Question

If \(y(t)=3 e^{t^{2}}\) is known to be the solution of the initial value problem $$ y^{\prime}+p(t) y=0, \quad y(0)=y_{0}, $$ what must the function \(p(t)\) and the constant \(y_{0}\) be?

Short answer

Answer: The function p(t) is -2t and the initial value y_0 is 3.

Step-by-step solution

Find y'(t)

Differentiate the given function \(y(t) = 3e^{t^2}\) with respect to \(t\): $$ y'(t) = \frac{d}{dt}(3e^{t^2}) = 3 \cdot \frac{d}{dt}(e^{t^2}) = 3e^{t^2} \cdot \frac{d}{dt}(t^2) = 3e^{t^2} \cdot 2t. $$ So, \(y'(t) = 6te^{t^2}\).

Find p(t)

Plug the values of \(y(t)\) and \(y'(t)\) into the given differential equation \(y'(t) + p(t) y(t) = 0\): $$ 6te^{t^2} + p(t)(3e^{t^2}) = 0. $$ We want to find \(p(t)\), so let's isolate it: $$ p(t) = -\frac{6te^{t^2}}{3e^{t^2}} = -2t. $$ Therefore, \(p(t) = -2t\).

Find y_0

Now, we'll use the initial condition \(y(0) = y_0\). Evaluate the given function \(y(t)\) at \(t = 0\): $$ y_0 = y(0) = 3e^{0^2} = 3e^0 = 3 \cdot 1 = 3. $$ Hence, \(y_0 = 3\). #Conclusion# The function \(p(t)\) must be \(-2t\) and the constant \(y_0\) must be \(3\) for \(y(t) = 3e^{t^2}\) to be the solution of the given initial value problem.

Key concepts

Differential Equations

A differential equation is a mathematical equation that relates some function with its derivatives. In applications, the functions usually represent physical quantities, the derivatives represent their rates of change, and the equation defines a relationship between the two. When you come across an initial value problem involving a differential equation, like the one in our exercise, you're given a differential equation to solve and an initial condition to satisfy.

In this case, the differential equation is linear of the form y' + p(t)y = 0, where y' is the derivative of y with respect to t, and p(t) is a function of the independent variable, t. The challenge lies not only in solving the equation but ensuring that the solution also fulfills the conditions specified at t = 0. This kind of problem is fundamental in various fields like physics, engineering, and economics because it describes a system based on an initial state and predicts its future behavior.

Exponential Functions

An exponential function is a mathematical function of the form f(x) = a^x, where the base a is a positive real number. Not merely confined to academic exercises, exponential functions are prevalent in real-world scenarios like population growth, radioactive decay, and interest calculations. The power of an exponential function is in its property of growth or decay at rates proportional to the value of the function itself.

For the student trying to grasp its essence, imagine putting money in a bank account with a constant interest rate; over time, the amount of money increases exponentially. Similarly, the given function in our problem, y(t) = 3e^{t^2}, suggests a scenario where, as time increases, the function's value rapidly grows at a rate which itself changes with time, an aspect characteristic of exponential growth.

Natural Exponential Function

Among all exponential functions, the natural exponential function is a unique one where the base a is the irrational number e, approximately equal to 2.71828. It is denoted by e^x or exp(x). This function arises frequently in mathematics and is a central function in the field of calculus because of its unique property: its derivative is equal to itself.

From a student’s perspective, understanding the natural exponential function is key to mastering many concepts in calculus and beyond since it's used as the foundational building block for continuous compounding, certain probability distributions, and more. In our exercise, the function y(t) = 3e^{t^2} takes advantage of the natural exponential function's properties. By differentiating it, the presence of e^{t^2} in both y(t) and its derivative y'(t) simplifies the process of solving the initial value differential equation.