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2.2 2nd order homogeneous linear differential equations with constant coefficients I

We recall that this type of equation has the form

\begin{eqnarray} y''(x) + py'(x) + qy(x) = 0,& & %&(2.8) \\ \end{eqnarray}

where p and q are real numbers, and y(x) is the function one would like to calculate. An example is the differential equation of the damped linear harmonic oscillator

\begin{eqnarray} \ddot{x}(t) + {γ\over m}\dot{x}(t) + {k\over m}x(t)& =& 0,\quad k > 0,γ > 0,%&(2.9) \\ \end{eqnarray}

cf. Eq.(2.1).

2.2.1 Undamped oscillator

Consider

y''(x) + qy(x) = 0,q > 0.

This is the case p = 0 of Eq. (2.8). An example for this is the differential equation of the undamped linear harmonic oscillator

\begin{eqnarray} \ddot{x}(t) + {k\over m}x(t)& =& 0,%&(2.10) \\ \end{eqnarray}

where k > 0 here, cf. Eq.(2.1). From our physical intuition, we know that the mass point described by Eq.(2.10) performs oscillations at an angular frequency ω. Therefore, we try \mathop{sin}\nolimits and \mathop{cos}\nolimits functions as solution: If we write

\begin{eqnarray} x(t)& =& {x}_{1}\mathop{ sin}\nolimits (ωt) ⇝\dot{ x}(t) = {x}_{1}ω\mathop{cos}\nolimits (ωt)%& \\ ⇝\ddot{ x}(t)& =& −{x}_{1}{ω}^{2}\mathop{ sin}\nolimits (ωt) = −{ω}^{2}x(t). %&(2.11) \\ \end{eqnarray}

Here, {x}_{1} is an arbitrary constant. The function x(t) = {x}_{1}\mathop{ sin}\nolimits (ωt) fulfills the differential equation Eq. (2.10), if

\begin{eqnarray}{ ω}^{2} = {k\over m}.& & %&(2.12) \\ \end{eqnarray}

If on the other hand we write

\begin{eqnarray} x(t)& =& {x}_{2}\mathop{ cos}\nolimits (ωt) ⇝\dot{ x}(t) = −{x}_{2}ω\mathop{sin}\nolimits (ωt)%& \\ ⇝\ddot{ x}(t)& =& −{x}_{2}{ω}^{2}\mathop{ cos}\nolimits (ωt) = −{ω}^{2}x(t), %&(2.13) \\ \end{eqnarray}

we again recognise that the functionx(t) = {x}_{2}\mathop{ cos}\nolimits (ωt) fulfills the differential equation Eq. (2.10), if {ω}^{2} = k∕m (same as before). Again, {x}_{2} is an arbitrary constant. Therefore, we find two solutions of the second order differential equation Eq. (2.10). Now we are a bit confused. Let us summarize what we have found so far, using our ‘mathematical notation’,

\begin{eqnarray} y''(x) + qy(x)& =& 0,\quad q > 0 ⇝ %& \\ y(x)& =& {y}_{1}(x) = {y}_{1}\mathop{ sin}\nolimits (\sqrt{q}x),\quad y(x) = {y}_{2}(x) = {y}_{2}\mathop{ cos}\nolimits (\sqrt{q}x).%&(2.14) \\ \end{eqnarray}

We now make an important observation:

THEOREM: With two solutions {y}_{1}(x) and {y}_{2}(x) of a linear homogeneous differential equation, also the sum {y}_{1}(x) + {y}_{2}(x) is a solution of the linear homogeneous differential equation.

PROOF:

\begin{eqnarray*}{ y}_{1}''(x) + p(x){y}_{1}'(x) + q(x){y}_{1}(x) = 0,{y}_{2}''(x) + p(x){y}_{2}'(x) + q(x){y}_{2}(x)& =& 0 ⇝ %& \\ \left [{y}_{1}''(x) + {y}_{2}''(x)\right ] + p(x)\left [{y}_{1}'(x) + {y}_{2}'(x)\right ] + q(x)\left [{y}_{1}(x) + {y}_{2}(x)\right ]& =& 0 ⇝ %& \\ \left [{y}_{1} + {y}_{2}\right ]''(x) + p(x)\left [{y}_{1} + {y}_{2}\right ]'(x) + q(x)\left [{y}_{1}(x) + {y}_{2}(x)\right ]& =& 0. %& \\ \end{eqnarray*}

We have used the fact that the sum of the derivatives of two functions is the derivative of the sum of the functions.

The general solution of y''(x) + qy(x) = 0,q > 0 can be written as the sum

\begin{eqnarray} y''(x) + qy(x) = 0 ⇝ y(x) = {y}_{1}\mathop{ sin}\nolimits (\sqrt{q}x) + {y}_{2}\mathop{ cos}\nolimits (\sqrt{q}x).& & %&(2.15) \\ \end{eqnarray}

2.2.2 Initial Value Problem

Here we consider

\ddot{x}(t) + {ω}^{2}x(t) = 0,

the equation of the undamped linear harmonic oscillator. Note that we write x(t) instead of y(x) here. We have found the general solution as

\begin{eqnarray} x(t)& =& {x}_{1}\mathop{ sin}\nolimits (ωt) + {x}_{2}\mathop{ cos}\nolimits (ωt),%&(2.16) \\ \end{eqnarray}

where x(t) is the position x at time t. As mentioned above, in order to know x(t) at all times later than, say, t = 0, we must specify the initial conditions, i.e. the initial position of the particle {x}_{0} = x(t = 0) and its initial velocity {v}_{0} =\dot{ x}(t = 0), i.e.

\begin{eqnarray}{ x}_{0} = x(t = 0)& =& {x}_{1}\mathop{ sin}\nolimits (ω0) + {x}_{2}\mathop{ cos}\nolimits (ω0) = {x}_{2} %& \\ {v}_{0} =\dot{ x}(t = 0)& =&{ \left .{x}_{1}ω\mathop{cos}\nolimits (ωt) − {x}_{2}ω\mathop{sin}\nolimits (ωt)\right |}_{t=0} = {x}_{1}ω.%&(2.17) \\ \end{eqnarray}

Therefore, we can express the parameters {x}_{1} and {x}_{2} by the given initial values {x}_{0} and {v}_{0} and obtain

\begin{eqnarray} x(t)& =& {{v}_{0}\over ω} \mathop{sin}\nolimits (ωt) + {x}_{0}\mathop{ cos}\nolimits (ωt).%&(2.18) \\ \end{eqnarray}

2.2.3 Exponential

Consider

y''(x) + qy(x) = 0,q < 0.

We notice that for q < 0 the argument in the \mathop{sin}\nolimits and \mathop{cos}\nolimits in Eq.(2.15) becomes imaginary since \sqrt{q} = \sqrt{−|q|} = i\sqrt{|q|} for q < 0. Let us find a solution by recalling that the exponential function f(x) =\mathop{ exp}\nolimits (x) fulfills

\begin{eqnarray} f(x) = {e}^{x} ⇝ f'(x) = {e}^{x} ⇝ f''(x) = {e}^{x} ⇝ ...& & %&(2.19) \\ \end{eqnarray}

More generally, we have

\begin{eqnarray} f(x)& =& {e}^{λx} ⇝ f'(x) = λ{e}^{λx} ⇝ f''(x) = {λ}^{2}{e}^{λx} ⇝ f''(x) = {λ}^{2}f(x) %&(2.20) \\ f(x)& =& {e}^{−λx} ⇝ f'(x) = −λ{e}^{−λx} ⇝ f''(x) = {(−λ)}^{2}{e}^{−λx} ⇝ f''(x) = {λ}^{2}f(x).%& \\ \end{eqnarray}

Comparing this to our differential equation,

\begin{eqnarray} y''(x) −|q|y(x) = 0 ⇔ y''(x) = |q|y(x),& & %&(2.21) \\ \end{eqnarray}

we recognize by comparing with Eq. (2.20) that two independent solutions of Eq. (2.21) are

\begin{eqnarray} y''(x) −|q|y(x)& =& 0,\quad q\mathrel{≠}0 ⇝ %& \\ {y}_{1}(x)& =& {y}_{1}{e}^{\sqrt{|q|}x},\quad {y}_{ 2}(x) = {y}_{2}{e}^{−\sqrt{|q|}x}.%&(2.22) \\ \end{eqnarray}

As above, the most general solution again is the sum of these two, i.e. the linear combination of {e}^{−\sqrt{|q|}x} and {e}^{\sqrt{|q|}x} with the two independent constants {y}_{1} and {y}_{2},

\begin{eqnarray} y''(x) −|q|y(x)& =& 0 ⇝ %& \\ y(x)& =& {y}_{1}{e}^{\sqrt{|q|}x} + {y}_{ 2}{e}^{−\sqrt{|q|}x}.%&(2.23) \\ \end{eqnarray}

2.2.4 Summary

We summarize the two pairs of solutions for q > 0 and q = −|q| < 0 of y''(x) + qy(x) = 0 in the following table:

y''(x) + qy(x) = 0,\quad q = {k}^{2} > 0 y''(x) + qy(x) = 0,\quad q = −{κ}^{2} < 0
two solutions two solutions
{y}_{1}(x) = {y}_{1}\mathop{ sin}\nolimits (kx), {y}_{2}(x) = {y}_{2}\mathop{ cos}\nolimits (kx) {y}_{1}(x) = {y}_{1}{e}^{κx}, {y}_{ 2}(x) = {y}_{2}{e}^{−κx}
general solution general solution
y(x) = {y}_{1}\mathop{ sin}\nolimits (kx) + {y}_{2}\mathop{ cos}\nolimits (kx) y(x) = {y}_{1}{e}^{κx} + {y}_{ 2}{e}^{−κx}
character: character:
oscillatory (\mathop{sin}\nolimits and \mathop{cos}\nolimits )exponential (decreasing and incr.)

Note that the sign of q makes all the difference!

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