Question

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

Short answer

The values of k for which y(t) = e^(kt) is a solution to the given differential equation are k = 1 and k = -1.

Step-by-step solution

Find the first derivative of y(t)

To find the first derivative of the given function \(y(t)=e^{k t}\), we can apply the chain rule: $$ y'(t) = \frac{\mathrm{d} }{\mathrm{d} t}(e^{k t}) = k \cdot e^{k t} $$

Find the second derivative of y(t)

Now, we should find the second derivative. We will find it by differentiating the first derivative \(y'(t)\) with respect to \(t\) using the chain rule again: $$ y''(t) = \frac{\mathrm{d} }{\mathrm{d} t}(k \cdot e^{k t}) = k^2 \cdot e^{k t} $$

Plug y(t), y'(t), and y''(t) into the differential equation

We will substitute the expressions we found for \(y(t)\), \(y'(t)\), and \(y''(t)\) into the given differential equation \(y^{\prime \prime}-y=0\): $$ (k^2 \cdot e^{k t}) - e^{k t} = 0 $$

Solve the equation for k

Now, we will solve this equation for \(k\). First, notice that we can factor out the term \(e^{kt}\) from both terms on the left side of the equation: $$ e^{k t}(k^2 - 1) = 0 $$ Since the term \(e^{kt}\) is never zero, the equation will only hold true when the term inside the parentheses is equal to zero: $$ k^2 - 1 = 0 $$ This is a simple quadratic equation, and upon solving this equation, we obtain two possible values for \(k\): $$ k = \pm 1 $$ Thus, the values of \(k\) for which \(y(t)=e^{k t}\) is a solution of the given differential equation are \(k=1\) and \(k=-1\).

Key concepts

Characteristic Equation

When tackling a linear homogeneous differential equation like \(y^{\prime \prime} - y = 0\), an essential tool is the characteristic equation. This equation is formed by replacing each derivative of \(y\) with a power of a variable (often \(r\) or \(\lambda\)). For instance, the second derivative \(y^{\prime \prime}\) correlates to \(r^2\), the first derivative \(y^{\prime}\) to \(r\), and \(y\) as itself.

Our differential equation translates to the characteristic equation \(r^2 - 1 = 0\). Solving for \(r\) gives us the roots that are crucial in determining the general solution to the differential equation. In our case, we seek the values of constant \(k\) that will make \(e^{kt}\) a particular solution, which connects to finding the roots of the characteristic equation.

In a broader context, characteristic equations describe the behavior of solutions to differential equations and are pivotal in finding both particular solutions and understanding the general behavior of dynamical systems.

Second Derivative

The second derivative of a function is the derivative of its derivative, giving insights into the function’s curvature or acceleration. In our problem, we found that the second derivative of \(y(t) = e^{kt}\) is \(y^{\text{\prime\prime}}(t) = k^2e^{kt}\).

Mathematically, the second derivative is represented as \(f^{\text{\prime\prime}}(x)\) or \(\frac{d^2}{dx^2}f(x)\). It plays an important role beyond just differential equations; in physics, it represents acceleration, whereas in geometry, it defines the concavity of a graph. Identifying \(y^{\text{\prime\prime}}\) as \(k^2e^{kt}\) and substituting it in the differential equation is a critical step in determining for which values of \(k\) the function \(e^{kt}\) satisfies the original equation.

Chain Rule

The chain rule is a formula to compute the derivative of a composite function. If you have a function \(h(x)\) that can be written as the composition of two functions \(f\) and \(g\), where \(h(x) = f(g(x))\), then by the chain rule, \(h’(x) = f’(g(x)) · g’(x)\).

In the context of our problem, the chain rule was utilized twice: once to find the first derivative \(y’(t) = ke^{kt}\) and then to find the second derivative \(y’’(t) = k^2e^{kt}\). These steps are fundamental in expressing each derivative in terms of the original function \(y(t)\), allowing for their substitution back into the differential equation. Understanding and applying the chain rule correctly is crucial in solving differential equations involving functions composed of exponentials and other functions.

Quadratic Equation

The quadratic equation is a second-order polynomial equation of the form \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are constants, and \(a eq 0\). Solutions to the quadratic equation can be found via factoring, completing the square, or using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}\).

After substituting \(y(t)\), \(y’(t)\), and \(y’’(t)\) into the differential equation, we arrived at the quadratic equation \(k^2 - 1 = 0\), which simplifies to \(k = \pm 1\). Recognizing and solving quadratic equations is a fundamental skill in algebra, appearing frequently in various areas of mathematics and science, including the analysis of differential equations.