Problem 28 - Calculus: An Integrated Approach to Functions and Their Rates of Change

Question

(a) There are two values of \(\lambda\) such that \(y=e^{\lambda^{t}}\) is a solution to \(y^{\prime \prime}+7 y^{\prime}+12 y=0\). Find them and label them \(\lambda\), and \(\lambda_{2}\). (b) Let \(y=C_{1} e^{\lambda_{1} I}+C_{2} e^{\lambda_{2} t}\), where \(C_{1}\) and \(C_{2}\) are arbitrary constants. Verify that \(y(t)\) is a solution to \(y^{\prime \prime}+7 y^{\prime}+12 y=0\).

Step-by-Step Solution

Answer

The values of \(\lambda\) that result in solutions to the differential equation are \(\lambda_{1} = -3\) and \(\lambda_{2} = -4\). The general solution \(y(t) = C_{1}e^{-3t} + C_{2}e^{-4t}\) was verified as a solution to this differential equation.

Step 1: Solve the Characteristic Equation

The characteristic equation for the given differential equation \(y^{\prime \prime}+7 y^{\prime}+12 y=0\) is \(\lambda^{2}+7\lambda+12=0\). Solving this quadratic equation for lambda will give the roots \(\lambda_{1}, \lambda_{2}\).

Step 2: Find the Roots of the Characteristic Equation

The roots of the equation \(\lambda^{2}+7\lambda+12=0\) can be found using the quadratic formula \(\lambda = \frac{-b \pm \sqrt{b^{2}-4ac}}{2a}\). Here a=1, b=7 and c=12. So, \(\lambda_{1} = \frac{-7 + \sqrt{49-48}}{2} = -3\) and \(\lambda_{2} = \frac{-7 - \sqrt{49-48}}{2} = -4\).

Step 3: Write the General Solution

The general solution of the differential equation is given by \(y(t) = C_{1}e^{\lambda_{1}t} + C_{2}e^{\lambda_{2}t}\) where \(C_{1}\) and \(C_{2}\) are arbitrary constants. Substituting the values of \(\lambda_{1}\) and \(\lambda_{2}\) gives \(y(t) = C_{1}e^{-3t} + C_{2}e^{-4t}\).

Step 4: Verify the General Solution

To verify this as a solution to the original differential equation, you must compute the first and second derivative of \(y(t)\), and substitute these as well as \(y(t)\) back into the differential equation.

Step 5: Compute the first and second derivatives of y

The first derivative of \(y(t)\) is \(y'(t) = -3C_{1}e^{-3t} - 4C_{2}e^{-4t}\) and the second derivative \(y''(t)\) is \(y''(t) = 9C_{1}e^{-3t} + 16C_{2}e^{-4t}\).

Step 6: Substitute y, y' and y'' back into the differential equation

Substituting these values into the original differential equation gives: \[9C_{1}e^{-3t} + 16C_{2}e^{-4t} + 7(-3C_{1}e^{-3t} - 4C_{2}e^{-4t}) + 12(C_{1}e^{-3t} + C_{2}e^{-4t}) = 0 \Rightarrow 0 = 0\] which verifies that \(y(t)\) is a solution to the differential equation.

Key Concepts

Characteristic Equation

When solving ordinary differential equations, one of the key steps is to find the characteristic equation. This is crucial for understanding the behavior and solutions of the differential equation. For a second-order linear homogeneous differential equation like \( y'' + 7y' + 12y = 0 \), the characteristic equation is derived by assuming a solution of the form \( y = e^{\lambda t} \). By substituting this assumed solution into the differential equation, we replace the derivatives with respect to \( t \). This manipulates the equation to involve \( \lambda \) instead of \( t \). Therefore, the characteristic equation formed is \( \lambda^2 + 7\lambda + 12 = 0 \). This is a quadratic equation in terms of \( \lambda \), where the roots \( \lambda_1 \) and \( \lambda_2 \) are critical because they represent the exponents in the exponential solutions of the differential equation. Understanding this gives insight into how the solutions behave and their overall structure.

Quadratic Formula

To solve the quadratic characteristic equation \( \lambda^2 + 7\lambda + 12 = 0 \), we use the quadratic formula. This formula is a straightforward method to find the roots of any quadratic equation given by \( ax^2 + bx + c = 0 \). For our specific equation, \( a = 1 \), \( b = 7 \), and \( c = 12 \). The quadratic formula is expressed as:

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

Plugging in the values, we get:

• \( \lambda = \frac{-7 \pm \sqrt{49 - 48}}{2} \)
• \( \lambda = \frac{-7 \pm 1}{2} \)

This gives us the roots \( \lambda_1 = -3 \) and \( \lambda_2 = -4 \). These roots are essential as they provide the exponential terms in the general solution of the differential equation. Utilizing the quadratic formula simplifies the problem of finding possible solutions significantly.

General Solution

After obtaining the roots from the characteristic equation, we move on to construct the general solution of the differential equation. For our differential equation \( y'' + 7y' + 12y = 0 \), the general solution is expressed as a linear combination of exponential functions corresponding to the roots:

• \( y(t) = C_1 e^{\lambda_1 t} + C_2 e^{\lambda_2 t} \)

Inserting our found roots \( \lambda_1 = -3 \) and \( \lambda_2 = -4 \), the general solution becomes:

• \( y(t) = C_1 e^{-3t} + C_2 e^{-4t} \)

Here, \( C_1 \) and \( C_2 \) are arbitrary constants that will be determined by initial or boundary conditions. This general solution captures the entirety of solutions that this differential equation might produce based on varying conditions. Understanding the components of the general solution helps illuminate how exponential decay and specific conditions impact the behavior of solutions. By mastering this concept, solving similar differential equations becomes much more manageable.