Problem 12 - Calculus Volume 3

Question

Find the general solution to the linear differential equation. $$ y^{\prime \prime}-7 y^{\prime}+12 y=0 $$

Step-by-Step Solution

Answer

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

Step 1: Identify the Differential Equation

The given equation is a second-order linear homogeneous differential equation with constant coefficients: \( y'' - 7y' + 12y = 0 \).

Step 2: Write the Characteristic Equation

To find the general solution, we substitute \( y = e^{rt} \) into the equation, then simplify to obtain the characteristic equation: \( r^2 - 7r + 12 = 0 \).

Step 3: Solve the Characteristic Equation

We solve the characteristic equation \( r^2 - 7r + 12 = 0 \) using the quadratic formula: \( r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1 \), \( b = -7 \), and \( c = 12 \). Calculate the discriminant: \( (-7)^2 - 4 \cdot 1 \cdot 12 = 49 - 48 = 1 \). The roots are \( r_1 = 4 \) and \( r_2 = 3 \).

Step 4: Write the General Solution

Since the roots \( r_1 \) and \( r_2 \) are real and distinct, the general solution to the differential equation is given by \( y(t) = C_1 e^{4t} + C_2 e^{3t} \), where \( C_1 \) and \( C_2 \) are arbitrary constants.

Key Concepts

Characteristic Equation

In the world of differential equations, the characteristic equation plays a vital role when dealing with linear homogeneous equations with constant coefficients. It is derived from the differential equation by assuming a solution of the form

• \( y = e^{rt} \)

where \( r \) is a constant that we need to determine. For our given equation

• \( y'' - 7y' + 12y = 0 \),

we replace \( y \), \( y' \), and \( y'' \) with \( e^{rt} \), leading us to the equation

• \( r^2 e^{rt} - 7r e^{rt} + 12 e^{rt} = 0 \).

Factoring out \( e^{rt} \), we simplify this to

• \( e^{rt}(r^2 - 7r + 12) = 0 \).

For non-trivial solutions \( e^{rt} eq 0 \), so the characteristic equation is

• \( r^2 - 7r + 12 = 0 \).

This quadratic equation is the key to finding solutions to our original differential equation.

Quadratic Formula

The quadratic formula is a powerful tool to solve quadratic equations, which are in the form

• \( ax^2 + bx + c = 0 \).

For our characteristic equation \( r^2 - 7r + 12 = 0 \), we summarized the terms with:

• \( a = 1 \)
• \( b = -7 \)
• \( c = 12 \)

The Quadratic Formula gives us:

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

Plugging into the formula, we first calculate the discriminant:

• \( (-7)^2 - 4 \cdot 1 \cdot 12 = 49 - 48 = 1 \).

The discriminant is positive, which means that our roots will be real and distinct. Substituting in our values, we compute the roots as:

• \( r_1 = \frac{7 + 1}{2} = 4 \)
• \( r_2 = \frac{7 - 1}{2} = 3 \)

These provide the fundamental structure of the solution for our differential equation.

Roots of Equations

Roots of equations are vital in solving differential equations, especially when dealing with characteristic equations derived from linear differential equations. The roots tell us the form of the solution.

• For the characteristic equation \( r^2 - 7r + 12 = 0 \), the roots are \( r_1 = 4 \) and \( r_2 = 3 \).

These are real and distinct roots, which lead to solutions of the form:

• \( y(t) = C_1 e^{r_1 t} + C_2 e^{r_2 t} \)
• \( y(t) = C_1 e^{4t} + C_2 e^{3t} \)

Here, \( C_1 \) and \( C_2 \) are constants dependent on initial conditions and ensure the general solution fits various particular solutions.
Roots have different impacts depending on their nature:

• Real and distinct roots lead to exponential solutions as in our example.
• Complex roots result in oscillatory solutions involving sine and cosine functions.
• Repeated real roots produce polynomial times exponential solutions.

By understanding the roots, we comprehend the behavior and form of the solution.