. Let us solve \(y'' + y' - 6y = \cos x\).
Note that \(F(t) = t^2 + t - 6 = (t - 2)(t + 3)\) so the given ODE is \((D - 2)(D+3)y = \cos x\). Solving the homogeneous ODE \((D - 2)(D + 3)y = 0\), we have
\(y = Ae^{2x} + Be^{-3x}\) where \(A, B\) are constants. Let \(z = (D+3)y\), then we have \((D-2)z = \cos x\) and its special solution is given by
\[z = e^{2x}\int e^{-2x}\cos x dx = -\frac{2}{5}\cos x + \frac{1}{5}\sin x.\]
Similarly, solving \((D+3)y = -\frac{2}{5}\cos x + \frac{1}{5}\sin x\), we have a special solution
\[y = e^{-3x}\int e^{3x}\left(-\frac{2}{5}\cos x + \frac{1}{5}\sin x\right)\,dx = -\frac{7}{50}\cos x + \frac{1}{50}\sin x.\]
Thus, the general solution is
\[y = -\frac{7}{50}\cos x + \frac{1}{50}\sin x + Ae^{2x} + Be^{-3x}\]
where \(A, B\) are constants. □
Let's consider the more general case where \(F(t) = t^2 + at + b = (t - \alpha)(t - \beta)\) with \(\alpha \neq \beta\). In particular, when \(a^2 - 4b < 0\), we have \(\beta = \bar{\alpha}\). We want to solve the following second-order ODE:
\[(D - \alpha)(D - \beta)y = q.\]
Let \(z = (D-\beta)y\), and we have
\[(D - \alpha)z = q.\tag{eq:odea}\]
We first solve the homogeneous case \((D - \alpha)z = 0\) which yields
\[z = Ae^{\alpha x}.\]
Now, regarding \(A\) as a function of \(x\), substitute it into (eq:odea), we have
\[A'(x) = e^{-\alpha x}q(x).\]
Thus, we have the special solution
\[\begin{eqnarray}z &=& e^{\alpha x}A(x)\\ &=& e^{\alpha x}\int e^{-\alpha x}q(x)\, dx. \tag{eq:odez}\end{eqnarray}\]
Next, we solve
\[(D - \beta)y = z.\tag{eq:odeb}\]
Solving the homogeneous case \((D - \beta)y = 0\), we have
\[y = Be^{\beta x}.\]
Regarding \(B\) as a function of \(x\) and substituting it into (eq:odeb), we have
\[B'(x) = e^{-\beta x}z(x).\]
Integrating this using (eq:odea),
\[\begin{eqnarray}
B(x) &=& \int e^{-\beta x}e^{\alpha x}A(x)\,dx\\
&=& \frac{e^{(\alpha - \beta)x}}{\alpha - \beta}A(x) - \int\frac{e^{(\alpha - \beta)x}}{\alpha - \beta}A'(x)\,dx ~ \text{(integration by parts)}\\
&=& \frac{e^{(\alpha - \beta)x}}{\alpha - \beta}A(x) +\int\frac{e^{(\alpha - \beta)x}}{\beta - \alpha}e^{-\alpha x}q(x)\,dx\\
&=& \frac{e^{(\alpha - \beta)x}}{\alpha - \beta}\int e^{-\alpha x}q(x)\,dx + \frac{1}{\beta - \alpha}\int e^{-\beta x}q(x)\,dx.
\end{eqnarray}\]
Thus, we have the special solution
\[\begin{eqnarray}y &=& e^{\beta x}B(x)\\& = &\frac{e^{\alpha x}}{\alpha - \beta}\int e^{-\alpha x}q(x)\,dx + \frac{e^{\beta x}}{\beta - \alpha}\int e^{-\beta x}q(x)\,dx.\end{eqnarray}\]
Note the symmetry between the two terms on the right-hand side. This means that swapping \(\alpha\) and \(\beta\) does not change the result. This is expected as the differential operators are commutative: \((D-\alpha)(D - \beta) = (D -\beta)(D - \alpha)\).
Example. Let us solve
\[y'' + y = \frac{1}{\cos x}.\]
The corresponding homogeneous equation is \(y'' + y = 0\). Noting \(D^2 + 1 = (D -i)(D+i)\), its solution is
\[y = Ae^{ix} + Be^{-ix} \tag{eg:hom}\]
where \(A, B \in \mathbb{C}\) are constants. Now, we use the method of variation of parameters by regarding \(A\) and \(B\) as functions of \(x\). According to the above result, we have
\[\begin{eqnarray}A(x) &=& \int \frac{e^{-ix}}{2i\cos x}\,dx \\
&=&\frac{1}{2i}\int\left(1 - i \frac{\sin x}{\cos x}\right)\,dx\\
&=&\frac{1}{2i}(x + i\log|\cos x|) + C_1
\end{eqnarray}\]
where \(C_1\) is a constant. Similarly,
\[B(x) = -\frac{1}{2i}(x - i\log|\cos x|) + C_2\]
where \(C_2\) is a constant. Thus, the general solution is
\[y = \left[\frac{1}{2i}(x + i\log|\cos x|) + C_1\right]e^{ix} + \left[-\frac{1}{2i}(x - i\log|\cos x|) + C_2\right]e^{-ix}.\]
Using Euler's formula (\(e^{\pm i x} = \cos x \pm i \sin x\)), this can be further "simplified" (exercise!) as
\[y = (\log|\cos x| + C)\cos x + (x + D)\sin x\]
where \(C\) and \(D\) are constants. Note that this is a real-valued function if \(C\) and \(D\) are real. □
Next, let's consider \((D-\alpha)^2y = q.\) Of course, \((t - \alpha)^2 = (t-\alpha)(t-\alpha)\) so that we can apply the technique above. Let \(y_1 = (D-\alpha)y\), and we solve \((D-\alpha)y_1 = q\). As we have seen above,
\[y_1 = e^{\alpha x}\int e^{-\alpha x}q(x)\,dx.\]
Next, we solve \((D-\alpha)y_2 = y_1\). This gives
\[y_2 = e^{\alpha x}\int e^{-\alpha x}y_1(x)\,dx.\]
We can continue the same process to solve \((D-\alpha)^my = q\) for any \(m = 3, 4, \cdots\). Let the solution of \((D-\alpha)^my = q\) be \(y_m\) for \(m = 0, 1, 2, \cdots\). We can see that
\[y_m = e^{\alpha x}\int e^{-\alpha x}y_{m-1}(x)\,dx\]
and
\[y_0(x) = q(x).\]
Example. Solve \(y'' -4y' +4y = \cos x\). It is an exercise to show that the general solution is
\[y = \frac{3}{25}\cos x -\frac{4}{25}\sin x + Ae^{2x} + Bxe^{2x}\]
where \(A\) and \(B\) are constants. □
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