Question

In each of Problems 1 through 10 find the general solution of the given differential equation. \(25 y^{\prime \prime}-20 y^{\prime}+4 y=0\)

Short answer

Answer: The general solution to the given differential equation is \(y(x) = C_1e^{\frac{2}{5}x} + C_2xe^{\frac{2}{5}x}\), where \(C_1\) and \(C_2\) are constants determined by any given initial conditions.

Step-by-step solution

Determine the Characteristic Equation

Rewrite the given differential equation in terms of algebraic operators, replacing the derivative terms with polynomial terms, and then finding the characteristic equation. Here, we replace \(y^{\prime\prime}\) with \(r^2\), \(y^{\prime}\) with \(r\), and \(y\) with 1. Doing so, we get: \(25r^2 - 20r + 4 = 0\)

Solve the Characteristic Equation

Now, we need to solve the quadratic equation for \(r\): \(25r^2 - 20r + 4 = 0\) We can factor the equation as: \((5r - 2)(5r - 2) = 0\) Thus, we have a repeated root: \(r = \frac{2}{5}\)

Write the General Solution

Since we have a repeated root, we'll write the general solution using the form: \(y(x) = C_1e^{r_1x} + C_2xe^{r_1x}\) Where \(r_1\) is the repeated root, and \(C_1\) and \(C_2\) are constants determined by any given initial conditions. In our case, \(r_1 = \frac{2}{5}\). So the general solution is: \(y(x) = C_1e^{\frac{2}{5}x} + C_2xe^{\frac{2}{5}x}\) And that's the general solution to the given differential equation.

Key concepts

Characteristic Equation

In the realm of differential equations, the characteristic equation is a key concept for solving linear homogeneous differential equations with constant coefficients. When faced with such an equation, the first step is to translate it into an algebraic form using a characteristic equation. For example, consider the differential equation \(25 y^{\prime \prime}-20 y^{\prime}+4 y=0\). To find its characteristic equation, we replace the derivatives with polynomials:

• Replace \(y^{\prime \prime}\) with \(r^2\)
• Replace \(y^{\prime}\) with \(r\)
• Replace \(y\) with \(1\)

This transforms the differential equation into a quadratic form: \(25r^2 - 20r + 4 = 0\). Solving this polynomial equation gives us potential solutions for \(r\), which are crucial for constructing the general solution.

Repeated Roots

Once you solve the characteristic equation, you may encounter repeated roots, which are specific solutions where the roots are equal. In the example equation, \(25r^2 - 20r + 4 = 0\) factors into \((5r - 2)(5r - 2) = 0\), leading directly to a repeated root of \(r = \frac{2}{5}\).
If you have repeated roots in a characteristic equation:

• The root appears more than once due to repeated factors in the equation.
• This influences the form of the general solution significantly.

Understanding how to handle repeated roots is important as it affects the structure of the general solution, ensuring that the solution captures all possible behaviors of the differential equation.

General Solution

Once the roots from the characteristic equation are found, it is time to write down the general solution. In cases of repeated roots, the general solution takes a special form to cover all solutions of the differential equation.
For instance, with a repeated root \(r_1 = \frac{2}{5}\), the general solution is:

• \(y(x) = C_1e^{\frac{2}{5}x} + C_2xe^{\frac{2}{5}x}\)
• \(C_1\) and \(C_2\) are arbitrary constants determined by initial conditions.

This structure ensures that the solution generalizes for all initial conditions and scenarios. Each component plays a role:

• \(C_1e^{\frac{2}{5}x}\) accounts for the simple exponential growth or decay.
• \(C_2xe^{\frac{2}{5}x}\) introduces the linear term, addressing the distinctiveness of having repeated roots.

By constructing this solution correctly, we ensure that every potential outcome of the differential equation is captured.