Problem 20 - Calculus Volume 3

Question

Find the general solution to the linear differential equation. $$ 8 y^{\prime \prime}+14 y^{\prime}-15 y=0 $$

Step-by-Step Solution

Answer

The general solution is \(y(t) = C_1 e^{\frac{3}{4} t} + C_2 e^{-\frac{5}{2} t}\).

Step 1: Write the characteristic equation

To solve the linear differential equation \(8 y'' + 14 y' - 15 y = 0\), we start by assuming a solution of the form \(y = e^{rt}\), where \(r\) is a constant. This assumption leads us to form the characteristic equation: \[8r^2 + 14r - 15 = 0.\]

Step 2: Solve the characteristic equation

Next, we solve the quadratic characteristic equation \(8r^2 + 14r - 15 = 0\). Using the quadratic formula \(r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 8\), \(b = 14\), and \(c = -15\), we find: \[r = \frac{-14 \pm \sqrt{14^2 - 4 \cdot 8 \cdot (-15)}}{2 \cdot 8}.\] Simplifying the expression under the square root gives us: \(b^2 - 4ac = 196 + 480 = 676\). Therefore, \[r = \frac{-14 \pm \sqrt{676}}{16} = \frac{-14 \pm 26}{16}.\]

Step 3: Calculate the roots

From the solved characteristic equation, calculate the roots: \[r_1 = \frac{-14 + 26}{16} = \frac{12}{16} = \frac{3}{4},\] \[r_2 = \frac{-14 - 26}{16} = \frac{-40}{16} = -\frac{5}{2}.\] These are distinct real roots, \(r_1 = \frac{3}{4}\) and \(r_2 = -\frac{5}{2}\).

Step 4: Write the general solution

With distinct real roots, the general solution to the differential equation is given by: \[y(t) = C_1 e^{r_1 t} + C_2 e^{r_2 t},\] where \(C_1\) and \(C_2\) are arbitrary constants. Substituting the roots \(r_1\) and \(r_2\), the solution becomes: \[y(t) = C_1 e^{\frac{3}{4} t} + C_2 e^{-\frac{5}{2} t}.\]

Key Concepts

Characteristic Equation

When dealing with a linear differential equation, such as the one given, the first step is to find what is known as the characteristic equation. This process transforms the differential equation into an algebraic one by proposing a trial solution of the form \( y = e^{rt} \). Here, \( r \) is a constant that we are solving for.- We substitute \( y = e^{rt} \), \( y' = re^{rt} \), and \( y'' = r^2e^{rt} \) into the original differential equation.- For our example: \[ 8 r^2 e^{rt} + 14 r e^{rt} - 15 e^{rt} = 0 \]- Factor out \( e^{rt} \) (only zero if it is the trivial solution): \[ e^{rt}(8r^2 + 14r - 15) = 0 \]The characteristic equation, therefore, is \( 8r^2 + 14r - 15 = 0 \). This equation is key in pinpointing the roots, which directly correspond to the types of solutions for the differential equation.

General Solution

The general solution of a differential equation arises once you have solved the characteristic equation and identified the roots. These roots, \( r_1 \) and \( r_2 \), dictate the behavior and form of the general solution. For the equation at hand, we found two distinct roots: \( r_1 = \frac{3}{4} \) and \( r_2 = -\frac{5}{2} \).- With the roots of the characteristic equation known, the general solution is expressed as: \[ y(t) = C_1 e^{r_1 t} + C_2 e^{r_2 t} \]- For distinct real roots like ours, the solution is a combination of exponentials: \[ y(t) = C_1 e^{\frac{3}{4} t} + C_2 e^{-\frac{5}{2} t} \]In this solution:- \( C_1 \) and \( C_2 \) are arbitrary constants, determined by initial conditions.- This form illustrates how solutions to differential equations are influenced directly by the nature of the roots.

Quadratic Formula

The quadratic formula is a fundamental tool used to solve quadratic equations, especially when the equation cannot be easily factored. In the context of solving a characteristic equation, this formula helps us find the roots that define our general solution. Here’s how we use the quadratic formula in our example:The characteristic equation from the original differential equation was \( 8r^2 + 14r - 15 = 0 \). To find the roots \( r_1 \) and \( r_2 \), we used:- Quadratic formula: \[ r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]- Plugging the values \( a = 8 \), \( b = 14 \), and \( c = -15 \) into the formula gives: \[ r = \frac{-14 \pm \sqrt{14^2 - 4 \times 8 \times (-15)}}{2 \times 8} \]- This simplifies to \( r = \frac{-14 \pm \sqrt{676}}{16} \), leading to: - \( r_1 = \frac{12}{16} = \frac{3}{4} \) - \( r_2 = \frac{-40}{16} = -\frac{5}{2} \)Using the quadratic formula ensures that even in complex or unfactorable scenarios, the roots (and hence the solutions) can be determined accurately.