Question

True or False If \(d y / d x=k y,\) then \(y=e^{k x}+C .\) Justify your answer.

Short answer

True. The solution to the differential equation \(dy/dx = ky\) is indeed \(y = e^{kx} + C\), after considering \(C\) as a general constant that allows for negative and non-integer values.

Step-by-step solution

Understand the differential equation

The given differential equation is \(dy/dx = ky\), where \(y\) is the function of \(x\) that we need to find. This is a first-order linear differential equation.

Separate the variables

Rearrange the equation to separate the variables \(y\) and \(x\): \(dy/y = kdx\)

Integrate both sides

The next step is to integrate both sides of the equation, \(\int dy/y = \int k dx\), which yields \(ln |y| = kx + C_1\), where \(C_1\) is the constant of integration.

Solve for y

By taking the exponential of both sides, the differential equation solution is found as \(y = e^{kx+C_1} = Ce^{kx}\), where \(C = e^{C_1}\) is an arbitrary constant, equivalent to the constant in the proposed solution.

Compare with given solution

The solution obtained is \(y = Ce^{kx}\), which is the same form as the given solution \(y = e^{kx}+C\) if we consider the constant \(C\) in general terms, allowing for negative and non-integer values.

Key concepts

Differential Equation

A differential equation is a mathematical equation that relates some function with its derivatives. In practice, these equations are used to model the behavior of complex systems, such as physical processes, biological systems, or economic models. The equation given in the exercise, \(dy/dx = ky\), is a classic example of a first-order linear differential equation. This means it involves a first derivative of the unknown function \(y\) and the equation is linear with respect to \(y\) and \(dy/dx\).

Understanding differential equations involves determining the function \(y\) that satisfies the relationship outlined by the derivative \(dy/dx\). The solution to such an equation can often describe how a physical quantity changes over time or space when subjected to some kind of consistent relationship or force, represented here by the constant \(k\).

Separable Variables

Separable variables is a method used to solve a specific type of differential equation. In essence, this method involves rearranging the equation so that each variable and its corresponding differentials are on opposite sides of the equation.

In our example, the equation \(dy/dx = ky\) is manipulated by dividing both sides by \(y\) and multiplying by \(dx\) to give \(dy/y = k dx\). This rearrangement means that all terms involving \(y\) are on one side of the equation and all terms involving \(x\) are on the other, thus they are 'separated'. The importance of this step is that it allows us to use integration on each side independently to find the function \(y\) that satisfies the original equation.

Integration

Integration is the process of finding the integral of a function, which is the reverse operation of differentiation. When we integrate, we are essentially finding the area under the curve of a graph of the function. With separable variables, after we have isolated \(y\) and \(x\) on different sides of the equation, we integrate both sides to solve for our unknown function \(y\).

For the given equation \(dy/y = k dx\), integrating both sides leads us to \(\ln |y| = kx + C_1\), where \(C_1\) represents the constant of integration—a value that we determine based on initial conditions or other criteria. This step moves us closer to expressing the function \(y\) explicitly in terms of \(x\).

Exponential Function

The exponential function is a mathematical function denoted as \(e^x\), where \(e\) is Euler's number, an irrational and transcendental number approximately equal to 2.71828. This function is unique because it is its own derivative and integral, which makes it extremely important in the calculus of differential equations.

Once we have our integrated result \(\ln |y| = kx + C_1\), we can solve for \(y\) by taking the exponential of both sides, resulting in \(y = e^{kx + C_1}\). It's critical to realize that the constant \(C_1\) within the exponent can be expressed as \(Ce^{kx}\), where \(C = e^{C_1}\) is a new constant. The exponential function in our solution indicates that \(y\) grows or decays exponentially based on the sign and value of the constant \(k\), an insight that applies across numerous fields, from physics to finance.