Chapter 8: Problem 40
Find the solution of the following initial value problems. $$y^{\prime}(t)=t e^{t}, y(0)=-1$$
Short Answer
Expert verified
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:
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.
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.
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.