Question

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

Short answer

Answer: The solution to the given initial value problem is $$y(t) = te^t - e^t$$.

Step-by-step solution

Identify the differential equation and initial condition

We are given the first order differential equation: $$y^{\prime}(t) = te^t$$ and the initial condition: $$y(0) = -1.$$

Solve the differential equation

To solve the differential equation, we need to find the function y(t) whose derivative is equal to $$te^t$$. We can do this via integration: $$\int y^{\prime}(t) dt = \int te^t dt.$$ We integrate the right hand side using integration by parts (set u=t, dv=e^t dt). For this, we need to calculate the following: du = dt v = ∫ e^t dt = e^t Now, apply the integration by parts formula: $$\int udv = uv - \int vdu.$$ Plugging in our values, we get: $$y(t) = te^t - \int e^t dt.$$ Now, we integrate the remaining integral: $$y(t) = te^t - e^t + C,$$ where C is the constant of integration.

Apply the initial condition

We now apply the initial condition y(0) = -1 to find the constant C: $$-1 = 0\cdot e^0 - e^0 + C$$ $$-1 = -1 + C$$ $$C = 0$$

Write the final solution

With the constant C found, we can now write the final solution to our initial value problem: $$y(t) = te^t - e^t.$$

Key concepts

Differential Equations

A differential equation is a type of equation that involves functions and their derivatives. It describes how a particular quantity changes with respect to another. In other words, it captures the relationships between varying quantities and their rates of change. Differential equations are fundamental in mathematical modeling for various scientific fields like physics, biology, and engineering.

The exercise involves a first order differential equation, which means the highest derivative in the equation is the first derivative. The given equation in this task is \( y'(t) = te^t \). It tells us that the rate of change of the function \( y \) with respect to \( t \) is equal to \( te^t \). Often, solving such equations reveals insights into the behavior of dynamic systems. You find these equations often in problems involving exponential growth, radioactive decay, or Newton's law of cooling.

Integration by Parts

Integration by parts is a technique used to integrate products of functions. It's derived from the product rule for differentiation and is particularly useful when dealing with integrals that are not straightforward. The formula for integration by parts is:

• \( \int u \, dv = uv - \int v \, du \)

Here, \( u \) and \( dv \) are chosen such that their differentiation and integration simplify the integral.
In the exercise, this method is used to solve the differential equation \( y'(t) = te^t \). By choosing \( u = t \) and \( dv = e^t dt \), the integration becomes possible:

• \( du = dt \)
• \( v = \int e^t dt = e^t \)

Substitute back into the formula to find \( y(t) = te^t - \int e^t dt \), which simplifies to \( y(t) = te^t - e^t + C \). This step helps break down a complex integral into manageable parts.

Initial Conditions

Initial conditions provide specific values for the solution of a differential equation at a particular point. They are crucial for determining unique solutions in problems involving differential equations. Think of initial conditions as starting points or benchmarks.
In this exercise, we are given the initial condition \( y(0) = -1 \). This means when \( t = 0 \), the value of \( y \) is \(-1\).

Applying initial conditions allows us to solve for any constants that appear during integration. In this exercise, after integrating, the general solution had a constant \( C \). By substituting \( y(0) = -1 \) into \( y(t) = te^t - e^t + C \), we determine that \( C = 0 \). Thus, the complete solution adheres to the given initial start point.

First Order Differential Equations

First order differential equations involve only first derivatives and are typically expressed as \( y'(t) = f(t, y) \). Their solutions aim to find a function \( y(t) \) that satisfies the given relation between its derivative and the function itself.
These equations frequently arise in real-world scenarios where a rate of change links directly to other variables. Examples include population modeling, electrical circuits, and cooling of objects.
For instance, in this exercise, the first order differential equation \( y'(t) = te^t \) presents a functional relationship between \( y'(t) \) and \( t \). The task is to determine the unknown function \( y(t) \) that ensures equality. Here, by using integration and initial conditions, the solution \( y(t) = te^t - e^t \) provides the specific behavior of the system described by the problem.