Problem 15 - Calculus Volume 3

Question

Find the general solution to the linear differential equation. $$ 2 y^{\prime \prime}-3 y^{\prime}-5 y=0 $$

Step-by-Step Solution

Answer

The general solution is \( y(t) = C_1 e^{2.5t} + C_2 e^{-t} \).

Step 1: Write the Characteristic Equation

The differential equation is a linear homogeneous equation with constant coefficients. To solve it, we start by assuming a solution of the form \( y = e^{rt} \), where \( r \) is a constant. The derivatives \( y' = re^{rt} \), and \( y'' = r^2 e^{rt} \) are substituted into the given equation. This leads to the characteristic equation: \[ 2r^2 - 3r - 5 = 0 \]

Step 2: Solve the Characteristic Equation

We solve the quadratic characteristic equation \( 2r^2 - 3r - 5 = 0 \) using the quadratic formula, \( r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 2 \), \( b = -3 \), and \( c = -5 \). Calculate the discriminant's value: \[ b^2 - 4ac = (-3)^2 - 4(2)(-5) = 9 + 40 = 49 \]Since the discriminant is positive, we have two distinct real roots. Compute the roots:\[ r = \frac{3 \pm \sqrt{49}}{4} = \frac{3 \pm 7}{4} \]

Step 3: Calculate the Roots

From \( r = \frac{3 \pm 7}{4} \), we find the roots to be:\[ r_1 = \frac{3 + 7}{4} = 2.5 \quad \text{and} \quad r_2 = \frac{3 - 7}{4} = -1 \]These are the roots for the characteristic equation, which are used to construct the general solution for the differential equation.

Step 4: Write the General Solution

Since the characteristic equation has two distinct real roots \( r_1 = 2.5 \) and \( r_2 = -1 \), the general solution of the differential equation is given by:\[ y(t) = C_1 e^{r_1 t} + C_2 e^{r_2 t} = C_1 e^{2.5 t} + C_2 e^{-t} \]where \( C_1 \) and \( C_2 \) are arbitrary constants that can be determined if initial conditions are provided.

Key Concepts

Homogeneous Differential Equations

Homogeneous differential equations are a special class of differential equations that play a crucial role in mathematical modeling. These are called "homogeneous" because all terms are proportional to the function and its derivatives. In simpler terms, every term in the equation depends solely on the solution function or its derivatives.

In our example, the equation is:

• \( 2 y'' - 3 y' - 5 y = 0 \)

This equation is linear and homogeneous because each term involves the dependent variable \( y \) or its derivatives \( y' \) and \( y'' \). There is no standalone constant or non-zero term on the right side of the equation.

Understanding homogeneous equations helps you recognize and solve problems where the solution likely involves exponential functions or trigonometric identities, leading to elegant mathematical solutions.

Characteristic Equation

The characteristic equation is a crucial tool when solving linear homogeneous differential equations with constant coefficients. It enables you to transform a complicated differential equation problem into a simpler algebraic equation.

For our differential equation:

• \( 2 y'' - 3 y' - 5 y = 0 \)

We assume a solution of the form \( y = e^{rt} \). By substituting \( y \), \( y' \), and \( y'' \) into the differential equation, we derive a polynomial equation known as the characteristic equation:

• \( 2r^2 - 3r - 5 = 0 \)

This equation is independent of the specific form of the solution and focuses only on the exponents \( r \), allowing us to find possible solutions (roots) for \( r \) that solve the original differential equation.

Quadratic Formula

The quadratic formula is an essential mathematical tool used to find solutions to quadratic equations of the general form \( ax^2 + bx + c = 0 \). This formula can solve any such quadratic equation, provided the coefficients \( a \), \( b \), and \( c \) are real numbers.

In our case, we apply it to the characteristic equation:

• \( 2r^2 - 3r - 5 = 0 \)

The quadratic formula is:

• \( r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)

Substituting \( a = 2 \), \( b = -3 \), \( c = -5 \) into the formula, we calculate the discriminant \( b^2 - 4ac = 49 \). Since the discriminant is positive, there are two distinct real roots. These roots are crucial for constructing the general solution of the differential equation.

General Solution

The general solution of a linear homogeneous differential equation describes all possible solutions, encompassing all initial conditions you might encounter. Once you've determined the roots of the characteristic equation, you can express the general solution using these roots.

For the characteristic equation \( 2r^2 - 3r - 5 = 0 \), we found two distinct real roots:

• \( r_1 = 2.5 \)
• \( r_2 = -1 \)

These roots give the general solution in the form:

• \( y(t) = C_1 e^{2.5t} + C_2 e^{-t} \)

Here, \( C_1 \) and \( C_2 \) are arbitrary constants determined by initial conditions. This form represents the complete set of solutions to the differential equation and is crucial for understanding how solutions evolve over time.