Chapter 3: Problem 7
In each of Problems 1 through 10 find the general solution of the given differential equation. \(4 y^{\prime \prime}+17 y^{\prime}+4 y=0\)
Short Answer
Expert verified
Answer: The general solution of the given differential equation is \(y(x) = C_1 e^{-1x} + C_2 e^{-4x}\), where \(C_1\) and \(C_2\) are arbitrary constants.
Step by step solution
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Characteristic Equation
Understanding the characteristic equation is fundamental to solving second-order linear homogeneous differential equations. The characteristic equation is derived from the original differential equation by replacing the derivative terms with powers of a variable (typically denoted as 'r'). For example, for a second derivative \( y^{\prime \prime} \) you would use \( r^2 \), and similarly \( y^{\prime} \) would be represented by \( r \). This transformation results in a polynomial equation that reflects the structure of the differential equation.
To illustrate, consider the exercise where the differential equation is \( 4y^{\prime \prime} + 17y^{\prime} + 4y = 0 \). By replacing the derivatives, we obtain the characteristic equation \( 4r^2 + 17r + 4 = 0 \) which is a quadratic equation in terms of 'r'. This equation can be solved by factoring, as shown in the steps provided, or by using the quadratic formula \( r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) as an alternative method if factoring is not straightforward.
To illustrate, consider the exercise where the differential equation is \( 4y^{\prime \prime} + 17y^{\prime} + 4y = 0 \). By replacing the derivatives, we obtain the characteristic equation \( 4r^2 + 17r + 4 = 0 \) which is a quadratic equation in terms of 'r'. This equation can be solved by factoring, as shown in the steps provided, or by using the quadratic formula \( r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) as an alternative method if factoring is not straightforward.
Second-Order Linear Homogeneous Differential Equation
A second-order linear homogeneous differential equation has the form \( a y^{\prime \prime} + b y^{\prime} + c y = 0 \), where \( a \), \( b \) and \( c \) are constants. In the context of this exercise, we are looking at a differential equation in which the right-hand side is zero, denoting it 'homogeneous'.
To solve this type of equation, we rely on the characteristic equation as previously discussed. The roots of the characteristic equation, which can be real or complex, dictate the form of the general solution. If the roots are real and distinct, as with \( r_1 \) and \( r_2 \) found in our example problem, the general solution is a linear combination of exponential functions: \( y(x) = C_1 e^{r_1 x} + C_2 e^{r_2 x} \). On the other hand, if the roots are real and repeated or complex, the form of the solution will differ, involving repeated exponential terms or sinusoids and cosines, respectively.
To solve this type of equation, we rely on the characteristic equation as previously discussed. The roots of the characteristic equation, which can be real or complex, dictate the form of the general solution. If the roots are real and distinct, as with \( r_1 \) and \( r_2 \) found in our example problem, the general solution is a linear combination of exponential functions: \( y(x) = C_1 e^{r_1 x} + C_2 e^{r_2 x} \). On the other hand, if the roots are real and repeated or complex, the form of the solution will differ, involving repeated exponential terms or sinusoids and cosines, respectively.
General Solution of Differential Equation
The general solution of a differential equation incorporates the entire family of solutions that satisfy the equation. It includes arbitrary constants, \( C_1 \) and \( C_2 \) in our case, because the second-order differential equation results in a solution space spanned by two linearly independent functions. These constants represent the initial conditions or specifics of a particular solution within the general family.
In the given exercise, after finding the roots of the characteristic equation, we construct the general solution by combining the exponential terms involving the roots and the arbitrary constants: \( y(x) = C_1 e^{-1x} + C_2 e^{-4x} \). This expression can fit infinite scenarios depending on the values assigned to \( C_1 \) and \( C_2 \), which are typically determined from initial value conditions or boundary conditions given in a specific problem. It's crucial for students to understand that these constants are essential in tailoring the general solution to the particular problem at hand.
In the given exercise, after finding the roots of the characteristic equation, we construct the general solution by combining the exponential terms involving the roots and the arbitrary constants: \( y(x) = C_1 e^{-1x} + C_2 e^{-4x} \). This expression can fit infinite scenarios depending on the values assigned to \( C_1 \) and \( C_2 \), which are typically determined from initial value conditions or boundary conditions given in a specific problem. It's crucial for students to understand that these constants are essential in tailoring the general solution to the particular problem at hand.
