/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 7 Solve each first-order linear di... [FREE SOLUTION] | 91Ó°ÊÓ

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Solve each first-order linear differential equation. $$ x y^{\prime}-y=x^{2} $$

Short Answer

Expert verified
The solution is \( y = x^2 + Cx \).

Step by step solution

01

Convert the Equation to Standard Linear Form

The given differential equation is \( x y^{\prime} - y = x^2 \). First, we rewrite it in the standard form of a first-order linear differential equation \( y' + P(x) y = Q(x) \). Divide all terms by \( x \) to get:\[ y' - \frac{y}{x} = x \].
02

Identify P(x) and Q(x)

In the standard linear form \( y' + P(x) y = Q(x) \), we identify \( P(x) = -\frac{1}{x} \) and \( Q(x) = x \).
03

Determine the Integrating Factor

The integrating factor \( \mu(x) \) is given by \( e^{\int P(x) dx} \). Calculate:\[ \mu(x) = e^{\int -\frac{1}{x} dx} = e^{ -\ln|x|} = \frac{1}{x} \].
04

Multiply the Entire Equation by the Integrating Factor

Multiply the entire equation by the integrating factor \( \mu(x) = \frac{1}{x} \):\[ \frac{1}{x} y' - \frac{y}{x^2} = 1 \].
05

Recognize the Left-Hand Side as a Derivative

The left-hand side of the equation is now the derivative of \( \frac{y}{x} \):\[ \frac{d}{dx} \left( \frac{y}{x} \right) = 1 \].
06

Integrate Both Sides

Integrate both sides of the equation with respect to \( x \):\[ \int \frac{d}{dx} \left( \frac{y}{x} \right) dx = \int 1 \, dx \].This gives:\[ \frac{y}{x} = x + C \], where \( C \) is a constant of integration.
07

Solve for y

Multiply both sides by \( x \) to solve for \( y \):\[ y = x^2 + Cx \].

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Key Concepts

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

Differential Equations
Differential equations are mathematical equations that relate some function with its derivatives. They are a key concept in calculus and are crucial for modeling various phenomena in fields like physics, engineering, and biology. A differential equation involves unknown functions and their rates of change. Simple examples can include describing how populations change over time or how heat flows through materials. There are several types of differential equations, but first-order linear differential equations, like the one given in the exercise, generally involve derivatives of only the first degree.These equations take the form:
  • \( y' + P(x)y = Q(x) \)
This exercise presented such an equation, but not in standard form. By rearranging, we identify functions \( P(x) \) and \( Q(x) \), which guide us in solving it systematically. First-order linear differential equations can often be solved using methods that involve straightforward algebraic manipulation, making them excellent starting points for learners new to calculus problem-solving.
Integrating Factors
The concept of an integrating factor makes solving linear differential equations more manageable. An integrating factor is a function you multiply through an equation, transforming it into a simpler (or solved) form. For first-order linear differential equations, we use the integrating factor to "un-differentiate" — or reverse the process of differentiation.To find an integrating factor for an equation of the form:
  • \( y' + P(x)y = Q(x) \)
we compute \( \mu(x) = e^{\int P(x) \, dx} \).For the equation in the exercise, where \( P(x) = -\frac{1}{x} \), the integrating factor is \( \frac{1}{x} \). When we multiply this factor across the entire equation, it simplifies the left-hand side into a derivative of \( \frac{y}{x} \), which is remarkably easier to integrate. This simplification is crucial in deriving the solution effortlessly, showcasing the power of integrating factors in calculus.
Calculus Problem Solving
Calculus problem solving is all about turning complex mathematical scenarios into a series of simpler, solvable steps. This includes identifying functions, integrating, differentiating, and employing various rules and theorems. The given problem involves transforming a differential equation into a solvable form and then integrating to find the solution. The exercise exemplifies typical calculus problem-solving techniques:
  • Begin by writing the equation in a standard form, allowing clearer paths for next steps.
  • Use methods like multiplying by integrating factors to reduce complexity.
  • Recognize patterns like derivatives or standard integrals that simplify the process.
  • Integrate, solve for constants, and simplify the result into a usable function.
Breaking problems down into these logical steps makes seemingly daunting calculus problems more approachable and solvable, emphasizing the importance of systematic thought processes in advanced mathematics.

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Most popular questions from this chapter

BIOMEDICAL: Dieting A person's weight \(w(t)\) after days of eating \(c\) calories per day can be modeled by the following differential equation $$ w^{\prime}+0.005 w=\frac{c}{3500} $$ where the 0.005 represents the proportional weight loss per day when eating nothing, and 3500 is the conversion rate for calories into pounds. a. If a person initially weighing 170 pounds goes on a diet of 2100 calories per day, find a formula for the person's weight after \(t\) days. b. Use your solution to find when the person will have lost 15 pounds. c. Find the "limiting weight" that will be a approached if the person continues on this diet indefinitely.

Solve each differential equation with the given initial condition. $$ \begin{array}{l} y^{\prime}-3 x^{2} y=6 x^{2} \\ y(0)=1 \end{array} $$

Solve each first-order linear differential equation. $$ y^{\prime}+x y=0 $$

For the initial value problem \(\left\\{\begin{array}{l}y^{\prime}=4 x y \\\ y(1)=3\end{array}\right.\) state the initial point \(\left(x_{0}, y_{0}\right)\) and calculate the slope of the solution at this point.

When you swallow a pill, the medication passes through your stomach lining into your bloodstream, where some is absorbed by the cells of your body and the rest continues to circulate for future absorption. The amount \(y(t)\) of medication remaining in the bloodstream after \(t\) hours can be modeled by the differential equation $$ \frac{d y}{d t}=a b e^{-b t}-c y $$ for constants \(a, b,\) and \(c\) (respectively the dosage of the pill, the dissolution constant of the pill, and the absorption constant of the medication). For the given values of the constants: a. Substitute the constants into the stated differential equation. b. Solve the differential equation (with the initial condition of having no medicine in the bloodstream at time \(t=0)\) to find a formula for the amount of medicine in the bloodstream at any time \(t\) (hours). c. Use your solution to find the amount of medicine in the bloodstream at time \(t=2\) hours. d. Graph your solution on a graphing calculator and find when the amount of medication in the bloodstream is maximized. \(a=10 \mathrm{mg}, \quad b=3, \quad c=0.2\)

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