Question

For what value(s) of the constant \(k\), if any, is \(y(t)\) a solution of the given differential equation? $$ y^{\prime}+2 y=0, \quad y(t)=e^{k t} $$

Short answer

Answer: The value of the constant k is -2.

Step-by-step solution

Find derivative of \(y(t)\)

Differentiate the given solution \(y(t) = e^{kt}\) with respect to \(t\). $$ y'(t) = \frac{d}{dt}(e^{kt}) = ke^{kt} $$

Substitute into differential equation

Substitute the expressions for \(y'(t)\) and \(y(t)\) into the given differential equation \(y'(t) + 2y(t) = 0\). $$ (ke^{kt}) + 2(e^{kt}) = 0 $$

Solve for \(k\)

To solve for \(k\), we can factor the common term \(e^{kt}\) from the equation. $$ e^{kt}(k+2) = 0 $$ Since the exponential function \(e^{kt}\) is never equal to 0, we can divide both sides by the exponential term without changing the solution. $$ k+2 = 0 $$ Now, solving for \(k\) yields $$ k = -2 $$ Thus, the value of the constant \(k\) that makes \(y(t) = e^{kt}\) a solution of the given differential equation is \(k = -2\).

Key concepts

Exponential Solutions

Exponential solutions are a powerful tool in solving differential equations, especially when the equations are linear and involve constant coefficients. In the exercise, we are given a potential solution of the form \( y(t) = e^{kt} \). This type of solution is characteristic of exponential growth or decay, depending on the value of \( k \). The exponential function \( e^{kt} \) acts as a solution to many fundamental differential equations because its unique property, the derivative of an exponential function, is proportional to the function itself. This makes it easy to manipulate mathematically and to fit in various contexts, especially in natural sciences, finance, and engineering where exponential growth or decay is observed.

Derivative Calculation

To check if \( y(t) = e^{kt} \) is a solution to the given differential equation, the first step is to calculate the derivative \( y'(t) \). Since the function is exponential, its derivative can be determined using the chain rule. The chain rule states that the derivative of \( e^{kt} \) with respect to \( t \) is \( ke^{kt} \). Thus, \( y'(t) = \frac{d}{dt}(e^{kt}) = ke^{kt} \). This derivative calculation is crucial as it provides the component needed to be replaced back into the original differential equation, allowing us to check if the equation holds true with this form of \( y(t) \).

Solving Differential Equations

Solving a differential equation involves finding a function that satisfies the equation. In this context, we insert the derived expressions of \( y'(t) \) and \( y(t) \) into the differential equation \( y'(t) + 2y(t) = 0 \). Substituting these yields \( (ke^{kt}) + 2(e^{kt}) = 0 \). Simplifying this gives \( e^{kt}(k+2) = 0 \). Because exponential functions never equal zero, the expression \( k+2 \) must be zero for the equation to be satisfied, leading directly to the need to solve \( k + 2 = 0 \). Solving this simple linear equation gives us the required value of \( k \).

Constant Determination

The final step in confirming the solution is determining the constant \( k \). In our simplified equation \( k + 2 = 0 \), it's clear that solving for \( k \) gives \( k = -2 \). This single constant frame not only determines the behavior of the entire solution but also sets the parameters for the exponential growth or decay inherent in \( y(t) = e^{kt} \). This process of constant determination is pivotal as it allows us to tailor the general solution \( y(t) = e^{kt} \) to the specific constraints of the differential equation given. By adjusting \( k \), we align with the condition that renders the differential equation valid, ensuring the solution is not just arbitrary but precise to the equation’s requirements.