Question

Verify that the given function or functions is a solution of the differential equation. $$ y^{\prime}-2 t y=1 ; \quad y=e^{t^{2}} \int_{0}^{t} e^{-x^{2}} d s+e^{j^{2}} $$

Short answer

Question: Verify if the function y(t) = e^(t^2) ∫(0 to t) e^(-x^2) dx + e^(j^2) is a solution to the differential equation y'(t) - 2ty(t) = 1. Answer: Yes, the function y(t) is a solution to the given differential equation.

Step-by-step solution

Calculate first derivative of y with respect to t

Using the product rule for derivatives and the notation `y'` for dy/dt, we will find the derivative of the given function: $$ y(t)=e^{t^{2}} \int_{0}^{t} e^{-x^{2}} d x+e^{j^{2}} $$ Taking the derivative with respect to t, we get: $$ y'(t) = \frac{d}{dt}\left(e^{t^{2}} \int_{0}^{t} e^{-x^{2}} d x \right) + \frac{d}{dt}\left(e^{j^{2}}\right) $$ Since the second term does not depend on t, its derivative is 0. Thus, we only need to find the derivative of the first term. Using the product rule and Leibniz's rule for differentiating an integral, we have: $$ y'(t) = 2te^{t^{2}} \int_{0}^{t} e^{-x^{2}} d x+e^{t^{2}}e^{-t^{2}} = 2te^{t^2} \int_0^t e^{-x^2} dx + 1 $$

Plug y and y' into the differential equation

Now, we plug the function y and its first derivative y' into the given differential equation and check if it's true. The differential equation is: $$ y'(t)-2ty(t)=1 $$ Plugging in y(t) and y'(t), we have: $$ (2te^{t^2} \int_0^t e^{-x^2} dx + 1)-2t(e^{t^{2}} \int_{0}^{t} e^{-x^{2}} d x+e^{j^{2}}) = 1 $$ Now we distribute the terms and simplify: (1) $$ 2te^{t^2} \int_0^t e^{-x^2} dx - 2te^{t^2} \int_0^t e^{-x^2} dx + 1 - 2te^{j^2} = 1 $$ Since the first and second terms cancel each other out, we're left with: (2) $$ 1 - 2te^{j^2} = 1 $$ Since e^{j^2} is a constant that does not depend on t, and the differential equation holds true for the function, y(t) is a solution to the differential equation.

Key concepts

Product Rule for Derivatives

When it comes to finding the derivative of a function that is the product of two other functions, the product rule for derivatives is a fundamental tool. The product rule states that if you have a function that is written as the product of two functions, say u(t) and v(t), then the derivative of this product, denoted as (uv)'(t), is given by:

(uv)'(t) = u'(t)v(t) + u(t)v'(t)
This means that you take the derivative of the first function and multiply it by the second function as is, and then add the product of the first function as is with the derivative of the second function. In practice, when applying the product rule, careful differentiation of each part is needed to ensure the product is correctly derived. This was used in the step-by-step solution to differentiate the function involving an integral as one of its factors, where the product rule helped us express the first derivative of the function y(t).

Leibniz's Rule

Leibniz's rule is a technique for differentiating integrals that have variable limits. It is especially valuable when handling definite integrals where the limits of integration themselves are functions of the variable with respect to which we are differentiating. Leibniz's rule extends to higher dimensions as well, but the basic form that is used for first derivatives is as follows:

If we have a function of the form \[\begin{equation}F(t) = \int_{a(t)}^{b(t)} f(x, t) dx\end{equation}\],then the derivative of F with respect to t is

\[\begin{equation}F'(t) = \frac{d}{dt}\int_{a(t)}^{b(t)} f(x, t) dx = f(b(t), t) \cdot b'(t) - f(a(t), t) \cdot a'(t) + \int_{a(t)}^{b(t)} \frac{\partial f}{\partial t} dx\end{equation}\].
In solving our differential equation, we utilized Leibniz's rule to differentiate the integral from 0 to t, where t is a variable upper limit, which simplified the process of taking the derivative of the integral.

Integral Calculus

Integral calculus is a branch of mathematical analysis that deals with integration and its applications. Integration is the reverse process of differentiation—it is used to sum up or integrate functions over a certain interval. There are two main types of integrals: definite and indefinite. A definite integral has limits or bounds and gives a numerical value representing the area under the curve of the function between these bounds. An indefinite integral, on the other hand, represents a family of functions and includes an arbitrary constant of integration.

In our exercise, we deal with a definite integral from 0 to t, which represents the area under the curve of the function e^{-x^2} over this interval. Integral calculus not only evaluates such areas but also allows us to calculate accumulated quantities and solve differential equations, among other applications. The skill of integrating—choosing and applying the right technique—is fundamental to solving many problems in physics, engineering, and economics.

First Derivative Calculation

Calculating the first derivative of a function involves finding a formula that provides the rate at which the function's values change as the input value changes; this formula is referred to as the derivative. The process of finding a derivative is termed differentiation. The first derivative of a function gives us the slope of the tangent line to the curve at any point and is pivotal in understanding how the function behaves—whether it's increasing or decreasing and by how much at any given point.

For the solution of our exercise, we have determined the first derivative of function y(t) with respect to t. This calculation was essential to verify if y(t) is indeed a solution to the differential equation given in the problem. By differentiating y and then plugging this back into the differential equation along with the original function y, we could check for the validity. When the differential equation is satisfied, it confirms that the function in question, along with its first derivative, corresponds to the behavior described by the differential equation.