DE-BOOK 2nd-Order Equations

Section 6.4 2nd-Order Equations

Now that we’ve mastered first-order LHCC equations, it’s time to explore second-order LHCC equations. These are slightly more complex but follow a similar approach. We’ll see how the characteristic equation helps us find solutions and look at the different cases that arise depending on the nature of the roots.

Consider the following second-order equation:

\begin{matrix} (27) & y^{″} - 5 y^{'} + 6 y = 0 \end{matrix}

Substituting

y = e^{r x}

into this equation, we get:

\begin{aligned} {[e^{r x}]}^{″} - 5 {[e^{r x}]}^{'} + 6 [e^{r x}] = & 0 \\ r^{2} e^{r x} - 5 r e^{r x} + 6 e^{r x} = & 0 \\ (r^{2} - 5 r + 6) e^{r x} = & 0 (e^{r x} \neq 0) \\ characteristic equation \to r^{2} - 5 r + 6 = & 0 \\ r_{1} = 2, r_{2} = & 3 . \end{aligned}

Therefore,

y = e^{2 x}

and

y = e^{3 x}

are solutions, and the general solution is the combination of these two:

y = C_{1} e^{2 x} + C_{2} e^{3 x} .

In general, for any second-order LHCC equation of the form:

\begin{matrix} (28) & a y^{″} + b y^{'} + c y = 0 \end{matrix}

The characteristic equation is:

\begin{matrix} (29) & a r^{2} + b r + c = 0 \end{matrix}

This quadratic equation will have two solutions, which could be either real or complex. The general solution depends on the nature of these solutions, and we will now explore the different cases. Before moving on, we highly recommend taking a minute to review some concepts related to solving quadratic equations.

Let

r_{1}

and

r_{2}

be the solutions to the characteristic equation. When

r_{1} \neq r_{2},

the general solution is:

\begin{matrix} (30) & y = C_{1} e^{r_{1} x} + C_{2} e^{r_{2} x} \end{matrix}

However, if

r_{1} = r_{2},

the terms combine into like terms,

y = C_{1} e^{r_{1} x} + C_{2} e^{r_{2} x} = \underset{like terms}{\underset{↖}{\underset{―}{C_{1} e^{r_{1} x}}} + \underset{↗}{\underset{―}{C_{2} e^{r_{1} x}}}} = (\underset{C}{\underset{⏟}{C_{1} + C_{2}}}) e^{r_{1} x},

which is an incomplete solution. To fix this, we multiply the second term by

x,

resulting in the correct general solution:

⁶

See Exercise 6.6.1 for an outline of why the extra

x

is required in the general solution of a second order LHCC equation when

r_{1} = r_{2} .

y = C_{1} e^{r_{1} x} + C_{2} x e^{r_{1} x}

Although, the above explanation covers the situation when

r_{1}

and

r_{2}

are equal or not, we will split the

r_{1} \neq r_{2}

case into two cases that treat real or complex values separately. The three cases are summarized below.

Summary of Cases.

Let

r_{1}

and

r_{2}

be the solutions of the characteristic equation associated with the second-order LHCC equation. The general solution is:

Case 1 (real & $\neq$ ): $y = C_{1} e^{r_{1} x} + C_{2} e^{r_{2} x} (r_{1} \neq r_{2})$
Case 2 (real & $=$ ): $y = (C_{1} + C_{2} x) e^{r_{1} x} (r_{1} = r_{2})$
Case 3 (complex): $y = e^{α x} (C_{1} \cos (β x) + C_{2} \sin (β x))$
$(r = α \pm i β ⟹ r_{1} = α + i β, r_{2} = α - i β)$

Let’s now practice solving some second-order LHCC equations.

Example 6.

Find the general solutions for each LHCC equation.

y^{″} - 5 y^{'} + 6 y = 0

Solution. Solution

First, write down and solve the characteristic equation:

\begin{aligned} r^{2} - 5 r + 6 = & 0 \\ (r - 2) (r - 3) = & 0 \\ r_{1} = 2 and r_{2} = & 3 . \end{aligned}

Since

r_{1}

and

r_{2}

are real and

r_{1} \neq r_{2},

we are in Case 1. Therefore, the general solution is:

y = C_{1} e^{2 x} + C_{2} e^{3 x} .

y^{″} + 4 y^{'} + 4 y = 0

Solution. Solution

The characteristic equation is:

\begin{aligned} r^{2} + 4 r + 4 = & 0 \\ (r + 2) (r + 2) = & 0 \\ r_{1} = - 2 and r_{2} = & - 2 \leftarrow Case 2 . \end{aligned}

Therefore, the general solution is:

y = (C_{1} + C_{2} x) e^{- 2 x} .

y^{″} + 2 y^{'} + 5 y = 0

Solution. Solution

The characteristic equation in this problem is:

\begin{aligned} r^{2} + 2 r + 5 = 0 \\ r = \underset{α}{- 1} \pm \underset{β}{2} i & \leftarrow Case 3 . \end{aligned}

Here,

α = - 1

and

β = 2 .

So the general solution is:

y = e^{- x} (C_{1} \cos (2 x) + C_{2} \sin (2 x)) .

Reading Questions Check-Point Questions

1. Match the DE to Its Characteristic Equation.

Match each LHCC differential equation on the left to its corresponding characteristic equation on the right.

Note: Each DE has a unique characteristic equation.

$2 y^{″} - 3 y^{'} + y = 0$
$2 r^{2} - 3 r + 1 = 0$
$y^{″} - 5 y^{'} + 6 y = 0$
$r^{2} - 5 r + 6 = 0$
$2 y^{'} + 7 y = 0$
$2 r + 7 = 0$
$y^{″} + 4 y^{'} = 0$
$r^{2} + 4 r = 0$

2. Solving the characteristic equation.

r^{2} - 4 r + 4 = 0,

$r = 2$ (double root)
Correct! The root $r = 2$ has a multiplicity of 2.
$r = 4$
Incorrect. Check the quadratic equation carefully.
$r = - 2$ and $r = 2$
Incorrect. Ensure you solve the quadratic equation correctly.
$r = 2 \pm 2 i$
Incorrect. These are not the correct roots for this equation.

3. Match the Characteristic Equation Roots to the General Solution.

Match each set of characteristic equation roots on the left to the corresponding general solution on the right.

Note: One of the differential equations is first-order, and the rest are second-order.

$r = 5$
$C e^{5 x}$
$r_{1} = 2, r_{2} = 3$
$C_{1} e^{2 x} + C_{2} e^{3 x}$
$r_{1} = - 1, r_{2} = - 4$
$C_{1} e^{- x} + C_{2} e^{- 4 x}$
$r_{1} = 3 + 2 i, r_{2} = 3 - 2 i$
$e^{3 x} (C_{1} \cos (2 x) + C_{2} \sin (2 x))$
$r_{1} = - 2, r_{2} = - 2$
$(C_{1} + C_{2} x) e^{- 2 x}$
$r_{1} = - 1 + i, r_{2} = - 1 - i$
$e^{- x} (C_{1} \cos (x) + C_{2} \sin (x))$

4. Discriminant of the characteristic equation.

- 4,

Real and equal solutions
Incorrect. The discriminant is the number under the square root in the quadratic formula.
Real and unequal solutions
Incorrect. The discriminant is the number under the square root in the quadratic formula.
Complex solutions
Correct! A negative discriminant indicates the square root of a negative number in the quadratic formula, which results in the complex number $i .$

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