Chapter 7: Problem 16
\(x^{\prime \prime}+4 x^{\prime}+5 x=0\)
Short Answer
Expert verified
The general solution to the differential equation is \( x(t) = e^{-2t} (C_1 \cos(t) + C_2 \sin(t)) \)
Step by step solution
01
Identify the Differential Equation
The given differential equation is a second-order linear homogeneous differential equation with constant coefficients: \[ x^{\rrime} + 4x^{\rpime} + 5x = 0 \]
02
Form the Characteristic Equation
To solve this differential equation, first form its characteristic equation by replacing \(x\) with \(r^2\), \(x'\) with \(r\), and \(x''\) with 1 in the equation: \[ r^2 + 4r + 5 = 0 \]
03
Solve the Characteristic Equation
Solve the quadratic characteristic equation: \[ r^2 + 4r + 5 = 0 \] Use the quadratic formula: \[ r = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{2a} \] where \(a = 1\), \(b = 4\), and \(c = 5\).
04
Calculate the Discriminant
The discriminant of the quadratic equation is: \[ b^2 - 4ac = 4^2 - 4(1)(5) = 16 - 20 = -4 \] The discriminant is negative, indicating complex roots.
05
Find the Complex Roots
Calculate the roots using the quadratic formula: \[ r = \frac{{-4 \pm \sqrt{{-4}}}}{2(1)} = \frac{{-4 \pm 2i}}{2} = -2 \pm i \] This gives the roots as \(r_1 = -2 + i\) and \(r_2 = -2 - i\).
06
Write the General Solution
The general solution for the differential equation is formed using the complex roots \(r_1\) and \(r_2\): \[ x(t) = e^{\alpha t} (C_1 \cos(\beta t) + C_2 \sin(\beta t)) \] where \( \alpha = -2\) and \( \beta = 1\). Therefore, the general solution is: \[ x(t) = e^{-2t} (C_1 \cos(t) + C_2 \sin(t)) \]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Characteristic Equation
To solve second-order linear homogeneous differential equations like \(x^{\rrime} + 4x^{\rpime} + 5x = 0\), we first need to form the characteristic equation. This step is important because it transforms a differential equation into a simpler algebraic equation.
The general form of a second-order linear homogeneous differential equation with constant coefficients is:
\(ax^{\rrime} + bx^{\rpime} + cx = 0\).
To form the characteristic equation, replace the second derivative \(x''\) with \(r^2\), the first derivative \(x'\) with \(r\), and the function \(x\) with 1.
For our given equation \(x^{\rrime} + 4x^{\rpime} + 5x = 0\), the characteristic equation becomes:
\(r^2 + 4r + 5 = 0\).
This characteristic equation is a quadratic equation which we can solve to find the roots.
The general form of a second-order linear homogeneous differential equation with constant coefficients is:
\(ax^{\rrime} + bx^{\rpime} + cx = 0\).
To form the characteristic equation, replace the second derivative \(x''\) with \(r^2\), the first derivative \(x'\) with \(r\), and the function \(x\) with 1.
For our given equation \(x^{\rrime} + 4x^{\rpime} + 5x = 0\), the characteristic equation becomes:
\(r^2 + 4r + 5 = 0\).
This characteristic equation is a quadratic equation which we can solve to find the roots.
Complex Roots
Solving the characteristic equation \(r^2 + 4r + 5 = 0\) involves using the quadratic formula:
\(r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)
Here, \(a = 1\), \(b = 4\), and \(c = 5\). To find the roots, first calculate the discriminant:
\(b^2 - 4ac = 16 - 20 = -4\).
The negative discriminant indicates that the roots are complex. Complex roots have the form:
\(r = \alpha \pm i\beta\).
Using the quadratic formula, we get the complex roots as:
\(r = \frac{-4 \pm \sqrt{-4}}{2} = \frac{-4 \pm 2i}{2} = -2 \pm i\).
Here, \(-2\) is the real part \(\alpha\) and \(1\) is the imaginary part \(\beta\). Complex roots relate to oscillatory solutions in the original differential equation.
\(r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)
Here, \(a = 1\), \(b = 4\), and \(c = 5\). To find the roots, first calculate the discriminant:
\(b^2 - 4ac = 16 - 20 = -4\).
The negative discriminant indicates that the roots are complex. Complex roots have the form:
\(r = \alpha \pm i\beta\).
Using the quadratic formula, we get the complex roots as:
\(r = \frac{-4 \pm \sqrt{-4}}{2} = \frac{-4 \pm 2i}{2} = -2 \pm i\).
Here, \(-2\) is the real part \(\alpha\) and \(1\) is the imaginary part \(\beta\). Complex roots relate to oscillatory solutions in the original differential equation.
General Solution
The general solution for a differential equation with complex roots is derived from the roots \(r_1 = -2 + i\) and \(r_2 = -2 - i\). When roots are complex, the solution includes exponential functions combined with sine and cosine functions.
Using the formula:
\(x(t) = e^{\alpha t} (C_1 \cos(\beta t) + C_2 \sin(\beta t))\),
we can write the solution where \(\alpha\) is the real part and \(\beta\) is the imaginary part. For our equation, \(\alpha = -2\) and \(\beta = 1\).
This gives us the general solution:
\(x(t) = e^{-2t} (C_1 \cos(t) + C_2 \sin(t))\).
Here, \(C_1\) and \(C_2\) are constants determined by initial conditions. This form of the solution signifies a function experiencing exponential decay coupled with oscillations, typical for systems with complex roots.
Using the formula:
\(x(t) = e^{\alpha t} (C_1 \cos(\beta t) + C_2 \sin(\beta t))\),
we can write the solution where \(\alpha\) is the real part and \(\beta\) is the imaginary part. For our equation, \(\alpha = -2\) and \(\beta = 1\).
This gives us the general solution:
\(x(t) = e^{-2t} (C_1 \cos(t) + C_2 \sin(t))\).
Here, \(C_1\) and \(C_2\) are constants determined by initial conditions. This form of the solution signifies a function experiencing exponential decay coupled with oscillations, typical for systems with complex roots.