Problem 6 Solve the following first-order ... [FREE SOLUTION]

Chapter 15: Problem 6

Solve the following first-order linear differential equations; if an initial condition is given, definitize the arbitrary constant: $$\frac{d y}{d t}+y=t$$

Short Answer

Expert verified

The solution of the differential equation is $ y = t - 1 + Ce^{-t} $.

Step by step solution

Identify the Standard Form of Linear Differential Equation

First, recognize that the given equation $ \frac{d y}{d t} + y = t $ is a first-order linear differential equation of the form $ \frac{d y}{d t} + P(t)y = Q(t) $. Here, $ P(t) = 1 $ and $ Q(t) = t $.

Find the Integrating Factor

For a linear differential equation, the integrating factor $ \mu(t) $ is given by $ e^{\int P(t) \, dt} $. Here, $ P(t) = 1 $, so $ \int P(t) \, dt = \int 1 \, dt = t $. Thus, $ \mu(t) = e^t $.

Multiply the Equation by the Integrating Factor

Multiply the entire differential equation by the integrating factor $ e^t $. This gives: \[ e^t \frac{d y}{d t} + e^t y = t e^t \].

Simplify to Find the Left-Hand Side

Notice that the left-hand side of the equation can be written as the derivative of a product: \[ \frac{d}{d t}(e^t y) = e^t \frac{d y}{d t} + e^t y \]. So the equation becomes: \[ \frac{d}{d t}(e^t y) = t e^t \].

Integrate Both Sides

Integrate both sides with respect to $ t $: \[ \int \frac{d}{d t}(e^t y) \, dt = \int t e^t \, dt \]. The left-hand side simplifies to $ e^t y $, and the integral on the right-hand side can be solved using integration by parts. Let $ u = t $ and $ dv = e^t \, dt $, then $ du = dt $ and $ v = e^t $. Integration by parts formula $ \int u \, dv = uv - \int v \, du $ gives: \[ \int t e^t \, dt = t e^t - \int e^t \, dt = t e^t - e^t + C \]. Thus, \[ e^t y = t e^t - e^t + C \].

Solve for y

Solve the equation $ e^t y = t e^t - e^t + C $ for $ y $ by dividing every term by $ e^t $: \[ y = t - 1 + Ce^{-t} \].

Apply Initial Condition (If Given)

If an initial condition is specified, such as $ y(0) = y_0 $, substitute $ t = 0 $ and $ y = y_0 $ into the solution to determine the constant $ C $. If no initial condition is provided, $ C $ remains arbitrary.

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Key Concepts

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

First-order linear differential equations

A first-order linear differential equation is a type of equation that involves two main components: a function and its derivative. These equations follow a standard form, which is crucial to identify when you begin solving. The form is given by \[ \frac{d y}{d t} + P(t)y = Q(t) \] Understanding this structure helps you apply systematic methods to solve the equation. In this form,

$ y $ is the function you want to find,
$ \frac{d y}{d t} $ is its derivative with respect to $ t $,
$ P(t) $ and $ Q(t) $ are known functions of $ t $.

In our example, $ P(t) = 1 $ and $ Q(t) = t $, which directly fits into this form, simplifying the process of finding solutions.

Integrating factor

One powerful technique to solve first-order linear differential equations is using an integrating factor. This method transforms a difficult equation into one that is much easier to solve.Here's how it works step-by-step:

Determine the integrating factor $ \mu(t) $ using the formula $ e^{\int P(t) \, dt} $.
In our example, $ P(t) = 1 $, so the integral is simply $ t $, leading to the integrating factor $ \mu(t) = e^t $.
Multiply every term in the differential equation by $ \mu(t) $. This rewrites the left-hand side as the derivative of a product.

This simplifies integration significantly, guiding you towards the solution with ease.

Integration by parts

Once the integrating factor simplifies the equation, you might be left with integrals that need special techniques to solve. For integration involving products of functions, **integration by parts** is a key tool.The integration by parts formula is: \[ \int u \, dv = uv - \int v \, du \] Where you choose:

$ u $ and $ dv $
Find their derivatives $ du $ and $ v $

In our solution:

Set $ u = t $ and $ dv = e^t \, dt $ leading to $ du = dt $ and $ v = e^t $.
This transforms the integral $ \int t e^t \, dt $ into $ t e^t - \int e^t \, dt $.

This method uncovers solutions to otherwise challenging problems.

Initial condition

An **initial condition** provides specific values for the function at a certain point, narrowing down the infinite possibilities into a single, unique solution.Given a general solution for a first-order linear differential equation:

The constant $ C $ appears in the solution due to the integration process.
If an initial condition is given, like $ y(0) = y_0 $, substitute these values to solve for $ C $.

Applying these initial conditions to our example:

If $ y(0) = y_0 $, plug in $ t = 0 $, and $ y = y_0 $ into the general solution $ y = t - 1 + Ce^{-t} $.
Isolating $ C $ will render a specific solution tailored to this initial scenario.

Initial conditions are essential when aiming for precise results with real-world applications.

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