Question

Find a differential equation whose general solution is \(y=c_{1} e^{-t / 2}+c_{2} e^{-2 t}\)

Short answer

Question: Find a differential equation whose general solution is given by \(y=c_{1} e^{-t/2}+c_{2} e^{-2 t}\). Answer: The differential equation is \(2\frac{dy}{dt} + \frac{d^2y}{dt^2} = 0\).

Step-by-step solution

Differentiate the function once

First, differentiate the given function y with respect to t: $$ \frac{dy}{dt} = \frac{d(c_{1} e^{-t/2}+c_{2} e^{-2 t})}{dt} $$ Using the chain rule for differentiation: $$ \frac{dy}{dt} = -\frac{1}{2} c_{1} e^{-t/2} - 2 c_{2} e^{-2 t} $$

Differentiate the function twice

Now, differentiate the function a second time with respect to t: $$ \frac{d^2y}{dt^2} = \frac{d(-\frac{1}{2} c_{1} e^{-t/2} - 2 c_{2} e^{-2 t})}{dt} $$ Using the chain rule for differentiation: $$ \frac{d^2y}{dt^2} = \frac{1}{4} c_{1} e^{-t/2} + 4 c_{2} e^{-2 t} $$

Obtain the differential equation

Now we have two equations: $$ \frac{dy}{dt} = -\frac{1}{2} c_{1} e^{-t/2} - 2 c_{2} e^{-2 t} \\ \frac{d^2y}{dt^2} = \frac{1}{4} c_{1} e^{-t/2} + 4 c_{2} e^{-2 t} $$ In order to eliminate constants \(c_{1}\) and \(c_{2}\), multiply the first equation (the first derivative) by 2 and add it to the second equation (the second derivative): $$ 2\frac{dy}{dt} + \frac{d^2y}{dt^2} = -c_{1} e^{-t/2} - 4 c_{2} e^{-2 t} + \frac{1}{4} c_{1} e^{-t/2} + 4 c_{2} e^{-2 t} $$ Notice that the terms containing \(c_{1}\) and \(c_{2}\) cancel out: $$ 2\frac{dy}{dt} + \frac{d^2y}{dt^2} = 0 $$ This is the desired differential equation, which has the general solution \(y=c_{1} e^{-t /2}+c_{2} e^{-2 t}\).

Key concepts

Chain Rule Differentiation

The chain rule is a crucial concept in calculus, particularly when dealing with differentiation of functions composed of other functions. Put simply, it helps us find the derivative of a function that is nested within another function. In the context of our exercise, we see the function of interest is an exponential function multiplied by a constant. Let's break down the process:

To differentiate the function y=c_{1} e^{-t/2}+c_{2} e^{-2 t}, we apply the chain rule. For the component c_{1} e^{-t/2}, the outer function is the exponential, and the inner function is -t/2. Likewise, for c_{2} e^{-2t}, the outer function is the exponential, and the inner is -2t.

The Process of Applying Chain Rule

During differentiation, each term is treated separately. For the first term, the derivative of the outer function (with respect to the inner function) is e^{-t/2}. We then multiply this by the derivative of the inner function, which is -1/2. The result is -1/2 c_{1} e^{-t/2}. We follow a similar process for the second term, which gives us -2 c_{2} e^{-2t}. Together, this provides the derivative of y with respect to t.

Second Derivative

The second derivative is just what it sounds like: the derivative of the derivative. It gives us information about the curvature or concavity of a function's graph and how a function's rate of change is itself changing. In practical terms, the second derivative tells us whether the slope of the original function is increasing or decreasing.

In the given exercise, after calculating the first derivative of y, we proceed to find the second derivative by differentiating once more. This step is crucial as it moves us towards developing a differential equation that does not involve the constants c_{1} and c_{2}, which are arbitrary and represent any number.

The Importance of Second Derivative in this Exercise

The second derivative in our exercise involves applying the chain rule differentiation again to the result of the first derivative. This shows the repeated application of differentiation techniques to model physical systems or express complex relationships through equations. The result of the second derivative in this context plays a pivotal role in formulating the differential equation that represents the general solution.

General Solution of Differential Equations

A general solution of a differential equation represents a family of solutions that encompasses all possible scenarios. In contrast to a particular solution, which is specific to initial conditions, the general solution includes arbitrary constants signifying that it contains various solutions depending on the initial conditions applied.

In our exercise, the goal is to find a differential equation compatible with the given general solution y=c_{1} e^{-t/2}+c_{2} e^{-2t}. The process involves taking derivatives and combining them to eliminate the constants c_{1} and c_{2}, leading us to the desired equation.

Obtaining the General Differential Equation

The manipulations performed with the first and second derivatives result in an equation free of the constants c_{1} and c_{2}. This ultimately leaves us with a differential equation in y and its derivatives only. The reason why we can eliminate c_{1} and c_{2} and still claim that the equation matches the initial general solution is due to the nature of differential equations: they are designed to work for an infinite number of functions that only vary by their initial conditions, represented by these constants.