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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 erx. However, not every value of r will work. The correct values of r arise from solving an important algebraic equation, obtained by substituting y=erx 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:
y5y=0
We want to find the value of r such that y=erx is a solution. Substituting y=erx into the differential equation gives:
[erx]5[erx]=0rerx5erx=0(r5)erx=0
Since erx is never zero, we must have:
r5=0r=5
This tells us that y=e5x is a solution. Therefore, the general solution is:
y=Ce5x
The equation r5=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:
y=erxay+by=0a[erx]+b[erx]=0arerx+berx=0(ar+b)erx=0
This results in the characteristic equation and general solution:
ar+b=0r=bay=Cebax
Now, let’s apply this method to a couple of examples.

Example 5.

Find the general solution for each LHCC equation.
3y9y=05y=πy
Solution. Solution
First, rewrite the equations in standard form:
y3y=05yπy=0
Now solve the characteristic equations:
3r9=0r=35rπ=0r=π5
Thus, the general solutions are:
y=Ce3xy=Ceπ5x
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.

    What is the characteristic equation for y5y=0?
  • r5=0
  • Correct! The characteristic equation is r5=0.
  • r+5=0
  • Incorrect. Check the sign of the coefficient of y.
  • r25=0
  • Incorrect. The characteristic equation for a first-order LHCC is linear, not quadratic.
  • 5r1=0
  • Incorrect. Make sure to use the correct coefficients from the original equation.

2. Select the characteristic equation.

    Select the characteristic equation for the first-order LHCC equation
    4y+3y=0?
  • 4r+3=1
  • Incorrect. Remember, the characteristic equation should be formed by setting the equation to zero.
  • 4r+3=0
  • Correct! The characteristic equation is found by substituting y=erx into the differential equation.
  • 4r3=0
  • Incorrect. Check the signs of the coefficients.
  • 3r+4=0
  • Incorrect. Ensure you have used the correct coefficients from the original equation.

3. Select the general solution.

    Give the general solution for the first-order LHCC equation
    y2y=0?
  • y=Ce2x
  • Incorrect. Check the sign of the exponent.
  • y=Ce2x
  • Correct! The solution has the form y=Cerx, where r is found from the characteristic equation.
  • y=Ce12x
  • Incorrect. Make sure to solve the characteristic equation correctly.
  • y=Ce12x
  • Incorrect. Check the coefficient of y in the original equation.

4. Identify the first-order LHCC equation.

    Which of the following is a first-order LHCC equation?
  • y+yy=0
  • Incorrect. This is a second-order equation.
  • 3y+5y=0
  • Correct! This is a first-order linear homogeneous equation with constant coefficients.
  • 2y+y=3
  • Incorrect. This equation is not homogeneous.
  • y+xy=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
    2y7y=0?
  • e7/2x
  • Correct! Solving 2r7=0 gives r=72. So e7/2x is the exponential term in the solution.
  • e7/2x
  • Incorrect. Check the signs when solving the characteristic equation.
  • e2/7x
  • Incorrect. Ensure you are solving the characteristic equation correctly.
  • e2/7x
  • Incorrect. Revisit the steps to solve the characteristic equation.
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