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Magazine Chapter 2: Problem 1
Find a general solution. Check your answer by substitution. $$y^{\prime \prime}-6 y^{\prime}-7 y=0$$
Short Answer
Expert verified
The general solution is \( y(t) = C_1 e^{7t} + C_2 e^{-t} \).
Step by step solution
01
Identify the type of differential equation
The given equation is a second-order linear homogeneous differential equation with constant coefficients: \( y'' - 6y' - 7y = 0 \).
02
Write the characteristic equation
For a differential equation of the form \( ay'' + by' + cy = 0 \), the characteristic equation is \( ar^2 + br + c = 0 \). Here, \( a = 1 \), \( b = -6 \), and \( c = -7 \). The characteristic equation is \( r^2 - 6r - 7 = 0 \).
03
Solve the characteristic equation
Solve the quadratic equation \( r^2 - 6r - 7 = 0 \) using the quadratic formula: \( r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). Substituting the values, we have \( r = \frac{6 \pm \sqrt{36 + 28}}{2} \), resulting in \( r = \frac{6 \pm \sqrt{64}}{2} \).
04
Find the roots of the characteristic equation
The roots are \( r_1 = \frac{6 + 8}{2} = 7 \) and \( r_2 = \frac{6 - 8}{2} = -1 \). These roots are real and distinct.
05
Write the general solution
For real and distinct roots \( r_1 \) and \( r_2 \), the general solution of the differential equation is \( y(t) = C_1 e^{r_1 t} + C_2 e^{r_2 t} \). Thus, the general solution is \( y(t) = C_1 e^{7t} + C_2 e^{-t} \).
06
Verify the solution by substitution
Substitute \( y(t) = C_1 e^{7t} + C_2 e^{-t} \) into the original differential equation. Compute the first derivative: \( y' = 7C_1 e^{7t} - C_2 e^{-t} \), and the second derivative: \( y'' = 49C_1 e^{7t} + C_2 e^{-t} \). Plug these into the differential equation: \( (49C_1 e^{7t} + C_2 e^{-t}) - 6(7C_1 e^{7t} - C_2 e^{-t}) - 7(C_1 e^{7t} + C_2 e^{-t}) \). Simplify to verify the equation equals zero, confirming the solution is correct.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Characteristic Equation
When dealing with second-order linear differential equations like the one in our exercise, understanding the characteristic equation is crucial.The original differential equation is given as \( y'' - 6y' - 7y = 0 \). The typical format for such equations is \( ay'' + by' + cy = 0 \), where \( a \), \( b \), and \( c \) are constants.The characteristic equation is formed by replacing \( y'' \), \( y' \), and \( y \) with powers of \( r \), resulting in \( ar^2 + br + c = 0 \).In this exercise:
- \( a = 1 \)
- \( b = -6 \)
- \( c = -7 \)
Real and Distinct Roots
Solving the characteristic equation provides the roots that are key to understanding the behavior of the differential equation.For the equation \( r^2 - 6r - 7 = 0 \), we apply the quadratic formula:\[ r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]Plugging in the known values:
- \( b = -6 \)
- \( a = 1 \)
- \( c = -7 \)
- \( r_1 = 7 \)
- \( r_2 = -1 \)
General Solution
Once the roots of the characteristic equation are identified, determining the general solution becomes straightforward.For real and distinct roots \( r_1 \) and \( r_2 \), the general solution of the differential equation is expressed as:\[ y(t) = C_1 e^{r_1 t} + C_2 e^{r_2 t} \]With our roots\( r_1 = 7 \) and \( r_2 = -1 \), the specific general solution is:\[ y(t) = C_1 e^{7t} + C_2 e^{-t} \]Here:
- \( C_1 \) and \( C_2 \) are arbitrary constants that can be determined with initial conditions.
- \( e^{7t} \) and \( e^{-t} \) are linearly independent solutions.
