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Chapter 9: Problem 18

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

Short Answer

Expert verified
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 \(ty'(t) - 6y(t) = 18\) and the initial condition \(y(1) = 5\).

Step by step solution

01

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, $$y'(t) = \frac{d}{dt} (8t^6 - 3) = 48t^5$$
02

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

Now let's substitute the given function \(y(t)\) and its derivative \(y'(t)\) into the given differential equation: $$ty'(t) - 6y(t) = 18$$ Replace \(y'(t)\) with \(48t^5\) and \(y(t)\) with \(8t^6 - 3\): $$t(48t^5) - 6(8t^6 - 3) = 18$$
03

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.
04

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.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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 \( t y'(t) - 6 y(t) = 18 \), connects the function \( y(t) \) with its derivative, \( y'(t) \).
  • 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 \( y'(t) = 48t^5 \), obtained by differentials such as \( \frac{d}{dt}(t^n) = nt^{n-1} \), 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 \( y'(t) = 48t^5 \) into the differential equation \( t y'(t) - 6 y(t) = 18 \).
  • 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.

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Most popular questions from this chapter

Orthogonal trajectories Two curves are orthogonal to each other if their tangent lines are perpendicular at each point of intersection. A family of curves forms orthogonal trajectories with another family of curves if each curve in one family is orthogonal to each curve in the other family. Use the following steps to find the orthogonal trajectories of the family of ellipses \(2 x^{2}+y^{2}=a^{2}\). a. Apply implicit differentiation to \(2 x^{2}+y^{2}=a^{2}\) to show that \(\frac{d y}{d x}=\frac{-2 x}{y}\) b. The family of trajectories orthogonal to \(2 x^{2}+y^{2}=a^{2}\) satisfies the differential equation \(\frac{d y}{d x}=\frac{y}{2 x} .\) Why? c. Solve the differential equation in part (b) to verify that \(y^{2}=e^{c}|x|\) and then explain why it follows that \(y^{2}=k x\) where \(k\) is an arbitrary constant. Therefore, the family of parabolas \(y^{2}=k x\) forms the orthogonal trajectories of the family of ellipses \(2 x^{2}+y^{2}=a^{2}\).

For the following separable equations, carry out the indicated analysis. a. Find the general solution of the equation. b. Find the value of the arbitrary constant associated with each initial condition. (Each initial condition requires a different constant.) c. Use the graph of the general solution that is provided to sketch the solution curve for each initial condition. $$y^{\prime}(t)=\frac{t^{2}}{y^{2}+1} ; y(-1)=1, y(0)=0, y(-1)=-1$$

For each of the following stirred tank reactions, carry out the following analysis. a. Write an initial value problem for the mass of the substance. b. Solve the initial value problem. A one-million-liter pond is contaminated by a chemical pollutant with a concentration of \(20 \mathrm{g} / \mathrm{L} .\) The source of the pollutant is removed, and pure water is allowed to flow into the pond at a rate of \(1200 \mathrm{L} / \mathrm{hr}\). Assuming the pond is thoroughly mixed and drained at a rate of \(1200 \mathrm{L} / \mathrm{hr}\), how long does it take to reduce the concentration of the solution in the pond to \(10 \%\) of the initial value?

Logistic equation for spread of rumors Sociologists model the spread of rumors using logistic equations. The key assumption is that at any given time, a fraction \(y\) of the population, where \(0 \leq y \leq 1,\) knows the rumor, while the remaining fraction \(1-y\) does not. Furthermore, the rumor spreads by interactions between those who know the rumor and those who do not. The number of such interactions is proportional to \(y(1-y) .\) Therefore, the equation that describes the spread of the rumor is \(y^{\prime}(t)=k y(1-y)\) for \(t \geq 0,\) where \(k\) is a positive real number and \(t\) is measured in weeks. The number of people who initially know the rumor is \(y(0)=y_{0},\) where \(0 \leq y_{0} \leq 1\). a. Solve this initial value problem and give the solution in terms of \(k\) and \(y_{0}.\) b. Assume \(k=0.3\) weeks \(^{-1}\) and graph the solution for \(y_{0}=0.1\) and \(y_{0}=0.7.\) c. Describe and interpret the long-term behavior of the rumor function, for any \(0 \leq y_{0} \leq 1\).

Give a geometrical explanation of how Euler's method works.
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