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.