Chapter 17: Problem 41
Find the general solution. $$y^{\prime \prime}-2 y^{\prime}-3 y=0$$
Short Answer
Expert verified
The general solution is \(y(x) = C_1e^{3x} + C_2e^{-x}\).
Step by step solution
01
Identify the Differential Equation
We are given the second-order linear homogeneous differential equation: \[ y'' - 2y' - 3y = 0 \] to solve for the general solution.
02
Find the Characteristic Equation
The characteristic equation for the given differential equation is formed by replacing \( y'' \) with \( r^2 \), \( y' \) with \( r \), and \( y \) with 1. This gives the quadratic equation: \[ r^2 - 2r - 3 = 0 \].
03
Solve the Characteristic Equation
We solve \( r^2 - 2r - 3 = 0 \) using the quadratic formula: \[ r = \frac{-b \pm \sqrt{b^2-4ac}}{2a} \]. Here, \( a=1 \), \( b=-2 \), and \( c=-3 \). Calculating gives: \[ r = \frac{2 \pm \sqrt{4+12}}{2} = \frac{2 \pm \sqrt{16}}{2} \].
04
Calculate the Roots
Solving the quadratic formula, we find: \[ r = \frac{2 \pm 4}{2} \]. This gives roots \( r_1 = 3 \) and \( r_2 = -1 \). These are distinct real roots.
05
Write the General Solution
For distinct real roots \( r_1 \) and \( r_2 \), the general solution of the differential equation is: \[ y(x) = C_1e^{r_1x} + C_2e^{r_2x} \]. Substituting the values of the roots, we get: \[ y(x) = C_1e^{3x} + C_2e^{-x} \].
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Characteristic Equation in Second-Order Differential Equations
When we deal with second-order linear homogeneous differential equations like \(y'' - 2y' - 3y = 0\), the characteristic equation plays a central role. This equation helps us transform a complex differential problem into an easier algebraic problem. We find the characteristic equation by substituting the derivatives in the differential equation with powers of \(r\):
- Replace \(y''\) with \(r^2\)
- Replace \(y'\) with \(r\)
- Replace \(y\) with 1
Using the Quadratic Formula to Solve Characteristic Equations
The quadratic formula is a powerful tool that solves any quadratic equation of the form \(ax^2 + bx + c = 0\). In our case, the characteristic equation is \(r^2 - 2r - 3 = 0\). Here, we identify the coefficients as \(a = 1\), \(b = -2\), and \(c = -3\). The quadratic formula is: \[ r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] Applying the formula, we replace and calculate:
- \(-b = 2\)
- \(b^2 - 4ac = 4 + 12 = 16\)
- The square root of 16 is 4
General Solution of Second-Order Homogeneous Differential Equations
Once the roots of the characteristic equation are known, we can construct the general solution to the differential equation. For equations with distinct real roots \(r_1\) and \(r_2\), the general solution takes the form:\[ y(x) = C_1e^{r_1x} + C_2e^{r_2x} \]Here, \(C_1\) and \(C_2\) are arbitrary constants, determined by the initial or boundary conditions of the specific problem you are working on. For our given equation, the roots we found are \(r_1 = 3\) and \(r_2 = -1\). Substituting these values, we arrive at:\[ y(x) = C_1e^{3x} + C_2e^{-x} \]This expression is the general solution, representing a family of functions that satisfy the original differential equation. Each specific solution within this family can be determined through additional information, such as initial conditions.