Curriculum Objectives
- recall the meaning of the terms ‘complementary function’ and ‘particular integral’ in the context of linear differential equations, and recall that the general solution is the sum of the complementary function and a particular integral;
- find the complementary function for a second order linear differential equation with constant coefficients;
- recall the form of, and find, a particular integral for a second order linear differential equation in the cases where a polynomial or ebx or a cos px + b sin px is a suitable form, and in other simple cases find the appropriate coefficient(s) given a suitable form of particular integral;
- use a substitution to reduce a given differential equation to a second order linear equation with constant coefficients;
Note: This chapter is written rather hypocritically. I am a firm believer of understanding over rote learning, yet in this chapter, often you will be required to learn the procedure, without being given an explanation, as to their need. This chapter requires some learning and will not over-complicate things by digressing into the proofs or reasoning behind theorems.
Introduction to the Differential Equation
You must already be acquainted with the general idea of a differential equation. Yet a concise introduction is presented here.
The fundamental definition of a differential equation is that it is an equation that involves differential coefficients. Consider the following:
$\begin{array}{l} i)\;\frac{{dy}}{{dx}} = 2x\\ ii)\,y\frac{{{d^2}y}}{{d{x^2}}} + x = 0\\ iii)y = x \cdot \frac{{dy}}{{dx}} + \frac{5}{{\frac{{dy}}{{dx}}}} \end{array}$
When we talk about solving differential equations, we wish to find a function of the independent variable that it satisfies the original equation. If you were given the differential equation:
$\frac{{dy}}{{dx}} = 2$
You can easily deduce that the solution to this differential equation would be such a function, y, that when differentiated yields, 2. The solution of course is an entire "family" defined by: $y = 2x + c$
We shall introduce here the Second Order Linear Differential Equation, with constant coefficients ( the co-efficient themselves can be functions ), for it's assumed you have already met equations of the first order in the Mathematics course. The order of a differential equation tells us the highest order of derivative that is present in the equation. A second order differential equation will, then, always involve a function twice differentiated. The general differential equation of this sort is of the form:
The fundamental definition of a differential equation is that it is an equation that involves differential coefficients. Consider the following:
$\begin{array}{l} i)\;\frac{{dy}}{{dx}} = 2x\\ ii)\,y\frac{{{d^2}y}}{{d{x^2}}} + x = 0\\ iii)y = x \cdot \frac{{dy}}{{dx}} + \frac{5}{{\frac{{dy}}{{dx}}}} \end{array}$
When we talk about solving differential equations, we wish to find a function of the independent variable that it satisfies the original equation. If you were given the differential equation:
$\frac{{dy}}{{dx}} = 2$
You can easily deduce that the solution to this differential equation would be such a function, y, that when differentiated yields, 2. The solution of course is an entire "family" defined by: $y = 2x + c$
We shall introduce here the Second Order Linear Differential Equation, with constant coefficients ( the co-efficient themselves can be functions ), for it's assumed you have already met equations of the first order in the Mathematics course. The order of a differential equation tells us the highest order of derivative that is present in the equation. A second order differential equation will, then, always involve a function twice differentiated. The general differential equation of this sort is of the form:
$a\frac{{{d^2}y}}{{d{x^2}}} + b\frac{{dy}}{{dx}} + cy = f(x)$
Here, a cannot be equal to zero, for then it will not be a second order differential equation. The other constants do not function under any such condition.
Obviously we will begin with the easier case, and only take differential equations of the form
$a\frac{{{d^2}y}}{{d{x^2}}} + b\frac{{dy}}{{dx}} + cy = 0$
Such equations are called homogeneous equations.
Ex. Solve the differential equation:
$16y - \frac{{{d^2}y}}{{dx}} = 0$
Sol.
By inspection and some ingenuity alone I can deduce that the following will be the solution to the differential equation, as we require a function, such that differentiated twice,, equals 16 times the original function:
$y = {e^{4x}};{e^{ - 4x}}$
These alone aren't the only possible solutions. All of the expressions of the form $y = {c_1}{e^{4x}} + {c_2}{e^{ - 4x}}$ will be a solution to this differential equation. You make check this by twice differentiating the expression and substituting for both y and y''.
This is called the Principle of Superposition. If you know ${y_1},\,{y_2}$ to be solutions of a linear, homogeneous ( these are the equations that we are considering, equations that are equated with zero as opposed to some function of x ), differential equation, then $y = {c_1}{y_1} + {c_2}{y_2}$ is also a solution. This fact is the very definition of a linear differential equation. The solutions of a linear differential equation form a linear space ( you shall meet this in the next chapter ).
The presence of constants in the solution suggests that it can model a variety of situations. For a specific case, however, we will have to put in the values of the constants. The values are found from boundary conditions. For eg, if we are given the following boundary conditions for this question:
$y = {c_1}{e^{4x}} + {c_2}{e^{ - 4x}}$
This solution is called the complementary function.
$\begin{array}{l} y = 2,\quad when\,x = 0\\ \frac{{dy}}{{dx}} = - 1,\quad when\,x = 0\\ {c_1} + {c_2} = 2\\ \frac{{dy}}{{dx}} = 4{c_1}{e^{4x}} - 4{c_2}{e^{ - 4x}}\\ 4{c_1} - 4{c_2} = - 1\\ Solving\,this\,system\,of\,equations,\,we\,have:\\ {c_1} = \frac{7}{8};\,{c_2} = \frac{9}{8}\\ y = \frac{7}{8}{e^{4x}} + \frac{9}{8}{e^{ - 4x}} \end{array}$
In the original example we were able to deduce the answer by inspection, this doesn't occur very often in the examination. So let us then generalise our method.
This solution is called the complementary function.
$\begin{array}{l} y = 2,\quad when\,x = 0\\ \frac{{dy}}{{dx}} = - 1,\quad when\,x = 0\\ {c_1} + {c_2} = 2\\ \frac{{dy}}{{dx}} = 4{c_1}{e^{4x}} - 4{c_2}{e^{ - 4x}}\\ 4{c_1} - 4{c_2} = - 1\\ Solving\,this\,system\,of\,equations,\,we\,have:\\ {c_1} = \frac{7}{8};\,{c_2} = \frac{9}{8}\\ y = \frac{7}{8}{e^{4x}} + \frac{9}{8}{e^{ - 4x}} \end{array}$
In the original example we were able to deduce the answer by inspection, this doesn't occur very often in the examination. So let us then generalise our method.
Let me assure you that the solution of these equations is of the form $y = {e^{mx}}$ ( This particular function is our best guess at the answer) . The question is then, what is m?
We employ the following method:
$\begin{array}{l}
a\frac{{{d^2}y}}{{d{x^2}}} + b\frac{{dy}}{{dx}} + cy = 0\\
y = {e^{mx}}\\
\frac{{dy}}{{dx}} = m{e^{mx}}\\
\frac{{{d^2}y}}{{d{x^2}}} = {m^2}{e^{mx}}\\
a\left( {{m^2}{e^{mx}}} \right) + b\left( {m{e^{mx}}} \right) + c{e^{mx}} = 0\\
{e^{mx}}\left( {a{m^2} + bm + c} \right) = 0\\
{e^{mx}} \ne 0\quad \therefore a{m^2} + bm + c = 0
\end{array}$
The roots of this quadratic are then our values of m. This quadratic is of great importance to the solution to second order linear ODE, and therefore warrants its own name. Its called the characteristic equation.
There are three possibilities then for m. Either it has two completely different roots as its values, a repeated root as its only value, or a complex root and its conjugate as its value. These cases differ in treatment, therefore will be discussed separately.
Ex. Solve
$\frac{{{d^2}y}}{{d{x^2}}} + 11\frac{{dy}}{{dx}} + 24y = 0$
Sol.
The Characteristic equation is then:
$\begin{array}{l} {m^2} + 11m + 24 = 0\\ \left( {m + 3} \right)\left( {m + 8} \right) = 0 \end{array}$
The general solution is then:
$y = {c_1}{e^{ - 3x}} + {c_2}{e^{ - 8x}}$
$\begin{array}{l}
\frac{{{d^2}y}}{{d{x^2}}} - 4\frac{{dy}}{{dx}} + 4y = 0;\quad {y_{x = 0}} = 12;\quad {\frac{{dy}}{{dx}}_{x = 0}} = - 3\\
{r^2} - 4r + 4 = 0\\
{\left( {r - 2} \right)^2} = 0\\
r = 2\\
y = {c_1}{e^{2x}} + {c_2}x{e^{2x}}\\
\frac{{dy}}{{dx}} = 2{c_1}{e^{2x}} + {c_2}{e^{2x}} + 2{c_2}x{e^{2x}}\\
{c_1} = 12;{c_2} = - 27\\
y = 12{e^{2x}} - 27x{e^{2x}}
\end{array}$The roots of this quadratic are then our values of m. This quadratic is of great importance to the solution to second order linear ODE, and therefore warrants its own name. Its called the characteristic equation.
There are three possibilities then for m. Either it has two completely different roots as its values, a repeated root as its only value, or a complex root and its conjugate as its value. These cases differ in treatment, therefore will be discussed separately.
Real roots
If we get two real roots to our characteristic equation ($m_1$ and $m_2$), then the solution will be of the form:
$y = {c_1}{e^{{m_1}x}} + {c_2}{e^{{m_2}x}}$
$\frac{{{d^2}y}}{{d{x^2}}} + 11\frac{{dy}}{{dx}} + 24y = 0$
Sol.
The Characteristic equation is then:
$\begin{array}{l} {m^2} + 11m + 24 = 0\\ \left( {m + 3} \right)\left( {m + 8} \right) = 0 \end{array}$
The general solution is then:
$y = {c_1}{e^{ - 3x}} + {c_2}{e^{ - 8x}}$
Repeated Roots
For repeated roots, the general solution is of the form:
$y = \left( {{c_1} + {c_2}x} \right){e^{rx}}$
where r is the repeated root of the equation.
Ex.
$\frac{{{d^2}y}}{{d{x^2}}} - 4\frac{{dy}}{{dx}} + 4y = 0;\quad {y_{x = 0}} = 12;\quad {\frac{{dy}}{{dx}}_{x = 0}} = - 3$
Sol.
$y = \left( {{c_1} + {c_2}x} \right){e^{rx}}$
where r is the repeated root of the equation.
Ex.
$\frac{{{d^2}y}}{{d{x^2}}} - 4\frac{{dy}}{{dx}} + 4y = 0;\quad {y_{x = 0}} = 12;\quad {\frac{{dy}}{{dx}}_{x = 0}} = - 3$
Sol.
Complex Roots
For complex roots the solution is of the form:
$y = {e^{{r_1}x}}\left( {{c_1}\cos \left( {{r_2}x} \right) + {c_2}\sin \left( {{r_2}x} \right)} \right)$
Where the complex roots of the characteristic equation are: ${r_1} \pm \iota {r_2}$
Ex. Solve the equation $y'' + 6y' + 9y = 0$ given that $\frac{{dy}}{{dx}} = 6;\;y = 0;\;and\;x = 0$
Sol.
Solving the characteristic equation I find:
$\begin{array}{l} {m^2} + 6m + 9 = 0\\ m = 1 \pm \iota \sqrt 2 \end{array}$
The general solution is then of the form:
$y = {e^x}\left( {A\cos \left( {\sqrt 2 \cdot x} \right) + B\sin \left( {\sqrt 2 \cdot x} \right)} \right)$
Differentiating this equation, gives me:
$y' = {e^x}\left( {A\cos \left( {\sqrt 2 \cdot x} \right) + B\sin \left( {\sqrt 2 \cdot x} \right) + \sqrt 2 \left( {B\cos \left( {\sqrt 2 \cdot x} \right) - A\sin \left( {\sqrt 2 \cdot x} \right)} \right)} \right)$
Using the boundary conditions to create a system of equations and then solving it, I get the following final result:
$y = 3\sqrt 2 \sin \left( {\sqrt 2 \cdot x} \right)$
The reader is, as always, encouraged to solve this on his own and convince his self of this answer.
Ex. Solve the equation $y'' + 6y' + 9y = 0$ given that $\frac{{dy}}{{dx}} = 6;\;y = 0;\;and\;x = 0$
Sol.
Solving the characteristic equation I find:
$\begin{array}{l} {m^2} + 6m + 9 = 0\\ m = 1 \pm \iota \sqrt 2 \end{array}$
The general solution is then of the form:
$y = {e^x}\left( {A\cos \left( {\sqrt 2 \cdot x} \right) + B\sin \left( {\sqrt 2 \cdot x} \right)} \right)$
Differentiating this equation, gives me:
$y' = {e^x}\left( {A\cos \left( {\sqrt 2 \cdot x} \right) + B\sin \left( {\sqrt 2 \cdot x} \right) + \sqrt 2 \left( {B\cos \left( {\sqrt 2 \cdot x} \right) - A\sin \left( {\sqrt 2 \cdot x} \right)} \right)} \right)$
Using the boundary conditions to create a system of equations and then solving it, I get the following final result:
$y = 3\sqrt 2 \sin \left( {\sqrt 2 \cdot x} \right)$
The reader is, as always, encouraged to solve this on his own and convince his self of this answer.
Differential Equations of the form $a\frac{{{d^2}y}}{{d{x^2}}} + b\frac{{dy}}{{dx}} + cy = f(x)$
For such equations the solution is found by working out the complementary function and particular integral. The general solution is then the sum of these two.
The complementary function of such equations, $a\frac{{{d^2}y}}{{d{x^2}}} + b\frac{{dy}}{{dx}} + cy = f(x)$, is found by isolating the left hand side, i.e. $a\frac{{{d^2}y}}{{d{x^2}}} + b\frac{{dy}}{{dx}} + cy = 0$ and then solving it as we have done previously. The solution we shall call the complementary function
The particular integral is found using the entire differential equation, but depends heavily upon the function on the right side of the equation. The following explains the method:
Consider the following the examples
Ex. Solve $y'' + 3y' - 4y = 8$
The complementary function, is found by solving the characteristic equation, The reader will find that the complementary function is:
$y = {c_1}{e^{ - 4x}} + {c_2}{e^x}$
For the particular integral we see that the function on the right hand side is a polynomial of the degree zero, which is to say that it is a constant. We assume that particular integral will also be of the form, and hence let $y = k$. Plugging this into the equation:
The particular integral is then, $y = -2$. The general solution is then:
$y = {c_1}{e^{ - 4x}} + {c_2}{e^x} - 2$
If boundary conditions were given in the question, it would be at this point, that you plug in that information to work out the constants
Ex. Solve $y'' + 3y' - 4y = 2 + 8{x^2}$
The complementary function is the same. For the particular integral, we note that there is a quadratic on the right side. Hence we use a quadratic of our own, to work out the particular integral. We let $y = a{x^2} + bx + c$
$\begin{array}{l} y' = 2ax + b\\ y'' = 2a \end{array}$
$\begin{array}{l} y'' + 3y' - 4y = 8{x^2} + 2\\ 2a + 3\left( {2ax + b} \right) - 4\left( {a{x^2} + bx + c} \right) = 8{x^2} + 2\\ 2a + 6ax + 3b - 4a{x^2} - 4bx - 4c = 8{x^2} + 2\\ \therefore - 4a{x^2} = 8{x^2} \Rightarrow a = - 2\\ \left( {6a - 4b} \right)x = 0x \Rightarrow b = - 3\\ 2a + 3b - 4c = 2 \Rightarrow c = - 4 \end{array}$
The general solution is of the form:
$y = {c_1}{e^{ - 4x}} + {c_2}{e^x} - 2{x^2} - 3x - 4$
So, generally, if there is a polynomial on the right hand side of order n, you must use the following to find out your particular integral, $y = {a_0} + {a_1}x + {a_2}{x^2} + ... + {a_n}{x^n}$.
For the right hand side, we still have two possibilities to exhaust. Consider further the examples:
$y'' + 3y' + 2y = 3{e^{2x}}$
The complementary function is left to the reader to work out. We shall focus on the particular integral. We shall here use as our particular integral, the general form: $y = k{e^{2x}}$.
$y = k{e^{2x}};\,y' = 2k{e^{2x}};\,y'' = 4k{e^{2x}}$
As before, we put in our guess at the particular integral into the equation to give:
$\begin{array}{l} 4k{e^{2x}} + 3\left( {2k{e^{2x}}} \right) + 2\left( {k{e^{2x}}} \right) = 3{e^{2x}}\\ 12k = 3 \Rightarrow k = {\textstyle{1 \over 4}} \end{array}$
The particular integral is therefore $y = {\textstyle{1 \over 4}}{e^{2x}}$
Lastly consider:
$y'' + 3y' + 2y = \cos \left( {2x} \right)$
Here we find cosine on the right hand side. If we had sine on the right hand side, or a combination of the two ratios, our guess would be the same, which is: $a\cos \left( {2x} \right) + b\sin \left( {2x} \right)$
$\begin{array}{l} y = a\cos \left( {2x} \right) + b\sin \left( {2x} \right)\\ y' = - 2a\sin \left( {2x} \right) + 2b\cos \left( {2x} \right)\\ y'' = - 4a\cos \left( {2x} \right) - 4\sin \left( {2x} \right) \end{array}$
We put this into the initial equation:
$ - 4a\cos \left( {2x} \right) - 4\sin \left( {2x} \right) + 3\left( { - 2a\sin \left( {2x} \right) + 2b\cos \left( {2x} \right)} \right) + 2\left( {a\cos \left( {2x} \right) + b\sin \left( {2x} \right)} \right) = \cos \left( {2x} \right)$
Comparing terms in, sin(2x) and cos(2x), on either sides yields:
$\begin{array}{l} - 4a + 6b + 2a = 1\\ - 4b - 6a + 2b = 0 \end{array}$
This system of equations can be solved to give:
$a = - {\textstyle{1 \over {20}}};\,b = {\textstyle{3 \over {20}}}$
The particular integral is then: $y = - {\textstyle{1 \over {20}}}\cos \left( {2x} \right) + {\textstyle{3 \over {20}}}\sin \left( {2x} \right)$
The following table summarises these important rules; you should commit it to memory.
It might happen that the general expression you choose to derive the particular integral with, may be excepting the constants, is equivalent to the complementary function you have already derived. In this case you must amend the general expression you were going to use, by multiplying the general expression by the independent variable. Consider the example:
Ex. Solve $y'' - 3y' = 6$
It is easily seen that the complementary function is:
$y = A + B{e^{3x}}$
For the particular integral, I take the general expression, y = k. It should have worked, yet putting this into the initial equation, one sees:
$\begin{array}{l} y = k;\,y' = 0;\,y'' = 0\\ 0 = 6 \end{array}$
This happens because our particular integral is already present in the complementary function. A, a constant, is present in the CF, and we have chosen a constant for our PI. Here we multiply our PI by the independent variable, in this case (x). Our particular integral is then $ y = kx $. It is left to the reader to ascertain that the PI, is $ y = - 2x $
If the function on, right hand side been assume, $7{e^{3x}}$, we could not have used as our particular integral, $k{e^{3x}}$, rather we would have used: $kx{e^{3x}}$. Had this also been present in the CF, we would have used: $k{x^2}{e^{3x}}$
Using a substitution to reduce a given differential equation to a second order linear equation with constant coefficients
In your examination, you may be given a differential equation involving, say y and x as variables, that can be reduced to a second order linear equation with constant coefficient of say, z and x, provided a relationship between the variables is given. In most cases, you will be given an intimidating differential equation, the equation it needs to be reduced to, and the substitution. Consider the following example.
Ex. Solve $\frac{1}{x}\frac{{{d^2}y}}{{d{x^2}}} + \left( {\frac{6}{x} - \frac{2}{{{x^2}}}} \right)\frac{{dy}}{{dx}} + \left( {\frac{9}{x} - \frac{6}{{{x^2}}} + \frac{2}{{{x^3}}}} \right)y = 169\sin \left( {2x} \right)$ given that $y = xz$
Sol:
We note that:
$\begin{array}{l} \frac{{dy}}{{dx}} = z + x\left( {\frac{{dz}}{{dx}}} \right)\\ \frac{{{d^2}y}}{{d{x^2}}} = \frac{{dz}}{{dx}} + \frac{{dz}}{{dx}} + x\left( {\frac{{{d^2}z}}{{d{x^2}}}} \right) \end{array}$
Substituting these into the original equation:
$\begin{array}{l} \frac{1}{x}\left( {2\left( {\frac{{dz}}{{dx}}} \right) + x\left( {\frac{{{d^2}z}}{{dx}}} \right)} \right) + \left( {\frac{6}{x} - \frac{2}{{{x^2}}}} \right)\left( {z + x\left( {\frac{{dz}}{{dx}}} \right)} \right) + \left( {\frac{9}{x} - \frac{6}{{{x^2}}} + \frac{2}{{{x^3}}}} \right)\left( {xz} \right) = 169\sin \left( {2x} \right)\\ \frac{2}{x}\left( {\frac{{dz}}{{dx}}} \right) + \frac{{{d^2}z}}{{d{x^2}}} + \frac{6}{x}\left( z \right) + 6\left( {\frac{{dz}}{{dx}}} \right) - \frac{{2z}}{{{x^2}}} - \frac{2}{x}\left( {\frac{{dz}}{{dx}}} \right) + 9z - \frac{6}{x}z + \frac{2}{{{x^2}}}z = 169\sin \left( {2x} \right)\\ \frac{{{d^2}z}}{{d{x^2}}} + 6\frac{{dz}}{{dx}} + 9z = 169\sin \left( {2x} \right) \end{array}$
Now you may easily solve the differential equation using the method already taught. Alternatively you could have rearranged it thusly: z = y/x. Differentiated it, and compared with the original equation.
Last I shall attempt a question from your past papers.
Ex. (May/June 2005, Question No. 11 OR)
It is given that:
$\frac{{{d^2}y}}{{d{x^2}}} + \left( {2a - 1} \right)\frac{{dy}}{{dx}} + a\left( {a - 1} \right)y = 2a - 1 + a\left( {a - 1} \right)x$
Find y in the terms of a nand x, given that y and y' are zero, when x is zero; a is a constant.
Sol.
The characteristic equation will be made thusly:
${m^2} + \left( {2a - 1} \right)m + {a^2} - a = 0$
M is then:
$\begin{array}{l} m = \frac{{\left( {1 - 2a} \right) \pm \sqrt {{{\left( {2a - 1} \right)}^2} - 4\left( 1 \right)\left( {{a^2} - a} \right)} }}{2}\\ m = \frac{{\left( {1 - 2a} \right) \pm \sqrt {\left( {4{a^2} - 4a + 1} \right) - 4{a^2} + 4a} }}{2} = \frac{{\left( {1 - 2a} \right) \pm 1}}{2}\\ m = 1 - a;\, - a \end{array}$
CF is then:
$y = {c_1}{e^{\left( {1 - a} \right)x}} + {c_2}{e^{ - ax}}$
For the particular integral, we take $y = {c_3}x + {c_4}$
$y = {c_3}x + {c_4};\,y' = {c_3};\,y'' = 0$
We substitute this into the initial equation:
$\left( {2a - 1} \right)\left( {{c_3}} \right) + \left( {{a^2} - a} \right)\left( {{c_3}x + {c_4}} \right) = 2a - 1 + \left( {{a^2} - a} \right)x$
$\begin{array}{l} \left( {{a^2} - a} \right){c_3}x = \left( {{a^2} - a} \right)x \Rightarrow {c_3} = 1\\ \left( {2a - 1} \right){c_3} + \left( {{a^2} - a} \right){c_4} = 2a - 1 \Rightarrow {c_4} = 0 \end{array}$
The particular integral then becomes: $y = x$
The general solution to the equation becomes:
$y = x + {c_1}{e^{\left( {1 - a} \right)x}} + {c_2}{e^{ - ax}}$
$\frac{{dy}}{{dx}} = \left( {1 - a} \right){c_1}{e^{\left( {1 - a} \right)x}} - a{c_2}{e^{ - ax}}$
Using boundary conditions, we have:
$\begin{array}{l} y = {c_1}{e^{\left( {1 - a} \right)x}} + {c_2}{e^{ - ax}} + x \Rightarrow 0 = {c_1} + {c_2} + 0\\ \frac{{dy}}{{dx}} = \left( {1 - a} \right){c_1}{e^{\left( {1 - a} \right)x}} - a{c_2}{e^{ - ax}} + 1 \Rightarrow 0 = \left( {1 - a} \right){c_1} - a{c_2} + 1\\ {c_1} = - {c_2}\\ 0 = {c_1} - a{c_1} - a{c_2} + 1 \Rightarrow {c_1} = - 1\\ y = x + {e^{ - ax}} - {e^{\left( {1 - a} \right)x}} \end{array}$
Ex. Hence show that $y \approx x$ as $x \to \infty $
Sol.
We note:
$y = x + {e^{ - ax}} - {e^{\left( {1 - a} \right)x}}$
If $a > 1$, both the terms - -a and (1 - a) - will be negative. As $x \to \infty$ the terms - ${e^{ - ax}}$ and ${e^{\left( {1 - a} \right)x}}$ - will approach zero, and we will be left with:
$y \approx x$
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