Mastering Mathematics Smartly
by Wee Wen Shih

A unique self-help website that provides comprehensive coverage of mathematics at A-level & beyond, written in a student-friendly style.

Section 1

Some differential equations may be solved easily with the method of Laplace transforms. We shall describe how that is possible in Section 4. First, we shall focus on basic ideas.

Let \text{f}(x) be some expression in x such that x \ge 0. We write L\{\text{f}(x)\} to mean the Laplace transform of \text{f}(x) and it is defined formally as:
L\{\text{f}(x)\} = \int_{0}^{\infty} \text{e}^{-sx} \text{f}(x) \text{d}x
where s is a variable chosen specially to ensure that the integral converges.

Next, we shall apply the above definition to derive the Laplace transform of a simple expression, say \text{f}(x) = k (which is some constant). Now we have

L\{k\} = \int_{0}^{\infty} \text{e}^{-sx}k \: \text{d}x
            = k\left[\frac{\text{e}^{-sx}}{-s}\right]_{0}^{\infty}
            = \frac{k}{s}, provided that s > 0.

In the same way, we can derive this result: L\{\text{e}^{-kx}\} = \frac{1}{s + k}, provided that s > -k.

Section 2

If \text{F}(s) is the Laplace transform of \text{f}(x), then \text{f}(x) is the inverse Laplace transform of \text{F}(s):
\text{f}(x) = L^{-1}\{\text{F}(s)\}.

The two results from Section 1 can thus be written as:
(a) L^{-1}\left\{\frac{k}{s}\right\} = k and
(b) L^{-1}\left\{\frac{1}{s + k}\right\} = \text{e}^{-kx}.

Before we look at some examples, we highlight two important properties of the Laplace transform and its inverse.

(1) L\{\text{f}(x) \pm \text{g}(x)\} = L\{\text{f}(x)\} \pm L\{\text{g}(x)\} and L^{-1}\{\text{F}(s) \pm \text{G}(s)\} = L^{-1}\{\text{F}(s)\} \pm L^{-1}\{\text{G}(s)\}.

(2) Let k be a constant. L\{k\text{f}(x)\} = kL\{\text{f}(x)\} and L^{-1}\{k\text{F}(s)\} = kL^{-1}\{\text{F}(s)\}.

Worked example 1: Find the inverse transform of \text{F}(s) = \frac{1}{2s-3}.

Now \text{F}(s) = \frac{1}{2}\left(\frac{1}{s-\frac{3}{2}}\right) and applying result (b) and property (2), we obtain \text{f}(x) = \frac{1}{2}\text{e}^{\frac{3}{2}x}.

Worked example 2: Find the inverse transform of \text{F}(s) = \frac{9}{s^2 + 3s}.

First, we express \text{F}(s) in partial fractions \frac{A}{s} + \frac{B}{s+3}, where A and B are constants we need to find.

Comparing the numerators, we have A(s+3) + Bs = 9.
Put s = 0, A = 3.
Put s = -3, B = -3.

Applying results (a) and (b) and property (1), we arrive at \text{f}(x) = 3 - 3\text{e}^{-3x}.

Section 3

We state two results involving the Laplace transforms of derivatives:

(1) If \text{F}(s) is the Laplace transform of \text{f}(x), then the Laplace transform of \text{f}\:'(x) is:

L\{\text{f}\:'(x)\} = s\text{F}(s) - \text{f}(0).
 
(2) If \text{F}(s) is the Laplace transform of \text{f}(x), then the Laplace transform of \text{f}\:''(x) is:
L\{\text{f}\:''(x)\} = s^2\text{F}(s) - s\text{f}(0) - \text{f}\:'(0).
 
We shall apply these results to help us generate new transforms.
 
Worked example 3: Use the fact that L\{sin \: kx\} = \frac{k}{s^2 + k^2} to find the Laplace transform of \text{f}(x) = cos^2\:3x.
 
We have \text{f}(0) = 1 and \text{f}\:'(x) = -6\:sin\:3x\:cos\:3x = -3\:sin\:6x.
By result (1), L\{-3\:sin\:6x\} = sL\{cos^2\:3x\} - 1
\Rightarrow \frac{-18}{s^2 + 36} = sL\{cos^2\:3x\} - 1
\Rightarrow L\{cos^2\:3x\} = \frac{s^2 + 18}{s(s^2 + 36)}.
 
Worked example 4: Find the Laplace transform of \text{f}(x) = x\:cos\:kx where k is a constant.
 
We have \text{f}(0) = 0, \text{f}\:'(x) = -kx\:sin\:kx + cos\:kx so that \text{f}\:'(0) = 1, and \text{f}\:''(x) = -2k\:sin\:kx - k^2x\:cos\:kx.
By result (2), L\{-2k\:sin\:kx - k^2x\:cos\:kx\} = s^2\:L\{x\:cos\:kx\} - 1
\Rightarrow -2k\left(\frac{k}{s^2 + k^2}\right) - k^2L\{x\:cos\:kx\} = s^2\:L\{x\:cos\:kx\} - 1, using the two properties in Section 2
\Rightarrow (s^2 + k^2)\:L\{x\:cos\:kx\} = \frac{-2k^2}{s^2 + k^2} + 1
\Rightarrow L\{x\:cos\:kx\} = \frac{s^2 - k^2}{(s^2 + k^2)^2}.

Section 4

Now we are poised to use the results in Section 3 to solve, with ease, first- and second-order DEs which are linear, constant-coefficient and inhomogeneous.

The key steps to follow are:
1. Apply the Laplace transform of both sides of the DE and use result (1) or (2) in Section 3. 

2. Find the expression \text{F}(s) = L\{\text{f}(x)\} in the form of an algebraic fraction.

3. Express \text{F}(s) in partial fractions.

4. Find the inverse Laplace transform L^{-1}\{\text{F}(s)\}, which is the solution to the DE.

Worked example 5: Solve the equation 3\text{f}\:'(x) - 2\text{f}(x) = x where \text{f}(0) = -2. Use the fact that L\{x\} = \frac{1}{s^2}.

By step 1, L\{3\text{f}\:'(x) - 2\text{f}(x)\} = L\{x\}.
Simplifying, we have 3L\{\text{f}\:'(x)\} - 2L\{\text{f}(x)\} = \frac{1}{s^2}.

By result (1) in Section 3, we obtain 3[s\text{F}(s) - \text{f}(0)] - 2\text{F}(s) = \frac{1}{s^2}.
Simplifying, we reach step 2 and we have \text{F}(s) = \frac{1-6s^2}{s^2(3s-2)}.

By step (3), \text{F}(s) = \frac{A}{s} + \frac{B}{s^2} + \frac{C}{3s-2}, where A, B and C are constants to be found.
Comparing numerators on both sides, we obtain As(3s-2) + B(3s-2) + Cs^2 = 1 - 6s^2.
Straightforward algebra yields A = -\frac{3}{4}, B = -\frac{1}{2}, C = -\frac{15}{4}.

By step (4), \text{f}(x) = L^{-1}\left\{-\frac{3}{4s} - \frac{1}{2s^2} - \frac{15}{4(3s-2)}\right\}

\Rightarrow \text{f}(x) = -\frac{3}{4} - \frac{x}{2} - \frac{5}{4}\text{e}^{\frac{2x}{3}}.

Worked example 6: Solve the equation \text{f}''(x) + 25\text{f}(x) = x where \text{f}(0) = 1 and \text{f}'(0) = \frac{1}{25}. Use the facts that L\{x\} = \frac{1}{s^2} and

L\{cos\:kx\} = \frac{s}{s^2 + k^2}.
By step 1, L\{\text{f}''(x)\} + 25L\{\text{f}(x)\} = \frac{1}{s^2}.
By result (2) in Section 3, we obtain s^2\text{F}(s) - s - 0.04 + 25\text{F}(s) = \frac{1}{s^2}.
At the end of step 2, we arrive at \text{F}(s) = \frac{1}{s^2(s^2 + 25)} + \frac{s}{s^2 + 25} + \frac{1}{25(s^2 + 25)}.
In step 3, only the term \frac{1}{s^2(s^2 + 25)} needs to be expressed in partial fractions \frac{A}{s} + \frac{B}{s^2} + \frac{Cs + D}{s^2 + 25}, where A, B, C and D are constants to be found.
By the algebraic method, A = C = 0, B = \frac{1}{25} and D = -\frac{1}{25}.
Now \text{F}(s) = \frac{1}{25s^2} + \frac{s}{s^2 + 25}.
By step 4, \text{f}(x) = \frac{1}{25}L^{-1}\left\{\frac{1}{s^2}\right\} + L^{-1}\left\{\frac{s}{s^2 + 25}\right\}
\Rightarrow \text{f}(x) = \frac{x}{25} + cos\:5x.