Question

Verify that the given function \(y\) is a solution of the initial value problem that follows it. $$y(t)=8 t^{6}-3 ; t y^{\prime}(t)-6 y(t)=18, y(1)=5$$

Short answer

Question: Verify that the function \(y(t) = 8t^6 - 3\) is a solution to the initial value problem \(ty'(t) - 6y(t) = 18\) with initial condition \(y(1) = 5\). Answer: After finding the first derivative \(y'(t) = 48t^5\), substituting the function and its first derivative into the initial value problem, and simplifying the resulting equation, the equation holds true. Moreover, the initial condition \(y(1) = 5\) is satisfied by the given function. Therefore, we can verify that \(y(t) = 8t^6 - 3\) is a solution to the initial value problem.

Step-by-step solution

Find the first derivative of the given function

First, we need to find the first derivative of the given function. The given function is: $$y(t) = 8t^6 - 3$$ We will differentiate this function with respect to \(t\). Using the power rule, we get: $$y'(t) = 6 \cdot 8t^5 = 48t^5$$

Substitute the function and its first derivative into the initial value problem

Now that we found the first derivative, we can plug \(y(t)\) and \(y'(t)\) back into the initial value problem: $$ty'(t) - 6y(t) = 18$$ By replacing \(y(t)\) by \(8t^6 - 3\) and \(y'(t)\) by \(48t^5\), we have: $$t(48t^5) - 6(8t^6 - 3) = 18$$

Simplify and check if the equation holds true

Let's simplify the equation and see if it holds true: \begin{align*} 48t^6 - 48t^6 + 18 &= 18 \\ 18 &=18 \end{align*} Since both sides of the equation are equal, the given function does satisfy the initial value problem.

Check the initial condition

Finally, we need to check if the initial condition \(y(1) = 5\) is also satisfied by the given function. Plugging \(t = 1\) into the given function, we get: $$y(1) = 8(1)^6 - 3 = 8 - 3 = 5$$ Thus, the initial condition is also satisfied. Since the given function satisfies both the initial value problem and the initial condition, we can now conclude that \(y(t) = 8t^6 - 3\) is indeed a solution to the initial value problem.

Key concepts

Differentiation

Differentiation is a fundamental concept in calculus that focuses on finding the derivative of a function. A derivative represents how a function changes as its input changes. In simpler terms, it's the mathematical way of understanding the rate of change or the slope of the function at any given point.
For example, if we take a function like \(y(t) = 8t^6 - 3\), differentiation allows us to find its derivative, \(y'(t)\). The derivative tells us how the value of \(y\) changes for a small change in \(t\).
To differentiate effectively, we use specific rules like the power rule, which helps us compute derivatives easily and accurately. Understanding differentiation is crucial because it not only helps verify solutions in problems like the initial value problem but is also essential for optimizing functions and modeling situations in physics, engineering, and economics.

Function Verification

Function verification involves checking whether a specified function indeed satisfies a particular condition or set of conditions. In the context of initial value problems, this means determining if our function is a valid solution to the given differential equation and initial conditions.
For the problem considered, \(y(t) = 8t^6 - 3\) must satisfy both the differential equation \(ty'(t) - 6y(t) = 18\) and the initial condition \(y(1) = 5\).
Verification involves substituting the function and its derivative into the equation to confirm equality. If both sides of the equation match, the function is indeed a valid solution. Moreover, the initial condition is verified by ensuring that substituting the initial condition into the function results in the provided initial value.

Power Rule

The power rule is a key technique in differentiation that simplifies the process of finding derivatives. This rule states that if you have a function of the form \(x^n\), its derivative is \(nx^{n-1}\).
This makes differentiating polynomial functions straightforward and quick. For instance, considering the term \(8t^6\) in our function, the power rule is applied to find the derivative. The exponent \(6\) is multiplied by the coefficient \(8\) to get \(48\), and the new exponent becomes \(5\) after subtracting one, resulting in \(48t^5\).
The power rule greatly reduces the complexity involved in manual calculations, helping you tackle problems more efficiently. It is essential for any calculus student to master this rule, as it's widely used in both academic exercises and real-world applications.