Differential Equations - Second Order: Repeated Roots

In this section we will be looking at the last case for the constant coefficient, linear, homogeneous second order differential equations. In this case we want solutions to

a y^{″} + b y^{'} + c y = 0

where solutions to the characteristic equation

a r^{2} + b r + c = 0

are double roots

r_{1} = r_{2} = r

This leads to a problem however. Recall that the solutions are

y_{1} (t) = e^{r_{1} t} = e^{r t} y_{2} (t) = e^{r_{2} t} = e^{r t}

These are the same solution and will NOT be “nice enough” to form a general solution. We do promise that we’ll define “nice enough” eventually! So, we can use the first solution, but we’re going to need a second solution.

Before finding this second solution let’s take a little side trip. The reason for the side trip will be clear eventually. From the quadratic formula we know that the roots to the characteristic equation are,

r_{1, 2} = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}

In this case, since we have double roots we must have

b^{2} - 4 a c = 0

This is the only way that we can get double roots and in this case the roots will be

r_{1, 2} = \frac{- b}{2 a}

So, the one solution that we’ve got is

y_{1} (t) = e^{- \frac{b t}{2 a}}

To find a second solution we will use the fact that a constant times a solution to a linear homogeneous differential equation is also a solution. If this is true then maybe we’ll get lucky and the following will also be a solution

\begin{matrix}  \end{matrix}

with a proper choice of

v (t)

. To determine if this in fact can be done, let’s plug this back into the differential equation and see what we get. We’ll first need a couple of derivatives.

\begin{aligned} {y^{'}}_{2} (t) & = v^{'} e^{- \frac{b t}{2 a}} - \frac{b}{2 a} v e^{- \frac{b t}{2 a}} \\ {y^{″}}_{2} (t) & = v^{″} e^{- \frac{b t}{2 a}} - \frac{b}{2 a} v^{'} e^{- \frac{b t}{2 a}} - \frac{b}{2 a} v^{'} e^{- \frac{b t}{2 a}} + \frac{b^{2}}{4 a^{2}} v e^{- \frac{b t}{2 a}} \\ = v^{″} e^{- \frac{b t}{2 a}} - \frac{b}{a} v^{'} e^{- \frac{b t}{2 a}} + \frac{b^{2}}{4 a^{2}} v e^{- \frac{b t}{2 a}} \end{aligned}

We dropped the

(t)

part on the

v

to simplify things a little for the writing out of the derivatives. Now, plug these into the differential equation.

a (v^{″} e^{- \frac{b t}{2 a}} - \frac{b}{a} v^{'} e^{- \frac{b t}{2 a}} + \frac{b^{2}}{4 a^{2}} v e^{- \frac{b t}{2 a}}) + b (v^{'} e^{- \frac{b t}{2 a}} - \frac{b}{2 a} v e^{- \frac{b t}{2 a}}) + c (v e^{- \frac{b t}{2 a}}) = 0

We can factor an exponential out of all the terms so let’s do that. We’ll also collect all the coefficients of

v

and its derivatives.

\begin{aligned} e^{- \frac{b t}{2 a}} (a v^{″} + (- b + b) v^{'} + (\frac{b^{2}}{4 a} - \frac{b^{2}}{2 a} + c) v) & = 0 \\ e^{- \frac{b t}{2 a}} (a v^{″} + (- \frac{b^{2}}{4 a} + c) v) & = 0 \\ e^{- \frac{b t}{2 a}} (a v^{″} - \frac{1}{4 a} (b^{2} - 4 a c) v) & = 0 \end{aligned}

Now, because we are working with a double root we know that that the second term will be zero. Also exponentials are never zero. Therefore,

(1)

will be a solution to the differential equation provided

v (t)

is a function that satisfies the following differential equation.

a v^{″} = 0 OR v^{″} = 0

We can drop the

a

because we know that it can’t be zero. If it were we wouldn’t have a second order differential equation! So, we can now determine the most general possible form that is allowable for

v (t)

v^{'} = \int v^{″} d t = c v (t) = \int v^{'} d t = c t + k

This is actually more complicated than we need and in fact we can drop both of the constants from this. To see why this is let’s go ahead and use this to get the second solution. The two solutions are then

y_{1} (t) = e^{- \frac{b t}{2 a}} y_{2} (t) = (c t + k) e^{- \frac{b t}{2 a}}

Eventually you will be able to show that these two solutions are “nice enough” to form a general solution. The general solution would then be the following.

\begin{aligned} y (t) & = c_{1} e^{- \frac{b t}{2 a}} + c_{2} (c t + k) e^{- \frac{b t}{2 a}} \\ = c_{1} e^{- \frac{b t}{2 a}} + (c_{2} c t + c_{2} k) e^{- \frac{b t}{2 a}} \\ = (c_{1} + c_{2} k) e^{- \frac{b t}{2 a}} + c_{2} c t e^{- \frac{b t}{2 a}} \end{aligned}

Notice that we rearranged things a little. Now,

c

k

c_{1}

, and

c_{2}

are all unknown constants so any combination of them will also be unknown constants. In particular,

c_{1} + c_{2} k

and

c_{2} c

are unknown constants so we’ll just rewrite them as follows.

y (t) = c_{1} e^{- \frac{b t}{2 a}} + c_{2} t e^{- \frac{b t}{2 a}}

So, if we go back to the most general form for

v (t)

we can take

c = 1

and

k = 0

and we will arrive at the same general solution.

Let’s recap. If the roots of the characteristic equation are

r_{1} = r_{2} = r

, then the general solution is then

y (t) = c_{1} e^{r t} + c_{2} t e^{r t}

Now, let’s work a couple of examples.

Example 1 Solve the following IVP.

y^{″} - 4 y^{'} + 4 y = 0 y (0) = 12 y^{'} (0) = - 3

The characteristic equation and its roots are.

r^{2} - 4 r + 4 = {(r - 2)}^{2} = 0 r_{1, 2} = 2

The general solution and its derivative are

\begin{aligned} y (t) & = c_{1} e^{2 t} + c_{2} t e^{2 t} \\ y^{'} (t) & = 2 c_{1} e^{2 t} + c_{2} e^{2 t} + 2 c_{2} t e^{2 t} \end{aligned}

Don’t forget to product rule the second term! Plugging in the initial conditions gives the following system.

\begin{aligned} 12 & = y (0) = c_{1} \\ - 3 & = y^{'} (0) = 2 c_{1} + c_{2} \end{aligned}

This system is easily solved to get

c_{1} = 12

and

c_{2} = - 27

. The actual solution to the IVP is then.

y (t) = 12 e^{2 t} - 27 t e^{2 t}

Example 2 Solve the following IVP.

16 y^{″} - 40 y^{'} + 25 y = 0 y (0) = 3 y^{'} (0) = - \frac{9}{4}

The characteristic equation and its roots are.

16 r^{2} - 40 r + 25 = {(4 r - 5)}^{2} = 0 r_{1, 2} = \frac{5}{4}

The general solution and its derivative are

\begin{aligned} y (t) & = c_{1} e^{\frac{5 t}{4}} + c_{2} t e^{\frac{5 t}{4}} \\ y^{'} (t) & = \frac{5}{4} c_{1} e^{\frac{5 t}{4}} + c_{2} e^{\frac{5 t}{4}} + \frac{5}{4} c_{2} t e^{\frac{5 t}{4}} \end{aligned}

Don’t forget to product rule the second term! Plugging in the initial conditions gives the following system.

\begin{aligned} 3 = y (0) & = c_{1} \\ - \frac{9}{4} = y^{'} (0) & = \frac{5}{4} c_{1} + c_{2} \end{aligned}

This system is easily solve to get

c_{1} = 3

and

c_{2} = - 6

. The actual solution to the IVP is then.

y (t) = 3 e^{\frac{5 t}{4}} - 6 t e^{\frac{5 t}{4}}

Example 3 Solve the following IVP

y^{″} + 14 y^{'} + 49 y = 0 y (- 4) = - 1 y^{'} (- 4) = 5

The characteristic equation and its roots are.

r^{2} + 14 r + 49 = {(r + 7)}^{2} = 0 r_{1, 2} = - 7

The general solution and its derivative are

\begin{aligned} y (t) & = c_{1} e^{- 7 t} + c_{2} t e^{- 7 t} \\ y^{'} (t) & = - 7 c_{1} e^{- 7 t} + c_{2} e^{- 7 t} - 7 c_{2} t e^{- 7 t} \end{aligned}

Plugging in the initial conditions gives the following system of equations.

\begin{aligned} - 1 & = y (- 4) = c_{1} e^{28} - 4 c_{2} e^{28} \\ 5 & = y^{'} (- 4) = - 7 c_{1} e^{28} + c_{2} e^{28} + 28 c_{2} e^{28} = - 7 c_{1} e^{28} + 29 c_{2} e^{28} \end{aligned}

Solving this system gives the following constants.

c_{1} = - 9 e^{- 28} c_{2} = - 2 e^{- 28}

The actual solution to the IVP is then.

\begin{matrix} y (t) & = - 9 e^{- 28} e^{- 7 t} - 2 t e^{- 28} e^{- 7 t} y (t) & = - 9 e^{- 7 (t + 4)} - 2 t e^{- 7 (t + 4)} \end{matrix}

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