Question

Solve the following initial-value problems with the initial condition \(y_{0}=0\) and graph the solution.\(\frac{d y}{d t}=y+1\)

Short answer

The solution is \(y = -1 + e^{t}\). Graph starts at \(-1\) and increases exponentially.

Step-by-step solution

Recognize the Differential Equation Type

The given differential equation is \(\frac{dy}{dt} = y+1\). This is a first-order linear differential equation, which can be solved by the method of integrating factors.

Rewrite the Equation in Standard Form

To use the integrating factor method, rewrite the equation in the form \(\frac{dy}{dt} - y = 1\). This gives us the standard form \(dy/dt + P(t)y = Q(t)\) where \(P(t) = -1\) and \(Q(t) = 1\).

Calculate the Integrating Factor

The integrating factor \(\mu(t)\) is calculated as \(e^{\int -1 \, dt} = e^{-t}\). This will be used to multiply through the differential equation.

Multiply the Standard Form by the Integrating Factor

Multiply the entire equation \(\frac{dy}{dt} - y = 1\) by the integrating factor \(e^{-t}\). This gives \(e^{-t} \frac{dy}{dt} - e^{-t}y = e^{-t}\).

Recognize the Left-Hand Side as a Derivative

The left-hand side of the equation can be rewritten as a derivative: \(\frac{d}{dt}(e^{-t}y) = e^{-t}\). This simplifies the equation into an easily integrable form.

Integrate Both Sides with Respect to t

Integrate both sides of \(\frac{d}{dt}(e^{-t}y) = e^{-t}\) with respect to \(t\). This yields \(e^{-t}y = -e^{-t} + C\), where \(C\) is the constant of integration.

Solve for y(t)

Isolate \(y\) by multiplying both sides of the equation by \(e^{t}\), resulting in \(y = -1 + Ce^{t}\).

Apply the Initial Condition

Use the initial condition \(y(0) = 0\) to find the constant \(C\). Substituting into the equation, we get \(0 = -1 + C\cdot 1\), so \(C = 1\). Thus, the solution is \(y = -1 + e^{t}\).

Graph the Solution

Graph the solution \(y = -1 + e^{t}\) using a suitable range of \(t\). This will be an exponential curve starting from \(-1\) at \(t=0\) and increasing as \(t\) increases.

Key concepts

First-Order Linear Differential Equations

Understanding first-order linear differential equations is essential for solving many real-world problems. These equations involve derivatives and appear often in fields like physics and engineering. They are called "first-order" because they involve the first derivative of a function. The general form of a first-order linear differential equation is \( \frac{dy}{dt} + P(t)y = Q(t) \). Here, \( P(t) \) and \( Q(t) \) are functions of \( t \).

In our original exercise, the equation \( \frac{dy}{dt} = y + 1 \) can be rewritten to match this standard form. By doing so, we identify \( P(t) \) as \(-1\) and \( Q(t) \) as \(1\). Recognizing this form is the first step in finding a solution efficiently.

• Key Point: First-order indicates the involvement of only the first derivative.
• The equation must be linear in \( y \).

With these principles, you can tackle any first-order linear differential equation, beginning with rearranging it into the standard form.

Integrating Factor Method

The integrating factor method is a powerful technique for solving first-order linear differential equations. It transforms the differential equation into a form that is easy to integrate. To find the integrating factor, use the formula \( \mu(t) = e^{\int P(t) \ dt} \). This factor simplifies the equation dramatically.

In our case, \( P(t) = -1 \) so the integrating factor becomes \( \mu(t) = e^{-t} \). With this integrating factor, you multiply through the entire differential equation. This multiplication allows the left-hand side to be expressed as the derivative of a product, simplifying the equation to a form that is readily integrable.

• The method makes the complex differential equation more manageable.
• The goal is to make the equation integrable as a directly solvable derivative.

This method allows you to convert and solve the equation in just a few progressive steps.

Exponential Solutions

Exponential solutions often arise when solving differential equations using the integrating factor method. These solutions typically involve expressions with the base \( e \), the natural exponent. Exponential functions have distinctive properties that make them suitable for modeling growth and decay processes in nature and science.

In our solution, after integrating the transformed equation, we derived a function of the form \( y = -1 + Ce^{t} \). This is an exponential solution because it involves the term \( e^{t} \). More specifically, this represents an exponential growth starting from an initial value displaced downward by 1.

• Exponential growth and decay are common in natural phenomena.
• Recognizing this form helps you predict and model behaviors over time.

Solutions involving \( e \) are aesthetically simple but powerful in applications.

Constants of Integration

The constant of integration, typically represented as \( C \), appears when integrating a differential equation. Its role is to generalize the solution, as it accounts for the varied initial conditions that solutions might need to satisfy.

When we isolated \( y(t) \) in our equation, we obtained \( e^{-t}y = -e^{-t} + C \). This generalized form includes \( C \), which was determined by applying the given initial condition \( y(0) = 0 \). By substitute this into the equation, we found \( C = 1 \).

• \( C \) ensures the solution fits specific boundary or initial conditions.
• The initial condition personalizes and specifies the solution for a particular scenario.

Always remember to solve for \( C \) when an initial condition is provided, as it tailors the mathematical model to its real-life application.