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Find the solution of the following initial value problems. $$y^{\prime}(t)=t e^{t}, y(0)=-1$$

Short Answer

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Answer: The solution to the initial value problem is \(y(t) = e^t (t - 1)\).

Step by step solution

01

Find the general solution

We will integrate the given differential equation: $$y^{\prime}(t) = te^{t}$$ To do this, let's apply integration by parts using the formula: $$\int{u dv} = uv - \int {v du}$$ Take: \(u = t\), so \(du = dt\) and \(dv = e^{t}dt\), so \(v = \int{e^t dt} = e^t\) Now, apply the integration by parts formula: $$\int{t e^t dt} = e^{t} t - \int{e^t dt} = e^t t - e^{t} + C$$ So, the general solution is: $$y(t) = e^t t - e^{t} + C$$
02

Apply the initial condition

To find the particular solution, we'll use the initial condition: $$y(0) = -1$$ Substitute \(t=0\) into the general solution: $$-1 = e^0 \cdot 0 - e^0 + C$$ Simplify and solve for C: $$-1 = -1 + C$$ $$C = 0$$
03

Write the particular solution

Substitute the value of C back into the general solution to obtain the particular solution: $$y(t) = e^t t - e^{t}$$ So, the solution to the initial value problem is: $$y(t) = e^t (t - 1)$$

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Key Concepts

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

Initial Value Problems
An initial value problem addresses a specific type of differential equation where, aside from determining the general solution, we need to find a particular solution that satisfies a given initial condition. This initial condition typically specifies the value of the unknown function at a particular point, such as in our problem where we know that \(y(0) = -1\). This information allows us to find the constant of integration, ensuring that our solution is not just any solution, but the unique one that meets the specified conditions.

In practical terms, solving an initial value problem involves two key steps:
  • **Solving the Differential Equation:** First, we find the general solution of the differential equation without considering the initial condition. This often involves techniques such as separation of variables, integration by parts, or others, depending on the structure of the differential equation.
  • **Applying the Initial Condition:** Next, we use the initial condition to find the specific value of the constant that makes our solution unique. We substitute the given initial value into the general solution and solve for the constant of integration.
Initial value problems are vital in many fields, such as physics and engineering, as they model real-world scenarios where the starting point of a phenomenon is known. This allows for a more accurate representation and prediction of system behaviors.
Integration by Parts
Integration by parts is a fundamental technique used in calculus to integrate products of functions. It's particularly useful when dealing with expressions where one function is easily differentiable and the other easily integrable, like in our exercise with \(te^t\). The formula for integration by parts is derived from the product rule for differentiation:\[\int{u \, dv} = uv - \int{v \, du}\]To effectively use this method, we select two parts from the integrand:
  • **Function Choices:** Choose \(u\) and \(dv\) from the expression you're integrating. In our example, we let \(u = t\), so that \(du = dt\), and \(dv = e^t dt\), giving \(v = e^t\) upon integration.
  • **Substitution:** Apply the formula by substituting \(u\), \(v\), \(du\), and \(dv\) into the integration by parts formula, allowing us to break down the complex integral into simpler parts.
After performing the integration by parts, further simplify and solve any remaining integrals to get closer to the solution. This technique is widely used across mathematics and engineering due to its ability to simplify and solve integrals that would otherwise be difficult to manage with standard methods.
Differential Equation Solutions
Finding the solutions to differential equations involves understanding the relationship between a function and its derivatives. Generally, solving a differential equation means finding a function or set of functions that satisfy the equation. These solutions often involve an arbitrary constant, known as the constant of integration, which represents the infinite family of solutions.
  • **General Solution:** This is the most comprehensive form of a solution, encompassing all potential solutions to a differential equation. It includes the constant of integration, \(C\), which accounts for any shift along the curve produced by the solutions.
  • **Particular Solution:** We derive this from the general solution by applying specific initial conditions or boundary conditions. In the context of initial value problems, the particular solution resolves the unique path the function will take given the specific initial condition.
By substituting the initial condition's values into the general solution, we solve for the constant of integration and formulate the particular solution. This technique plays an essential role in predicting and understanding behaviors in systems described by differential equations, such as population dynamics in biology or motion under gravity in physics.

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Most popular questions from this chapter

Consider the following pairs of differential equations that model a predator- prey system with populations \(x\) and \(y .\) In each case, carry out the following steps. a. Identify which equation corresponds to the predator and which corresponds to the prey. b. Find the lines along which \(x^{\prime}(t)=0 .\) Find the lines along which \(y^{\prime}(t)=0\) c. Find the equilibrium points for the system. d. Identify the four regions in the first quadrant of the xy-plane in which \(x^{\prime}\) and \(y^{\prime}\) are positive or negative. e. Sketch a representative solution curve in the xy-plane and indicate the direction in which the solution evolves. $$x^{\prime}(t)=-3 x+x y, y^{\prime}(t)=2 y-x y$$

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. $$y^{2} y^{\prime}(t)=t^{2}+\frac{2}{3} t ; y(-1)=1, y(1)=0, y(-1)=-1$$

Explain why or why not Determine whether the following statements are true and give an explanation or counterexample. a. If the growth rate function for a population model is positive, then the population is increasing. b. The solution of a stirred tank initial value problem always approaches a constant as \(t \rightarrow \infty\) c. In the predator-prey models discussed in this section, if the initial predator population is zero and the initial prey population is positive, then the prey population increases without bound.

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 and graph the solution to be sure that \(m(0)\) and \(\lim _{t \rightarrow \infty} m(t)\) are correct. A 2000 -L tank is initially filled with a sugar solution with a concentration of \(40 \mathrm{g} / \mathrm{L} .\) A sugar solution with a concentration of \(10 \mathrm{g} / \mathrm{L}\) flows into the tank at a rate of \(10 \mathrm{L} / \mathrm{min} .\) The thoroughly mixed solution is drained from the tank at a rate of \(10 \mathrm{L} / \mathrm{min}\).

Suppose the solution of the initial value problem \(y^{\prime}(t)=f(t, y), y(a)=A\) is to be approximated on the interval \([a, b]\). a. If \(N+1\) grid points are used (including the endpoints), what is the time step \(\Delta t ?\) b. Write the first step of Euler's method to compute \(u_{1}\). c. Write the general step of Euler's method that applies, for \(k=0,1, \ldots, N-1\).

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