Contents

2. Structure of solutions

Definition 2.7 (Second-order ODE)

A second-order ODE is an equation of the form

\[ x'' = F(t,x,x'), \]

to be solved for \(x(t)\).

Note

Technically, this is not the most general form possible, since an equation in the implicit form \(F(t,x,x',x'')=0\) might be impossible to manipulate into an explicit expression for \(x''\). We won’t be considering such equations.

As with first-order problems, we can learn a little from the trivial case \(x''=0\). This ODE implies that \(x'\) is constant, so \(x(t)=c_1 t + c_2\) for arbitrary constants \(c_1,c_2\). While a generalization of this observation is far from obvious at this point, it is nonetheless true.

Note

The general solution of second-order ODE has two integration constants.

As a consequence, we note that two initial conditions are now needed to specify the integration constants to get a unique solution.

Definition 2.8 (Second-order IVP)

A second-order IVP (initial-value problem) is

\[x'' = F(t,x,x'), \quad x(a) = x_0,\, x'(a) = v_0.\]

2.1. Linear problems

We will focus on linear problems,

(2.5)\[x'' + p(t)x' + q(t) x = f(t).\]

Equation (2.5) is in the form of \(\opA[x]=f\) for the linear operator

\[\opA[x] = \ddd{x}{t} + p \dd{x}{t} + q x.\]

We now reap more rewards for using the abstract language of operators with first-order problems, because the connections between general solutions and homogeneous solutions is exactly the same as before. In particular, we still have a theorem and an algorithm that provide a blueprint for the solution:

  1. Find the general solution \(x_h\) of the associated homogeneous problem \(\opA[x]=0\).

  2. Find any particular solution \(x_p\) of the original \(\opA[x]=f\).

  3. Add them.

  4. If given, use initial conditions to solve for the integration constants.

2.2. Reduction of order

There is, however, a major new wrinkle regarding the homogeneous part of the solution. Suppose that \(x_1(t)\) is one homogeneous solution, and we set \(x_2(t)=v(t)x_1(t)\) for an undetermined \(v(t)\). A little elbow grease shows that

\[ \opA[vx_1] = x_1 v'' + 2 x_1' v' + p x_1 v' + \opA[x_1]. \]

We assumed that \(\opA[x_1]=0\). So we will also get \(\opA[x_2]=0\) if

\[ x_1 v'' + 2x_1' v' + p x_1 v' = 0. \]

This might not look important until we make one more definition, \(w=v'\), and rearrange:

\[ w' + \left( 2 \frac{x_1'}{x_1} + p\right) w = 0. \]

That is a linear, first-order equation for \(w\)! We can solve that for \(w\), then find \(v\) by integrating \(w\), and then we have a new homogeneous solution, \(x_2\). This technique is called reduction of order, because we pulled a first-order equation out of a second-order one.

Example

One solution of

\[ t^2 x'' + 2t x' + 2x = 0 \]

is \(x_1(t)=t\). Find another solution.

Solution

Rather than going through the entire process again, let’s plug into the formulas. We have \(p(t)=2/t\) and \(q(t)=2/t^2\). The equation for \(w\) is

\[ w' + \left( 2 \frac{1}{t} + \frac{2}{t} \right)w = 0. \]

Hence

\[ w(t) = \exp \left[ \int -\frac{4}{t}\, dt \right] = Ct^{-4}. \]

Then we get

\[ v(t) = C\int t^{-4}\, dt = k_1 t^{-3} + k_2. \]

We only need to get some example of \(x_2\), so let’s go easy on ourselves and choose \(v(t)=t^{-3}\). We finally get

\[ x_2(t) = v(t) x_1(t) = t^{-2}. \]

You can check that this is indeed a solution to the original homogeneous ODE.

The reduction of order technique suggests an important fact.

Theorem 2.9

The general solution of a homogeneous, linear, second-order ODE is of the form

(2.6)\[x_h(t) = c_1 x_1(t) + c_2 x_2(t),\]

for arbitrary constants \(c_1\) and \(c_2\).

It’s too deep for us to go into a true proof of this theorem. You might be tempted to do another reduction of order starting from \(x_2\) in order to get a third solution. Nice try, but you don’t get any new terms that change (2.6).

1. Overview 3. Homogeneous solutions