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Chapter 4: Problem 241

Solve the following initial-value problems by using integrating factors. $$ y^{\prime}+y=x, y(0)=3 $$

Short Answer

Expert verified
The solution to the initial-value problem is \( y = x - 1 + \frac{4}{e^x} \).

Step by step solution

01

Identify the Standard Form

The given differential equation is of the form \( y' + p(x)y = q(x) \) with \( p(x) = 1 \) and \( q(x) = x \). This is a linear first-order differential equation.
02

Find the Integrating Factor

The integrating factor \( \,\mu(x) \,\) is given by the formula \( \,e^{\int p(x)dx} \). For this problem, \( p(x) = 1 \), so \( \,\mu(x) = e^{\int 1 \cdot dx} = e^{x} \,\).
03

Multiply by the Integrating Factor

Multiply both sides of the differential equation by the integrating factor \( \,e^x \,\): \[ e^x y' + e^x y = e^x x. \]
04

Write the Left Side as a Derivative

Notice that the left side of the equation \( e^x y' + e^x y \) can be rewritten as the derivative of \( (e^x y) \): \[ \frac{d}{dx}(e^x y). \] So the equation becomes: \[ \frac{d}{dx}(e^x y) = e^x x. \]
05

Integrate Both Sides

Integrate both sides with respect to \( x \): \[ \int \frac{d}{dx}(e^x y) \, dx = \int e^x x \, dx. \] The left-hand side becomes \( e^x y \), and the right-hand side requires integration by parts. Let \( u = x \) and \( dv = e^x dx \). Then \( du = dx \) and \( v = e^x \). Using integration by parts formula \( \int u \, dv = uv - \int v \, du \), we find: \[ \int e^x x \, dx = e^x x - \int e^x \, dx = e^x x - e^x + C. \]
06

Substitute Back y and Solve for C

Substitute back the results of the integration: \[ e^x y = e^x x - e^x + C. \] Using the initial condition \( y(0) = 3 \), substitute \( x = 0 \) and \( y = 3 \) to find \( C \): \[ e^0 \cdot 3 = e^0 \cdot 0 - e^0 + C \Rightarrow 3 = -1 + C \Rightarrow C = 4. \]
07

Solve for y

Finally, solve for \( y \) by dividing through by \( e^x \): \[ y = x - 1 + \frac{4}{e^x}. \]

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Key Concepts

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

Linear Differential Equations
Linear differential equations are a type of differential equation where the unknown function and its derivatives appear linearly, meaning each term is either a constant, a function of the independent variable, or the product of these with the unknown function. A linear first-order differential equation can be expressed in the form:\[ y' + p(x)y = q(x) \] where:
  • \( y' \) is the derivative of \( y \) with respect to \( x \)
  • \( p(x) \) is a function of \( x \), called the coefficient of \( y \)
  • \( q(x) \) is the non-homogenous part or forcing function
Identifying and rewriting your equation in this form is crucial for applying methods like integrating factors. This equation's linearity ensures that the solution method is flexible and powerful, often allowing straightforward integration and algebraic manipulation.
Initial-Value Problems
Initial-value problems are differential equations that also include conditions specifying the value of the unknown function at a given point, often of the form \( y(x_0) = y_0 \). This additional information allows us to find a specific solution that fits these conditions.Given our exercise, the initial-value problem is defined by:
  • Differential equation: \( y' + y = x \)
  • Initial condition: \( y(0) = 3 \)
Integrating factors are particularly useful for solving initial-value problems because they guide us in finding the general solution of the differential equation, after which we use the initial condition to determine any constants of integration and hence, the specific solution.
Integration by Parts
Integration by parts is a key technique in calculus used to integrate products of functions. Based on the product rule for differentiation, the formula is:\[ \int u \, dv = uv - \int v \, du \]In our exercise, integration by parts is required to solve \( \int e^x x \, dx \). The method involves:
  • Choosing \( u = x \), meaning \( du = dx \)
  • Choosing \( dv = e^x \, dx \), meaning \( v = e^x \)
  • Solving as \( \int e^x x \, dx = e^x x - \int e^x \, dx \)
  • Finally resolving \( \int e^x \, dx \) to yield \( e^x x - e^x + C \)
This method of integration is vital when dealing with problems where simple integration is not directly possible, enhancing the flexibility of solving complex integral calculus problems.
First-Order Differential Equations
First-order differential equations involve the first derivative of an unknown function and are crucial for modeling various real-life phenomena. These equations have the form:\[ y' + p(x)y = q(x) \] or simply \( y'(x) = f(x, y) \).Key characteristics include:
  • The presence of only the first derivative \( y' \)
  • Potential applicability in physics, engineering, biology, economics
In the context of solving an equation such as \( y' + y = x \), the goal is to determine a function \( y(x) \) that satisfies this relationship and any given initial conditions. The integrating factor method renders solving these equations manageable by transforming them into easier-to-solve expressions, utilizing integration for calculating precise solutions.

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Most popular questions from this chapter

What are the integrating factors for the following differential equations? $$ \frac{d y}{d x}=\tanh (x) y+1 $$

Consider the logistic equation in the form \(P^{\prime}=C P-P^{2}\). Draw the directional field and find the stability of the equilibria. $$ C=3 $$

Find the general solution to the differential equations. $$ y^{\prime}=e^{-y} \sin x $$

Torricelli's law states that for a water tank with a hole in the bottom that has a cross-section of \(A\) and with a height of water \(h\) above the bottom of the tank, the rate of change of volume of water flowing from the tank is proportional to the square root of the height of water, according to \(\frac{d V}{d t}=-A \sqrt{2 g h},\) where \(g\) is the acceleration due to gravity. Note that \(\frac{d V}{d t}=A \frac{d h}{d t} .\) Solve the resulting initial-value problem for the height of the water, assuming a tank with a hole of radius \(2 \mathrm{ft}\). The initial height of water is \(100 \mathrm{ft}\).

[T] Rabbits in a park have an initial population of 10 and grow at a rate of \(4 \%\) per year. If the carrying capacity is 500 , at what time does the population reach 100 rabbits?
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