Problem 4 - Advanced Engineering Mathematics

Question

Find the general solution of the given second-order differential equation. $$ y^{\prime \prime}-3 y^{\prime}+2 y \quad 0 $$

Step-by-Step Solution

Answer

The general solution is \( y(t) = C_1 e^{t} + C_2 e^{2t} \).

Step 1: Write the Differential Equation

The given second-order differential equation is \( y^{\prime \prime}-3 y^{\prime}+2 y = 0 \). Our task is to find its general solution.

Step 2: Find the Characteristic Equation

To solve the differential equation, we first write the characteristic equation. The characteristic equation is found by assuming \( y = e^{rt} \) and substituting it into the differential equation, which gives us \( r^2 - 3r + 2 = 0 \).

Step 3: Solve the Characteristic Equation

Now we solve the characteristic equation \( r^2 - 3r + 2 = 0 \) by factoring. We find that it factors to \((r - 1)(r - 2) = 0\). Hence, the solutions are \( r = 1 \) and \( r = 2 \).

Step 4: Write the General Solution

Since we have two distinct roots \( r = 1 \) and \( r = 2 \), the general solution of the differential equation is \( y(t) = C_1e^{t} + C_2e^{2t} \), where \( C_1 \) and \( C_2 \) are arbitrary constants.

Key Concepts

Characteristic Equation

When faced with a second-order differential equation like \( y^{\prime \prime}-3 y^{\prime}+2 y = 0 \), the characteristic equation is an essential tool for finding solutions. The process begins by assuming a trial solution of the form \( y = e^{rt} \), where \( r \) is a constant to be determined. This approach is governed by the linearity of the exponential function, making it a fitting candidate for constant coefficient equations.

Substituting \( y = e^{rt} \) and its derivatives into the differential equation gives us a polynomial in \( r \). For our given equation, we end up with the characteristic equation \( r^2 - 3r + 2 = 0 \). This resulting quadratic equation provides insight into the behavior and possible solutions of the original differential equation.

Finding the roots of this characteristic equation is crucial since they help us directly construct the general solution to the differential equation.

General Solution

Once the characteristic equation \( r^2 - 3r + 2 = 0 \) has been solved, and the roots have been determined, we can move on to writing the general solution. For our equation, we found the roots \( r = 1 \) and \( r = 2 \). Since these roots are distinct, each corresponds to a natural exponential function that forms a part of the general solution.

The general solution in this case is built from the distinct solutions \( e^{r_1t} \) and \( e^{r_2t} \), where \( r_1 = 1 \) and \( r_2 = 2 \). Thus, we write the solution as:

• \( y(t) = C_1e^{t} + C_2e^{2t} \)

where \( C_1 \) and \( C_2 \) are arbitrary constants determined by initial or boundary conditions.

This general solution represents a family of curves that include all possible solutions to the differential equation.

Distinct Roots

Distinct roots occur when the characteristic equation associated with a second-order linear differential equation has two different solutions. In the quadratic \( r^2 - 3r + 2 = 0 \), the term "distinct" refers to how the roots \( r_1 = 1 \) and \( r_2 = 2 \) differ from each other. This is an important scenario because it significantly affects the structure of the general solution.

When the roots are distinct, as in this example, each root \( r_i \) contributes an independent exponential term \( e^{r_it} \) to the solution.

• This results in a general solution of the form \( y(t) = C_1 e^{r_1t} + C_2 e^{r_2t} \).

The distinctness of the roots ensures that the solution is a combination of two exponential functions, each weighted by arbitrary constants \( C_1 \) and \( C_2 \).

This offers a full representation of all possible behaviors described by the differential equation under consideration.