Question

\(y^{\prime}=1+2 y / t\), with \(y(1)=-3 / 4\), on \([1,4]\).

Short answer

The solution is \( y = -t + \frac{1}{4}t^2 \).

Step-by-step solution

Identify the Type of Differential Equation

The given differential equation is \( y' = 1 + \frac{2y}{t} \). This equation is of the form \( y' = f(t, y) \), indicating that it is a first-order ordinary differential equation.

Check for a Method of Solution

Since \( y' = 1 + \frac{2y}{t} \) is a linear first-order differential equation, but doesn't appear easily separable, we can consider using an integrating factor or checking for exactness. Integrating factor seems suitable here.

Rearrange and Identify Linear Form Components

Re-arrange the equation to the standard linear form \( y' + P(t)y = Q(t) \). The equation becomes \( y' - \frac{2}{t}y = 1 \). Here, \( P(t) = -\frac{2}{t} \) and \( Q(t) = 1 \).

Find the Integrating Factor

The integrating factor, \( \mu(t) \), is given by \( e^{\int P(t) \, dt} \). Calculate \( \mu(t) = e^{-2 \int \frac{1}{t} \, dt} = e^{-2 \ln |t|} = t^{-2} \).

Multiply Through by the Integrating Factor

Multiply every term in the equation \( y' - \frac{2}{t}y = 1 \) by \( t^{-2} \), yielding \( t^{-2}y' - \frac{2}{t^3}y = t^{-2} \).

Simplify to Solve for \( y \) by Integration

The left side is now the derivative of \( t^{-2}y \): \( \frac{d}{dt}(t^{-2}y) = t^{-2} \). Integrate both sides with respect to \( t \). The integration yields \( t^{-2}y = \int t^{-2} \, dt = -t^{-1} + C \), where \( C \) is a constant.

Solve for \( y \)

Multiply through by \( t^2 \) to get \( y = -t + Ct^2 \).

Solve for \( C \) Using Initial Condition

Use the initial condition \( y(1) = -\frac{3}{4} \). Substitute \( t = 1 \) and \( y = -\frac{3}{4} \) into \( y = -t + Ct^2 \), yielding \( -\frac{3}{4} = -1 + C \). Solve for \( C \) to get \( C = \frac{1}{4} \).

Write the Final Solution

Substitute \( C = \frac{1}{4} \) back into \( y = -t + Ct^2 \) to get \( y = -t + \frac{1}{4}t^2 \). This is the solution to the differential equation given the initial condition.

Key concepts

First-order differential equation

A first-order differential equation involves derivatives of a function with respect to one variable, and the highest derivative is the first derivative. In the given problem, the equation is \( y' = 1 + \frac{2y}{t} \), which is an example of such an equation. First-order differential equations take the general form \( y' = f(t, y) \), where \( f(t, y) \) is a function of the variable \( t \) and the unknown function \( y \).

These equations often model real-world processes like population growth, radioactive decay, or cooling liquids. Understanding and solving them is crucial as they provide insights into how systems change over time.

In many cases, finding the solution involves techniques such as separation of variables, integrating factors, and checking for exactness among others.

Linear differential equation

A linear differential equation is one where the unknown function and its derivatives appear linearly, meaning the power of \( y \) (or its derivatives) does not exceed one. Our equation, \( y' - \frac{2}{t}y = 1 \), is linear because it can be rewritten in the standard form \( y' + P(t)y = Q(t) \). Here, \( P(t) = -\frac{2}{t} \) and \( Q(t) = 1 \).

Linear equations are significant because they often have unique solutions determined by initial or boundary conditions. These solutions can describe predictable systems, making linear equations invaluable in engineering and physics.

To solve a linear differential equation, we typically use methods like integrating factors or variation of parameters, both of which simplify the equation to allow easier integration.

Integrating factor

The integrating factor is a technique used to solve linear first-order differential equations. It involves transforming an equation into a simpler form that can be easily integrated. For our equation \( y' - \frac{2}{t}y = 1 \), the integrating factor \( \mu(t) \) is found by computing \( e^{\int P(t) \, dt} \).

In this case, we compute \( \mu(t) = e^{-2 \int \frac{1}{t} \, dt} = e^{-2 \ln |t|} = t^{-2} \). This multiplier effectively turns the left side of the differential equation into the derivative of a product, \( \frac{d}{dt}(t^{-2}y) \).

After applying the integrating factor, integrate both sides with respect to \( t \) to find a solution. The process is straightforward once familiar with exponential and logarithmic functions, making integrating factors a powerful method for solving linear equations.

Initial value problem

An initial value problem specifies the value of the solution function at a particular point, which is crucial for determining a unique solution for the differential equation. Here, we are given \( y(1) = -\frac{3}{4} \), meaning the function \( y(t) \) at \( t = 1 \) equals \(-\frac{3}{4} \).

This initial condition helps us find the specific constant \( C \) after integrating the equation. By substituting \( t = 1 \) and \( y = -\frac{3}{4} \) into the general solution \( y = -t + Ct^2 \), we determine \( C = \frac{1}{4} \).

Initial value problems are vital in applications where the system's state at a specific time influences the entire solution. Such precise information guides the model's behavior and ensures the solution's relevance to the given context.