/*! 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 3 Solve the differential equation.... [FREE SOLUTION] | 91Ó°ÊÓ

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Solve the differential equation. $$ x y y^{\prime}=x^{2}+1 $$

Short Answer

Expert verified
The solution is \( y = \pm \sqrt{x^2 + 2 \ln|x| + C} \).

Step by step solution

01

Convert to Separable Form

Start with the given differential equation: \(x y y' = x^2 + 1\). To separate the variables, express \(y'\) in terms of \(x\) and \(y\): \(y' = \frac{x^2 + 1}{xy}\). This can be rewritten as \(y' = \frac{x}{y} + \frac{1}{xy}\).
02

Separate Variables

Now, rearrange the equation to get all terms involving \(y\) on one side and terms involving \(x\) on the other side. Multiply both sides by \(y\): \(y \, dy = \left( x + \frac{1}{x} \right) dx\).
03

Integrate Both Sides

Integrate both sides: \[\int y \, dy = \int \left( x + \frac{1}{x} \right) dx.\]The left side integrates to \(\frac{y^2}{2}\). The right side integrates to \(\frac{x^2}{2} + \ln|x| + C\), where \(C\) is the integration constant.
04

Combine and Solve for General Solution

Combine the integration results: \[ \frac{y^2}{2} = \frac{x^2}{2} + \ln|x| + C.\]Multiply every term by 2 to simplify, getting: \[ y^2 = x^2 + 2 \ln|x| + 2C.\]
05

Final Answer

Express the general solution for \(y\): \[ y = \pm \sqrt{x^2 + 2 \ln|x| + C}.\] Where \(C\) is the constant of integration after multiplying 2 into the constant \(C\) from the integration.

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Key Concepts

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

Separable Differential Equations
A separable differential equation is a type of differential equation where the variables can be separated on different sides of the equation. This means you can organize all terms involving one variable (like \(y\)) on one side, and all terms involving another variable (like \(x\)) on the other. The goal is to make the equation easier to solve by focusing on one variable at a time.

In the exercise, the original equation \(x y y' = x^2 + 1\) was rewritten to separate the variables: \(y' = \frac{x^2 + 1}{xy}\). This allows us to rearrange terms to eventually arrive at \(y \, dy = \left( x + \frac{1}{x} \right) dx\). Each side contains only one variable, making it ready for integration, a key step in solving separable differential equations.
Integration Techniques
Once you have a separable differential equation, the next step typically involves integrating both sides of the equation. Integration is a process that helps find a function when its derivative is known.

In our case, after separating the variables, we obtained \(\int y \, dy = \int \left( x + \frac{1}{x} \right) dx\). Here, you need to integrate the \(y\) terms with respect to \(y\) and the \(x\) terms with respect to \(x\).
  • The integration of \(y\) gives \(\frac{y^2}{2}\).
  • The integration for \(x\) includes two parts: \(\int x \, dx = \frac{x^2}{2}\) and \(\int \frac{1}{x} \, dx = \ln|x|\).
The result is a general form involving a constant \(C\), which we denote as the integration constant. These integration techniques are crucial tools when working with differential equations.
General Solution
The general solution of a differential equation represents all possible solutions, given by the integration process, typically involving an arbitrary constant known as the integration constant.

After solving the differential equation, we arrived at \(\frac{y^2}{2} = \frac{x^2}{2} + \ln|x| + C\). To simplify, multiply by 2, resulting in \(y^2 = x^2 + 2 \ln|x| + 2C\). Solving for \(y\) gives us the general solution: \(y = \pm \sqrt{x^2 + 2 \ln|x| + C}\).

This expression includes all possible solutions for the differential equation, although only specific solutions are valid based on initial conditions supplied or additional constraints, such as domain restrictions.
Integration Constant
The integration constant is crucial when finding the general solution to a differential equation. It accounts for the family of solutions due to the indefinite integration process.

During integration, each side was assigned a constant since the original boundaries or initial conditions are unknown. Here, it appeared as \(C\) in \(\int y \, dy = \int \left( x + \frac{1}{x} \right) dx + C\). After simplification of the general solution, the integration constant was adjusted through multiplication by 2, but it's important to note that the constant \(C\) remains arbitrary until further specified by additional conditions.
  • This constant allows for flexibility in solutions and adapts when initial values are introduced.
  • This concept ensures that the solution is adaptable to changes, representing a broad range of potential solutions.
Understanding the integration constant is vital in differential equations for handling various scenarios that satisfy the initial differential equation.

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Most popular questions from this chapter

Find the solution of the differential equation that satisfies the given initial condition. $$ x \ln x=y(1+\sqrt{3+y^{2}}) y^{\prime}, \quad y(1)=1 $$

When a raindrop falls, it increases in size and so its mass at time \(t\) is a function of \(t,\) namely, \(m(t) .\) The rate of growth of the mass is \(k m(t)\) for some positive constant \(k .\) When we apply Newton's Law of Motion to the raindrop, we get \((m v)^{\prime}=g m\) where \(v\) is the velocity of the raindrop (directed downward) and \(g\) is the acceleration due to gravity. The terminal velocity of the raindrop is \(\lim _{t \rightarrow \infty} v(t)\). Find an expression for the terminal velocity in terms of \(g\) and \(k .\)

The table gives the midyear population of Japan, in thousands, from 1960 to 2010 . $$ \begin{array}{|c|c|}\hline \text { Year } & {\text { Population }} \\ \hline 1960 & {94,092} \\ {1965} & {98,883} \\ {1970} & {104,345} \\ {1975} & {111,573} \\ {1980} & {116,807} \\ {1985} & {120,754} \\ \hline\end{array} $$ $$ \begin{array}{|c|c|}\hline \text { Year } & {\text { Population }} \\ \hline 1990 & {123,537} \\ {1995} & {125,327} \\ {2000} & {126,776} \\ {2005} & {127,715} \\ {2010} & {127,579} \\ \hline\end{array} $$ Use a calculator to fit both an exponential function and a logistic function to these data. Graph the data points and both functions, and comment on the accuracy of the models. [Hint: Subtract \(94,000\) from each of the population figures. Then, after obtaining a model from your calculator, add \(94,000\) to get your final model. It might be helpful to choose \(t=0 \text { to correspond to } 1960 \text { or } 1980 .]\) Use a calculator to fit both an exponential function and a logistic function to these data. Graph the data points and both functions, and comment on the accuracy of the models. [Hint: Subtract \(94,000\) from each of the population figures. Then, after obtaining a model from your calculator, add \(94,000\) to get your final model. It might be helpful to choose \(t=0 \text { to correspond to } 1960 \text { or } 1980 .]\)

A tank contains \(1000 \mathrm{L}\) of brine with \(15 \mathrm{kg}\) of dissolved salt. Pure water enters the tank at a rate of \(10 \mathrm{L} / \mathrm{min}\). The solution is kept thoroughly mixed and drains from the tank at the same rate. How much salt is in the tank (a) after \(t\) minutes and (b) after 20 minutes?

(a) Use a computer algebra system to draw a direction field for the differential equation. Get a printout and use it to sketch some solution curves without solving the differential equation. (b) Solve the differential equation. (c) Use the CAS to draw several members of the family of solutions obtained in part (b). Compare with the curves from part (a). $$ y^{\prime}=y^{2} $$

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