Question

The solution of the differential equation satisfying initial condition \(y(0)=1\) is given. \(y^{\prime}=2 t y ; \quad y(t)=e^{t^{2}}\)

Short answer

Based on the given steps: Question: Verify if the given solution \(y(t) = e^{t^2}\) satisfies the first-order linear differential equation \(y^{\prime} = 2ty\) and initial condition \(y(0) = 1\). Answer: Yes, the given solution \(y(t) = e^{t^2}\) does satisfy the differential equation \(y^{\prime} = 2ty\) and the initial condition \(y(0) = 1\).

Step-by-step solution

Differentiate the given solution with respect to t

Compute the derivative of the given solution \(y(t) = e^{t^2}\) with respect to \(t\). We will use the chain rule to differentiate. The derivative of \(e^{t^2}\) with respect to \(t\) is \(e^{t^2} \cdot (t^2)^{\prime}\).

Calculate the derivative of \(t^2\) with respect to t

The derivative of \(t^2\) with respect to \(t\) is simply \(2t\). Thus, substituting this back into the expression for the derivative of y(t), we get: \(y^{\prime}(t) = e^{t^2} \cdot 2t\)

Plug the given solution into the original differential equation

Now we need to check if the computed derivative y'(t) satisfies the given differential equation, \(y^{\prime} = 2ty\). Substitute y'(t) = \(2te^{t^2}\) and y(t) = \(e^{t^2}\) into the given equation: \(2te^{t^2} = 2t \cdot e^{t^2}\) This equation holds true, and we can conclude that the given solution, \(y(t) = e^{t^2}\), indeed satisfies the differential equation.

Check the initial condition

Finally, we need to check if the given solution, y(t) = \(e^{t^2}\), satisfies the initial condition, y(0) = 1. Substitute t = 0 into the solution: \(y(0) = e^{0^2} = e^0 = 1\) This matches the initial condition exactly. Therefore, the given solution, y(t) = \(e^{t^2}\), satisfies both the differential equation and the initial condition, making it the correct solution.

Key concepts

Understanding Initial Conditions

In the context of differential equations, initial conditions play a crucial role in determining the specific solution to a differential equation. A differential equation often has an infinite number of solutions that describe a family of curves, but when we apply an initial condition, it helps us to identify the unique curve that corresponds to the physical context of the problem.

For instance, the initial condition given as y(0) = 1 implies that when the independent variable, which is typically time (t), is zero, the dependent variable (y) must be one. This particular point (0, 1) must lie on the curve that represents the solution to the differential equation. Checking if our purported solution satisfies this condition is often the final step to verifying its correctness.

So, in our exercise, we ensure that our solution y(t) = e^{t^2} passes through the point specified by the initial condition. When t = 0, our solution becomes y(0) = e^0 = 1, matching perfectly with the initial condition provided, thus confirming that we've found the correct solution.

Applying the Chain Rule in Derivatives

The chain rule is an essential tool in calculus, particularly when dealing with the derivative calculation of composite functions. It is a method for finding the derivative of a function that is the composition of two or more functions.

The general form of the chain rule states that if you have a function y that is dependent on u, which in turn is dependent on t, namely, y = f(u) and u = g(t), the derivative of y with respect to t is given by dy/dt = (dy/du) * (du/dt).

In our exercise, the function y(t) = e^{t^2} is a composition of e^u where u = t^2. Using the chain rule, we differentiate the outer function, leave the inner function as is, and then multiply by the derivative of the inner function. This process simplifies the derivative calculation and ensures the accuracy of our results.

Calculating Derivatives

The calculation of a derivative is at the heart of differential calculus. It involves finding the rate at which one quantity changes with respect to another. In the case of our example, we're interested in how the function y(t) changes as t changes.

The process of calculating a derivative is guided by rules of differentiation, such as the power rule, product rule, quotient rule, and chain rule, as mentioned earlier. The power rule, for example, tells us that the derivative of t^n with respect to t is nt^{n-1}. This rule is particularly relevant to our problem where we find the derivative of t^2 to be 2t.

Derivative calculation is not just a mechanical process; it requires an understanding of the function behavior and the application of the appropriate rules. With practice, recognizing which rule to apply and performing the calculation becomes a more intuitive process. Always remember that the derivative gives us a powerful way to understand change and motion, and is widely applied in physics, engineering, economics, biology, and more.