Question

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

Short answer

In conclusion, the given function $y(t)=8t^6-3$ is a solution to the initial value problem as it satisfies both the differential equation $t{y}^{\prime }(t)-6y(t)=18$ and the initial condition $y(1)=5$.

Step-by-step solution

Calculate the first derivative of y(t)

First, we need to differentiate the given function with respect to t: $$y(t)=8t^6-3$$ Taking the derivative, $$y^{\prime }(t)=\frac{d}{dt}(8t^6-3)=48t^5$$

Substitute y(t) and y'(t) into the differential equation

Now let's substitute the given function $y(t)$ and its derivative $y^{\prime }(t)$ into the given differential equation: $$t{y}^{\prime }(t)-6y(t)=18$$ Replace $y^{\prime }(t)$ with $48t^5$ and $y(t)$ with $8t^6-3$: $$t(48t^5)-6(8t^6-3)=18$$

Check if the equation holds true

Now let's simplify the equation to see if it holds true: $$48t^6-48t^6+18=18$$ This equation holds true, as both sides equal 18.

Verify the initial condition

Finally, let's check if the initial condition, $y(1)=5$, is satisfied by the given function $y(t)$: $$y(1)=8(1{)}^6-3=8-3=5$$ The initial condition is satisfied. By following these steps, we have verified that the given function $y(t)=8t^6-3$ is a solution to the initial value problem.

Key concepts

Differential Equations

Differential equations are mathematical expressions that relate a function with its derivatives. These equations are used to describe various phenomena such as change in population, heat conduction, or motion of particles over time. In the context of our exercise, we deal with an initial value problem which involves finding a specific solution to a differential equation that satisfies certain predefined conditions, or initial values.

• The differential equation in this exercise, given by $t{y}^{\prime }(t)-6y(t)=18$, connects the function $y(t)$ with its derivative, $y^{\prime }(t)$.
• An initial value problem involves not only solving the differential equation but also fulfilling a condition like $y(1)=5$, which constrains the solution in such a way that it passes through this particular point.

By solving such problems, we learn how to interpret changes and behaviors of dynamic systems, which is critical in fields like physics, engineering, and economics. Even simple-looking differential equations can describe complex reality, making them an essential tool in modeling and simulation.

Derivative Calculation

Calculating a derivative is crucial to understanding differential equations as it describes the rate at which a function changes. In calculus, a derivative represents the slope of the tangent line to a function at any point.

• For our function $y(t)=8t^6-3$, calculating the derivative involves applying basic rules of differentiation.
• The derivative $y^{\prime }(t)=48t^5$, obtained by differentials such as $\frac{d}{dt}(t^n)=n{t}^{n-1}$, reveals how steeply the function changes as $t$ varies.

The derivative here is a preparatory step that allows us to substitute back into the differential equation, helping verify if the proposed function truly fulfills the requirements set by the differential equation. Understanding how to correctly calculate derivatives is fundamental in all areas where rates of change need to be measured, such as physics and economics.

Verification of Solutions

The process of verifying solutions involves substituting the proposed function and its derivatives back into the original differential equation to confirm it satisfies the equation and any given initial conditions.

• This exercise requires checking if $y(t)=8t^6-3$ is a solution by substituting it along with $y^{\prime }(t)=48t^5$ into the differential equation $t{y}^{\prime }(t)-6y(t)=18$.
• After substituting, simplifying both sides to ensure they are equal concludes the verification of the differential equation. Here, the simplification resulted in $18=18$.
• The initial condition $y(1)=5$ is checked separately by plugging $t=1$ into the function $y(t)$ and ensuring the output matches $5$.

Successfully verifying a solution acts like a confirmation flag, ensuring that our obtained function behaves exactly according to the specifications of both the equation and initial conditions, making the mathematical model accurate and reliable.