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. □


Comments

Post a Comment

Popular posts from this blog

Applications of multiple integrals

-
Image
We can use multiple integrals to compute areas and volumes of various shapes. Area of a planar region Definition (Area) Let \(D\) be a bounded closed region in \(\mathbb{R}^2\). \(D\) is said to have an area if the multiple integral of the constant function 1 over \(D\), \(\iint_Ddxdy\), exists. Its value is denoted by \(\mu(D)\): \[\mu(D) = \iint_Ddxdy.\] Example . Let us calculate the area of the disk \(D = \{(x,y)\mid x^2 + y^2 \leq a^2\}\). Using the polar coordinates, \(x = r\cos\theta, y = r\sin\theta\), \(dxdy = rdrd\theta\), and the ranges of \(r\) and \(\theta\) are \([0, a]\) and \([0, 2\pi]\), respectively. Thus, \[\begin{eqnarray*} \mu(D) &=& \iint_Ddxdy\\ &=&\int_0^a\left(\int_0^{2\pi}rd\theta\right)dr\\ &=&2\pi\int_0^a rdr\\ &=&2\pi\left[\frac{r^2}{2}\right]_0^a = \pi a^2. \end{eqnarray*}\] □ Volume of a solid figure Definition (Volume) Let \(V\) be a solid figure in the \((x,y,z)\) space \(\mathbb{R}^3\). \(V\) is...
Read more

Birth process

-
Image
  Defining the birth process  Consider a colony of bacteria that never dies. We study the following process known as  the birth process , also known as  the Yule process . The colony starts with \(n_0\) cells at time \(t = 0\). Assume that the probability that any individual cell divides in the time interval \((t, t + \delta t)\) is proportional to \(\delta t\) for small \(\delta t\). Further assume that each cell division is independent of others. Let \(\lambda\) be the  birth rate.  The probability of a cell division for a population of \(n\) cells during \(\delta t\) is \(\lambda n \delta t\). We assume that the probability that two or more births take place in the time interval \(\delta t\) is \(o(\delta t)\). That is, it can be ignored. Consequently, the probability that no cell divides during \(\delta t\) is \(1 - \lambda n \delta t - o(\delta t)\). Note that this process is an example of the Markov chain with states \({n_0}, {n_0 + 1}, {n_0 + 2}...
Read more

Improper multiple integrals

-
Image
Consider integrating a function \(f(x,y)\) over a region \(D\) which may not be bounded or closed. In the case of a univariate function, this corresponds to the improper integral where we took the limits of the endpoints of a closed interval. In the case of multiple integrals, we adopt the notion of a "sequence of regions." Consider a sequence of regions \(\{K_n\}\) where each \(K_n\) is a subset of \(\mathbb{R}^2\) that satisfies the following conditions: (a) \(K_1 \subset K_2\)\(\subset \cdots \subset\) \(K_n \subset K_{n+1} \subset \cdots\). (b) For all \(n\in \mathbb{N}\), \(K_n \subset D\). (c) For all \(n \in\mathbb{N}\), \(K_n\) is bounded and closed. (d) For any bounded closed set \(F\) that is included in \(D\) (i.e., \(F \subset D\)), if \(n\) is sufficiently large, then \(F \subset K_n\).  In other words: for all bounded closed \(F \subset D\), there exists some \(N\in \mathbb{N}\) such that, for all \(n\in \mathbb{N}\), if \(n \geq N\) then \(F \subset K_...
Read more