Question

In each of Problems 15 through 18 determine the values of \(r\) for which the given differential equation has solutions of the form \(y=e^{t} .\) $$ y^{\prime}+2 y=0 $$

Short answer

Question: For the first-order linear differential equation \(y^{\prime} + 2y = 0\), find the value of r for which the solution has the form \(y=e^{rt}\). Answer: The value of r for which the given differential equation has a solution of the form \(y=e^{rt}\) is \(r=-2\).

Step-by-step solution

Write the differential equation and given solution form

We are given the differential equation \(y^{\prime} + 2y = 0\) and the solution form \(y=e^{rt}\).

Find the derivative of the solution form

Differentiate the solution form with respect to t: $$\frac{dy}{dt} = re^{rt}$$

Substitute the solution form and its derivative into the differential equation

Substitute the function and its derivative into the given equation: $$re^{rt} + 2e^{rt} = 0$$

Factor out the common term

Factor out the term \(e^{rt}\) to simplify the equation: $$e^{rt}(r+2) = 0$$

Determine values of r

Since \(e^{rt}\) is never equal to 0, we can divide both sides of the equation by it. This results in the expression $$r+2=0$$ Solving for r, we get: $$r = -2$$

State the final answer

The value of r for which the given differential equation has a solution of the form \(y=e^{rt}\) is \(r=-2\).