Solving Second Order Linear D.E.'s With Constant Coefficients

Consider the homogeneous D.E.

ay'' + by' + cy = 0

Let us suppose that we have a solution of the form
y(t) = e^rt. What would r have to be?

$$\begin{eqnarray*} ay'' + by' + cy & = & ar^2e^{rt} + bre^{rt} + ce^{rt}\\ & = & (ar^2 + br + c)e^{rt} = 0\\ &\Longrightarrow & ar^2 + br + c = 0 \end{eqnarray*}$$

The equation ar² + br + c = 0 is called the
characteristic equation of the D.E.

There are several cases to consider when solving the
characteristic equation:

1.

The equation has two real, distinct roots

2.

The equation has one real, repeated root

3.

The equation has two complex, distinct roots

Case 1: Real, distinct roots

Suppose r₁ and r₂ are roots of the characteristic equation ar² + br + c = 0.

We will show that y₁(t) = e^r₁t and y₂(t) = e^r₂t are both solutions to the D.E.

ay'' + by' + cy = 0.

Let's try y₁(t):

a(e^r₁t)'' + b(e^r₁t)' + c(e^r₁t) = (ar₁² + br₁ + c) e^r₁t = 0

since r₁ is a root of ar² + br + c.

Hence, y₁(t) = e^r₁t is a solution (and similarly,

y₂(t) = e^r₂t is also a solution).

Example:

Find the general solution to the equation

y'' + 5y' + 6y = 0.

We observe that the characteristic equation is

r² + 5r + 6 = (r + 2)(r + 3),

hence

$$y_1(t) = e^{-2t} \quad \textrm{and} \quad y_2(t) = e^{-3t}.$$

So to construct the general solution, we invoke the
Principle of Superposition and take a linear combination of y₁ and y₂ to obtain

y = c₁ e^-2t + c₂ e^-3t

Note: the solutions y₁(t) and y₂(t) possess a certain relationship called linear independence that allows us to state that all solutions are of the form c₁y₁(t) + c₂y₂(t) - more on this later...

Case 2: repeated root

Again, we have the second order homogeneous equation

ay'' + by' + cy = 0.

Suppose that when we examine the characteristic equation ar² + br + c = 0, we find that it has one real root, r₁ of multiplicity 2.

Note: in order to obtain repeated roots, the discriminant b² - 4ac must be zero; hence when we use the quadratic formula we obtain

$$r_1 = \frac{-b \pm \sqrt{b^2-4ac}}{2a} = \frac{-b}{2a}$$

Using the same derivation as for the previous case, we can show that y₁(t) = e^r₁t is a solution.

However, second order D.E.'s generally have two ``different'' solutions: what can we do to obtain a second solution?

We guess.

It turns out that y₂(t) = te^r₁t is also a solution. Let's verify that this is the case:

$$\begin{eqnarray*} a(te^{r_1t})'' & + & b(te^{r_1t})' + c(te^{r_1t})\\ [.1in] & =... ... + [2a\left(\frac{-b}{2a}\right) + b] e^{r_1t}\\ [.1in] & = & 0. \end{eqnarray*}$$

Clever guess, eh?

So now we can write the general solution in the case where the characteristic equation has repeated roots:

y(t) = c₁e^r₁t + c₂te^r₁t

Example:

Find the solution to the initial value problem

$$4y'' + 36y' + 36y = 0, \hspace{.5in} y(0) = 5 \hspace{.5in} y'(0) = -22$$

First, we identify the solutions to the characteristic equation:

$$\begin{eqnarray*} % latex2html id marker 864r^2 + 64r + 64 & = & 0\\ \Longrightarrow r^2 + 16r + 16 & = & 0\\ \therefore \quad r & = & - 4. \end{eqnarray*}$$

So our general solution is

y(t) = c₁e^-4t + c₂te^-4t

Now, using the first initial condition y(0) = 5, we obtain

y(0) = c₁e⁰ + c₂(0)e⁰ = c₁ = 5

So our solution now looks like

y(t) = 5e^-4t + c₂te^-4t.

Now, in order to use the second initial condition, we must first compute the derivative of our solution, y(t) (plugging in the value of c₁ we just computed).

$$\begin{eqnarray*} y(t) & = & 5e^{-4t} + c_2te^{-4t}\\ y'(t) & = & -20e^{-4t} + c_2e^{-4t} - 4c_2te^{-4t} \end{eqnarray*}$$

Now, apply the second initial condition, y'(0) = 1 to obtain

y'(0) = -20e⁰ + c₂e⁰ -4c₂(0)e⁰ = -20 + c₂ = -22

$$% latex2html id marker 191 \therefore \quad c_2 = -2$$

So, the solution to the initial value problem is

y(t) = 5e^-4t - 2te^-4t.