DE-BOOK 1st-Order Equations

Section 6.3 1st-Order Equations

In the previous section, we learned that solutions to Linear Homogeneous Constant Coefficient (LHCC) equations often involve terms of the form

e^{r x} .

However, not every value of

r

will work. The correct values of

r

arise from solving an important algebraic equation, obtained by substituting

y = e^{r x}

into the differential equation. In this section, we will focus on this process for first-order equations before building up to higher-order cases.

Let’s start with a simple first-order equation:

y^{'} - 5 y = 0

We want to find the value of

r

such that

y = e^{r x}

is a solution. Substituting

y = e^{r x}

into the differential equation gives:

\begin{aligned} {[e^{r x}]}^{'} - 5 [e^{r x}] = & 0 \\ r e^{r x} - 5 e^{r x} = & 0 \\ (r - 5) e^{r x} = & 0 \end{aligned}

Since

e^{r x}

is never zero, we must have:

\begin{matrix} r - 5 = 0 \Rightarrow r = 5 \end{matrix}

This tells us that

y = e^{5 x}

is a solution. Therefore, the general solution is:

y = C e^{5 x}

The equation

r - 5 = 0,

which gave us the value of

r,

is called the characteristic equation. Every LHCC equation has one. The characteristic equation gives us the values of

r

that we need to construct the general solution.

Applying the same approach to the general first-order LHCC equation gives us:

\begin{aligned} y = e^{r x} \to a y^{'} + b y = 0 \Rightarrow a {[e^{r x}]}^{'} + b [e^{r x}] = & 0 \\ a r e^{r x} + b e^{r x} = & 0 \\ (a r + b) e^{r x} = & 0 \end{aligned}

This results in the characteristic equation and general solution:

a r + b = 0 \Rightarrow r = - \frac{b}{a} \Rightarrow y = C e^{- \frac{b}{a} x}

Now, let’s apply this method to a couple of examples.

Example 5.

Find the general solution for each LHCC equation.

3 y^{'} - 9 y = 0 5 y^{'} = π y

Solution. Solution

First, rewrite the equations in standard form:

y^{'} - 3 y = 0 5 y^{'} - π y = 0

Now solve the characteristic equations:

3 r - 9 = 0 \Rightarrow r = 3 5 r - π = 0 \Rightarrow r = \frac{π}{5}

Thus, the general solutions are:

y = C e^{3 x} y = C e^{\frac{π}{5} x}

This straightforward method works for any first-order LHCC equation by using the characteristic equation. In the next sections, we’ll extend this technique to higher-order equations.

Reading Questions Check-Point Questions

1. Characteristic equation for first-order LHCC.

y^{'} - 5 y = 0 ?

$r - 5 = 0$
Correct! The characteristic equation is $r - 5 = 0 .$
$r + 5 = 0$
Incorrect. Check the sign of the coefficient of $y .$
$r^{2} - 5 = 0$
Incorrect. The characteristic equation for a first-order LHCC is linear, not quadratic.
$5 r - 1 = 0$
Incorrect. Make sure to use the correct coefficients from the original equation.

2. Select the characteristic equation.

4 y^{'} + 3 y = 0 ?

$4 r + 3 = 1$
Incorrect. Remember, the characteristic equation should be formed by setting the equation to zero.
$4 r + 3 = 0$
Correct! The characteristic equation is found by substituting $y = e^{r x}$ into the differential equation.
$4 r - 3 = 0$
Incorrect. Check the signs of the coefficients.
$3 r + 4 = 0$
Incorrect. Ensure you have used the correct coefficients from the original equation.

3. Select the general solution.

y^{'} - 2 y = 0 ?

$y = C e^{2 x}$
Incorrect. Check the sign of the exponent.
$y = C e^{- 2 x}$
Correct! The solution has the form $y = C e^{r x},$ where $r$ is found from the characteristic equation.
$y = C e^{\frac{1}{2} x}$
Incorrect. Make sure to solve the characteristic equation correctly.
$y = C e^{- \frac{1}{2} x}$
Incorrect. Check the coefficient of $y$ in the original equation.

4. Identify the first-order LHCC equation.

$y^{″} + y^{'} - y = 0$
Incorrect. This is a second-order equation.
$3 y^{'} + 5 y = 0$
Correct! This is a first-order linear homogeneous equation with constant coefficients.
$2 y + y^{'} = 3$
Incorrect. This equation is not homogeneous.
$y^{'} + x y = 0$
Incorrect. This is not a constant coefficient equation.

5. What exponential term is in the solution.

What exponential term in the solution of the equation

- 2 y^{'} - 7 y = 0 ?

$e^{- 7 / 2 x}$
Correct! Solving $- 2 r - 7 = 0$ gives $r = - \frac{7}{2} .$ So $e^{- 7 / 2 x}$ is the exponential term in the solution.
$e^{7 / 2 x}$
Incorrect. Check the signs when solving the characteristic equation.
$e^{- 2 / 7 x}$
Incorrect. Ensure you are solving the characteristic equation correctly.
$e^{2 / 7 x}$
Incorrect. Revisit the steps to solve the characteristic equation.

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