Question

Solve the following differential equations. \(y^{\prime \prime}-2 y^{\prime}=0\)

Short answer

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

Step-by-step solution

Write the Differential Equation

Begin with the given differential equation: \[ y'' - 2y' = 0 \]

Obtain the Characteristic Equation

Rewrite the differential equation in terms of the characteristic equation. Let the solution be of the form \(y = e^{rt}\), where \(r\) is a constant. This yields the characteristic equation:\[ r^2 - 2r = 0 \]

Solve the Characteristic Equation

Solve the characteristic equation for \(r\): \[ r(r - 2) = 0 \]This gives two solutions: \[ r = 0 \] and \[ r = 2 \]

Form the General Solution

Based on the roots of the characteristic equation, form the general solution of the differential equation: \[ y(t) = C_1 e^{0t} + C_2 e^{2t} \]Since \(e^{0t} = 1\), simplify to: \[ y(t) = C_1 + C_2 e^{2t} \]

Write the Final Answer

Combine the results into the final answer:\[ y(t) = C_1 + C_2 e^{2t} \]

Key concepts

Second-Order Differential Equations

In mathematics, a second-order differential equation involves the second derivative of a function. These types of equations are common in various fields like physics, engineering, and economics. A second-order differential equation generally has the form: \[ a y'' + b y' + c y = 0 \].Here, \( y'' \) is the second derivative, \( y' \) is the first derivative, and \( y \) is the function of the variable (often time \( t \)).
Understanding the general form allows us to solve many different problems, as the steps and methods used can apply to any second-order differential equation. We typically begin by rewriting the equation, then moving on to finding its characteristic equation.

Characteristic Equation

The characteristic equation is crucial in solving second-order differential equations. It transforms the problem into solving a polynomial equation. Based on the given differential equation: \[ y'' - 2y' = 0 \],
we assume solutions of the form \( y = e^{rt} \), where \( r \) is a constant. By substituting \( y \) into the differential equation, we get: \[ r^2 e^{rt} - 2r e^{rt} = 0 \].Factor out the common term \( e^{rt} \), leading to: \[ r^2 - 2r = 0 \],
which is the characteristic equation. Solving the characteristic equation for \( r \) gives us the necessary roots to form the general solution of the original differential equation.

Homogeneous Differential Equation

A homogeneous differential equation has the form where all the terms depend on the function \( y \) and its derivatives. The equation \( y'' - 2y' = 0 \) is an example of a homogeneous second-order differential equation.
This means there are no terms independent of \( y \) or its derivatives. A homogeneous differential equation can be solved using methods like the characteristic equation, because the equation maintains a consistent form.

• The characteristic equation helps transform it into a simpler polynomial form.
• We then solve for the roots, leading us to find the general solution.

This method efficiently handles the homogeneous nature by focusing on the function's inherent properties.

Exponential Solutions

Exponential solutions are a common approach to solving differential equations due to their useful properties. For the equation \( y'' - 2y' = 0 \), we assume a solution of the form \( y = e^{rt} \). This assumption exploits the fact that the derivative of an exponential function is also an exponential function, simplifying the equation significantly.

• Substitute \( y = e^{rt} \) into the differential equation.
• Obtain the characteristic equation and solve for \( r \).
• Use the solutions of \( r \) to form the general solution.

The resulting general solution for our example is \( y(t) = C_1 + C_2e^{2t} \), where \( C_1 \) and \( C_2 \) are constants. This approach not only provides the solution but also demonstrates how exponential functions are powerful tools in analyzing complex differential equations.