# A calculus of the absurd

##### 14.4.3 Why two linearly independent solutions?

^{101}^{101} If two vectors (which we can call \(v_1\) and \(v_2\)) are linearly independent, then the only values of \(\alpha _1\) and \(\alpha _2\) which solve the equation \(\alpha _1 v_1 + \alpha _2 v_2 =
0\) are \(\alpha _1 = \alpha _2 = 0\). If there were a different combination, then we could write that \(v_1 = -\frac {\alpha _2}{\alpha _1} v_2\) And thus they wouldn’t be linearly independent, as one of the vectors is a multiple of the other. This can be generalised to
\(n\) vectors.

Let’s suppose we have a differential equation involving \(y(x)\) and its first and second derivative. We may want to solve this subject to the initial conditions

\(\seteqnumber{0}{14.}{10}\)\begin{equation} \label {initial conditions} y(x_0) = y_0 \text { and } \frac {dy}{dx}\Bigr |_{\substack {x=x_0}} = \dot {y}_0 \end{equation}

In order to be able to solve for *every* possible initial condition, we need a linear combination of two "linearly independent" solutions, using which we can satisfy any possible initial condition. This means that if we have two solutions \(y_1(x)\) and \(y_2(x)\) and neither can
be written as a multiple of the other, then the "general solution" (i.e. the one with the constants in it, like \(+c\) for first-order differential equations when we integrate them) is of the form

\begin{equation} y(x) = \alpha y_1(x) + \beta y_2(x) \end{equation}

The good news is that for homogenous second-order differential equations we do have two linearly independent solutions!^{102}^{102} Except in the case where \(\Delta =0\), but there’s a way to get around that (more
on that later). For example if our roots of the auxiliary are \(\alpha \pm \beta \), then we would have that the two solutions are

^{103}^{103} Which are linearly independent, as one of them cannot be written as a multiple (of something linear) of the other.

\begin{equation} y(x) = e^{(\alpha + \beta )x} \text { and } y(x) = e^{(\alpha - \beta )x} \end{equation}

**Proof that we need two linearly independent solutions, and that two linearly independent solutions are sufficient to solve any initial condition (note: no A Level exam board examines this).** The reason we *need* two linearly independent solutions is that if we
have two sets of linearly independent initial conditions we certainly can’t write both of them the as linear combinations of a single solution. ^{104}^{104} This is a *lot* like how we can write any vector in a 2D space
in terms of two vectors, so long as those vectors are not parallel (same for 3D space, except with three vectors). For example, suppose we have two possible initial conditions, one where

\begin{equation} \label {initial conditions a} y(x_0) = 1 \text { and } \frac {dy}{dx}\Bigr |_{\substack {x=x_0}} = 0 \end{equation}

and another where

\(\seteqnumber{0}{14.}{14}\)\begin{equation} \label {initial conditions b} y(x_0) = 0 \text { and } \frac {dy}{dx}\Bigr |_{\substack {x=x_0}} = 1 \end{equation}

These are linearly independent solutions, and we can’t write them both as a linear combination of a single solution. Thus, there is at least one case where we need at least two linearly independent solutions.

The next thing to prove is that if we have two solutions (\(y_1(x)\) and \(y_2(x)\)) which are linearly independent, then for suitable values of \(\alpha \) and \(\beta \) we can satisfy any initial condition using a solution of the form

\(\seteqnumber{0}{14.}{15}\)\begin{equation} \label {general solution with alpha and beta} y(x) = \alpha y_1(x) + \beta y_2(x) \end{equation}

Does this even solve the differential equation though? Yes! We can prove this with a bunch of algebra (there’s an easier way to do this by introducing some new notation, but that’s for later^{105}^{105} Note: I have yet
to write about this new notation)

\begin{align} y = \alpha y_1(x) + \beta y_2(x) & \implies \frac {dy}{dx} = \alpha \frac {dy_1}{dx} + \beta \frac {dy_2}{dx} \\ & \implies \frac {d^2y}{dx^2} = \alpha \frac {d^2y_1}{dx^2} + \beta \frac {d^2y_2}{dx^2} \end{align}

and thus that

^{106}^{106} Note that both \(\alpha [a\frac {d^2y_1}{dx^2} + b\frac {dy_1}{dx} + cy_1]\) and \(\beta [a\frac {d^2y_2}{dx^2} + b\frac {dy_2}{dx} + cy_2]\) are zero, because we know that
\(y_1(x)\) and \(y_2(x)\) solve the equation \(a\frac {d^2y}{dx^2} + b\frac {dy}{dx} + cy = 0\).

And thus \(y=\alpha y_1(x)+\beta y_2(x)\) is a solution to Equation 14.9

If that was messy, it gets worse. ^{107}^{107} I’ve had nightmares about drowning in a differential equation algebra soup. Literal soup made of algebra - it was a *very* strange dream.

From Equation 14.16 we know that the derivative of our solution will be

\(\seteqnumber{0}{14.}{22}\)\begin{equation} \frac {dy}{dx} = \alpha \frac {dy_1}{dx} + \alpha \frac {dy_2}{dx} \end{equation}

as we want to show that this can satisfy any set of initial conditions, where \(y(x_0) = y_0\) and \(\dot {y}(x_0) = \dot {y}_0\), we can start by writing a set of simultaneous equations

\(\seteqnumber{0}{14.}{23}\)\begin{equation} \begin{cases} \alpha y_1(x_0) + \beta y_2(x_0) = y_0 \\ \alpha \dot {y}_1(x_0) + \beta \dot {y}_2(x_0) = \dot {y}_0 \end {cases} \end{equation}

we can write these in matrix form and obtain that

\(\seteqnumber{0}{14.}{24}\)\begin{equation} \begin{pmatrix} y_1(x_0) & y_2(x_0) \\ \dot {y}_1(x_0) & \dot {y}_2(x_0) \end {pmatrix} \begin{pmatrix} \alpha \\ \beta \end {pmatrix} = \begin{pmatrix} y_0 \\ \dot {y}_1 \end {pmatrix} \end{equation}

These equations can be solved whenever the determinant of the 2x2 matrix above is not equal to zero, i.e. whenever

\(\seteqnumber{0}{14.}{25}\)\begin{equation} y_1(x_0) \dot {y}_2(x_0) - y_2(x_0) \dot {y}_1(x_0) \ne 0 \end{equation}

Then the two equations have a unique solution. The easiest way to go from here is to prove this by contradiction (as we have lots of techniques for dealing with equalities (\(=\)) and not very many for dealing with inequalities involving \(\ne \)). We can proceed by assuming that
two linearly independent solutions are *not* sufficient to determine the general solution of any second-order differential equation, and write that

\begin{equation} y_1(x_0) \dot {y}_2(x_0) - y_2(x_0) \dot {y}_1(x_0) = 0 \end{equation}

We can now manipulate this a little

\(\seteqnumber{0}{14.}{27}\)\begin{equation} y_1(x_0) \dot {y}_2(x_0) = y_2(x_0) \dot {y}_1(x_0) \end{equation}

Dividing through leads to the formula

\(\seteqnumber{0}{14.}{28}\)\begin{equation} \frac {y_1(x_0)}{y_2(x_0)} = \frac {\dot {y}_1(x_0)}{\dot {y}_2(x_0)} \end{equation}

Here, though it looks like we have a contraction. Why? Let’s set

\(\seteqnumber{0}{14.}{29}\)\begin{equation} c = \frac {y_1(x_0)}{y_2(x_0)} \end{equation}

and

\(\seteqnumber{0}{14.}{30}\)\begin{equation} d = \frac {\dot {y}_1(x_0)}{\dot {y}_2(x_0)} \end{equation}

Then we can write that

\(\seteqnumber{0}{14.}{31}\)\begin{equation} \label {but they're not linearly independent} y_1(x_0) = c y_2(x_0) \end{equation}

and that

\(\seteqnumber{0}{14.}{32}\)\begin{equation} \dot {y}_1(x_0) = d \dot {y}_2(x_0) \end{equation}

But we specified earlier that \(y_1(x)\) and \(y_2(x)\) are linearly independent! And now we’ve found that in Equation 14.32 that they’re *not* linearly
independent, and thus we’ve found that assuming that two linearly independent solutions *doesn’t* solve any set of initial conditions for a second-order differential equation leads to a contradiction. Hence, two linearly independent solutions *are* sufficient.