Question

find the general solution of the given differential equation. $$ y^{\prime \prime}+y^{\prime}+1.25 y=0 $$

Short answer

Answer: The general solution is \(y(x) = e^{-\frac{1}{2}x}(A \cos(\frac{\sqrt{3}}{2}x) + B \sin (\frac{\sqrt{3}}{2}x))\), where A and B are arbitrary constants.

Step-by-step solution

Identify the coefficients

Identify the coefficients of the given differential equation, which are: a = 1, b = 1, and c = 1.25. The equation is: $$ y^{\prime \prime} + y^{\prime} + 1.25 y=0 $$

Find the characteristic equation

Form the characteristic equation by replacing the derivatives with their corresponding powers of the variable r: $$ r^2 + r + 1.25 = 0 $$

Solve the characteristic equation

Now, solve for r to find the roots. We can use the quadratic formula: $$ r = \frac{-b\pm \sqrt{b^2-4ac}}{2a} $$ Substitute the known coefficients into the formula: $$ r = \frac{-1 \pm \sqrt{1^2-4(1)(1.25)}}{2(1)} $$

Simplify and determine roots

Upon simplification, we find that the discriminant, \(1-5 = -4\) is negative. Thus, we have complex roots: $$ r = \frac{-1 \pm \sqrt{-4}}{2} $$ Which simplifies to: $$ r_1 = -\frac{1}{2} + \frac{\sqrt{3}}{2}i, \quad r_2 = -\frac{1}{2} - \frac{\sqrt{3}}{2}i $$

Form the general solution based on roots

Since we have two distinct complex roots, \(r_1\) and \(r_2\), we can express the general solution of the given second-order differential equation as: $$ y(x) = e^{\alpha x}(A \cos(\beta x) + B \sin (\beta x)) $$ Using the real and imaginary parts of \(r_1\) and \(r_2\), \(\alpha = -\frac{1}{2}\) and \(\beta = \frac{\sqrt{3}}{2}\): $$ y(x) = e^{-\frac{1}{2}x}(A \cos(\frac{\sqrt{3}}{2}x) + B \sin (\frac{\sqrt{3}}{2}x)) $$

Write down the general solution

The general solution of the given second-order linear homogeneous differential equation is: $$ y(x) = e^{-\frac{1}{2}x}(A \cos(\frac{\sqrt{3}}{2}x) + B \sin (\frac{\sqrt{3}}{2}x)) $$ Here, A and B are arbitrary constants.

Key concepts

Characteristic Equation

The characteristic equation is a vital tool when solving differential equations, especially second-order linear homogeneous ones. When given a differential equation like \( y'' + y' + 1.25y = 0 \), the characteristic equation helps us determine the roots that dictate the behavior of the solution.
The characteristic equation is created by replacing each derivative in the differential equation with a power of \( r \), a variable often used for this purpose. Therefore, for our example:
- Replace \( y' \) with \( r \)- Replace \( y'' \) with \( r^2 \)
This gives us the characteristic equation \( r^2 + r + 1.25 = 0 \). Assessing these roots enables us to analyze if they are real or complex, which then affects the form of the general solution.

Complex Roots

Complex roots occur when the discriminant of the characteristic equation is negative. In our example, the discriminant \((b^2 - 4ac)\) equals \(-4\) because \(1^2 - 4(1)(1.25) = -4\). This negative result indicates complex roots.
To actually find these roots, we apply the quadratic formula \( r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). Inserting the coefficients from our characteristic equation, we get:
- \( r = \frac{-1 \pm \sqrt{-4}}{2} \)
Which simplifies to:
- \( r_1 = -\frac{1}{2} + \frac{\sqrt{3}}{2}i \)- \( r_2 = -\frac{1}{2} - \frac{\sqrt{3}}{2}i \)
Complex roots involve both real and imaginary parts, which we use in forming the general solution.

General Solution

For differential equations with complex roots, like \( r_1 \) and \( r_2 \) found earlier, the general solution takes on a specific form. This form accommodates the oscillatory nature of complex solutions, combining exponential decay or growth with sine and cosine functions.
The general solution for complex roots \( \alpha \pm \beta i \) is expressed as:
- \( y(x) = e^{\alpha x}(A \cos(\beta x) + B \sin(\beta x)) \)
Where:- \( \alpha \) is the real part of the roots- \( \beta \) is the imaginary part
For our specific equation, with \( \alpha = -\frac{1}{2} \) and \( \beta = \frac{\sqrt{3}}{2} \), the solution follows:
- \( y(x) = e^{-\frac{1}{2}x}(A \cos(\frac{\sqrt{3}}{2}x) + B \sin(\frac{\sqrt{3}}{2}x)) \)
Here, \( A \) and \( B \) represent constants determined by initial conditions.

Second-Order Linear Homogeneous Differential Equation

Second-order linear homogeneous differential equations are a fundamental class of differential equations. They typically look like \( a y'' + b y' + c y = 0 \), where:- \( y'' \) is the second derivative of \( y \)- \( y' \) is the first derivative- \( y \) is the function itself- \( a, b, \) and \( c \) are constants
These equations are homogeneous, meaning that every term includes the function \( y \) or its derivatives directly, without any standalone functions or constants.
To solve these equations, one derives a characteristic equation from which the roots are calculated. The nature of these roots—whether real or complex—guides the formulation of the general solution. This forms the basis for understanding and solving higher-order differential equations comprehensively.