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Chapter 8: Problem 32

Solve the initial-value problem. State an interval on which the solution exists. $$ t^{2} y^{\prime}+y=1 ; y(1)=4 $$

Short Answer

Expert verified
The solution is \( y(t) = 1 + \frac{3}{e} e^{1/t} \). The solution exists for \( t > 0 \).

Step by step solution

01

Convert to Standard Form

Rewrite the given differential equation to the form \( y' + p(t)y = g(t) \) First, divide every term by \( t^2 \): \[ y' + \frac{y}{t^2} = \frac{1}{t^2} \]
02

Find Integrating Factor

The integrating factor \( \mu(t) \) is found using the expression: \[ \mu(t) = e^{\int p(t) dt } \] Here, \( p(t) = \frac{1}{t^2} \). So, \[ \mu(t) = e^{\int \frac{1}{t^2} dt } = e^{-\frac{1}{t}} \]
03

Multiply Through by Integrating Factor

Multiply the entire differential equation by the integrating factor: \[ e^{-1/t} y' + \frac{1}{t^2}e^{-1/t} y = \frac{1}{t^2} e^{-1/t} \] The left-hand side simplifies to: \[ \frac{d}{dt}(e^{-1/t} y) = \frac{1}{t^2} e^{-1/t} \]
04

Integrate Both Sides

Integrate both sides with respect to \( t \): \[ \int \frac{d}{dt}(e^{-1/t} y) \ dt = \int \frac{1}{t^2} e^{-1/t} \ dt \] The left side becomes \[ e^{-1/t} y \] The right side requires an integrating factor solution: Let \( u = -\frac{1}{t} \), thus \( du = \frac{1}{t^2} dt \). So the integral becomes \[ \int e^{u} du = e^{u} = e^{-1/t} \]
05

Solve for y

We have now: \[ e^{-1/t} y = e^{-1/t} + C \] Multiply through by \( e^{1/t} \): \[ y = 1 + Ce^{1/t} \]
06

Apply Initial Condition

Use the initial condition \( y(1) = 4 \): \[ 4 = 1 + C \cdot e^{1/1} \implies 4 = 1 + C \cdot e \implies C = 3/e \]
07

State Solution on Interval

The solution to the differential equation is: \[ y(t) = 1 + \frac{3}{e} e^{1/t} \] The interval on which the solution exists is given by the domain of the original differential equation: \( t > 0 \).

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Key Concepts

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

differential equation
Differential equations are mathematical equations that relate a function with its derivatives. They describe various phenomena such as motion, growth, decay, and heat transfer. In this problem, we have a first-order linear differential equation: \[ t^{2} y^{\text{'} }+ y = 1 \; y(1) = 4 \] This equation involves an unknown function \( y(t) \) and its derivative \( y^{\text{'}}(t) \). Our goal is to find this function \( y(t) \) that satisfies both the differential equation and the given initial condition.
integrating factor
An integrating factor is a function used to simplify the process of solving a linear differential equation. It transforms the original equation into a form where the left-hand side becomes the derivative of a product of functions.To find an integrating factor, we start by writing the equation in standard form: \[ y^{\text{'}} + p(t) y = g(t) \]Here, we found\[ p(t) = \frac{1}{t^2} \]The integrating factor \( \mu(t) \) is given by:\[ \mu(t) = e^{ \int p(t) dt } = e^{ \int \frac{1}{t^2} dt } = e^{ - \frac{1}{t} } \]Multiplying through by this integrating factor aids in simplifying the differential equation.
initial condition
An initial condition specifies the value of the function at a particular point. This helps determine the specific solution to a differential equation from a family of potential solutions.In our exercise:\[ y(1) = 4 \]We use this condition after finding the general solution to solve for the constants. Applying it in our case:\[ y = 1 + C e^{ 1/t } \]Given:\[ 4 = 1 + C e^{ 1/1 } \implies 4 = 1 + C e \implies C = \frac{3}{e} \]
standard form conversion
Converting to standard form is a crucial step when solving linear differential equations. It involves rewriting the equation to match the form:\[ y^{\text{'}} + p(t)y = g(t) \]For our given equation:\[ t^{2} y^{\text{'}} + y = 1 \]We divide every term by \( t^2 \). This yields:\[ y^{\text{'}} + \frac{y}{t^2} = \frac{1}{t^2} \]This transformation makes it easier to identify the functions \( p(t) \) and \( g(t) \) and apply the method of integrating factors to solve the differential equation.

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Most popular questions from this chapter

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