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Find the general solution of the indicated differential equation. If possible, find an explicit solution. $$ x^{2} y^{\prime}=y \ln y-y^{\prime} $$

Short Answer

Expert verified
The solution is \(y = e^{Ce^{\tan^{-1}(x)}}\) for constant \(C\).

Step by step solution

01

Rearrange the Equation

Given the differential equation:\[x^{2} y^{\prime}=y \ln y-y^{\prime}\]We can start by rearranging the terms to group the derivatives together:\[x^{2} y^{\prime} + y^{\prime} = y \ln y\]Factor out \(y^{\prime}\) on the left side:\[y^{\prime}(x^2 + 1) = y \ln y\]
02

Separate Variables

We can separate the variables by dividing both sides by \((x^2 + 1)\) and \(y \ln y\) (assuming \(y eq 0\) and \(\ln y eq 0\)):\[\frac{1}{y \ln y} dy = \frac{1}{x^2 + 1} dx\]Now, each side of the equation is only in terms of one variable.
03

Integrate Both Sides

Integrate both sides of the equation:\[\int \frac{1}{y \ln y} dy = \int \frac{1}{x^2 + 1} dx\]The integration on the left side requires a substitution \(v = \ln y\), so \(dv = \frac{1}{y} dy\).We then integrate:\[\int \frac{1}{v} dv = \ln |v| + C_1 = \ln |\ln y| + C_1\]
04

Integrate the Right Side

For the right side, use the standard integral formula:\[\int \frac{1}{x^2 + 1} dx = \tan^{-1}(x) + C_2\]
05

Form the General Solution

Combine the integration results from both sides:\[\ln |\ln y| + C_1 = \tan^{-1}(x) + C_2\]You can combine the constants \(C_1\) and \(C_2\) into a single constant \(C\):\[\ln |\ln y| = \tan^{-1}(x) + C\]
06

Solve for y

Exponentiate both sides to solve for \(\ln y\):\[|\ln y| = e^{\tan^{-1}(x) + C}\]Taking the exponential gets rid of the absolute value and allows us to solve for \(y\):\[\ln y = \pm e^{\tan^{-1}(x) + C} = Ce^{\tan^{-1}(x)}\]Thus, solve for \(y\):\[y = e^{Ce^{\tan^{-1}(x)}}\]
07

Simplify the Explicit Solution

Since \(y = e^{Ce^{\tan^{-1}(x)}}\) can be a bit cumbersome, it is often presented simply as the general solution with constant \(C\), implying multiple solutions based on the choice of \(C\).

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Key Concepts

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

Separation of Variables
Separation of variables is a fundamental technique for solving differential equations. This method is used when you can rearrange the equation to isolate one variable on each side of the equation along with their respective derivatives. In this context, we transformed the given differential equation to:
  • \( \frac{1}{y \ln y} dy = \frac{1}{x^2 + 1} dx \)
This separation allows us to integrate both sides independently. It is essential to verify that you correctly separate the variables and ensure that all terms involving the same variable are on one side of the equation. Doing so often reveals a pathway to solving otherwise complicated equations, offering us straightforward integration tasks.
Integration Techniques
Once separated, the equation required us to integrate each side. The left-hand side, \( \int \frac{1}{y \ln y} dy \), necessitated a substitution integration technique, where you let \( v = \ln y \) so that \( dv = \frac{1}{y} dy \). This substitution simplifies the left-hand side, transforming it into a standard form \( \int \frac{1}{v} dv \) which can be directly integrated.
On the right-hand side, \( \int \frac{1}{x^2 + 1} dx \), we applied a standard integration formula yielding \( \tan^{-1}(x) + C_2 \). These two processes, substitution and recognizing standard forms, are crucial integration techniques often used when dealing with differential equations, helping to translate complex expressions into integrable forms.
General and Explicit Solutions
The ultimate goal of solving differential equations is to find solutions that satisfy them. In this problem, we first sought the general solution. After integrating both sides, we arrived at:
  • \( \ln |\ln y| = \tan^{-1}(x) + C \)
This equation represents the general solution, encompassing a family of curves based on different values of the constant \( C \).
To proceed from a general to an explicit solution, we solve for \( y \). We achieve this by exponentiating both sides to eliminate the logarithm, moving step by step towards isolating \( y \). The explicit solution obtained here is \( y = e^{Ce^{\tan^{-1}(x)}} \), a specific expression for \( y \) in terms of \( x \).
Differential Equations
Differential equations involve functions and their derivatives. They are pivotal in modeling a wide range of phenomena in physics, engineering, and other disciplines. The objective is to discover a function that satisfies the given equation.
In this instance, we dealt with a first-order differential equation involving both \( y \) and \( y' \). Techniques such as rearranging terms, separating variables, and integration are part of the toolkit used to find the solution.
  • These solutions often offer insights into the behavior of dynamic systems.
  • Solutions may be implicit or explicit depending on derivation and manipulation ease.
Understanding and solving differential equations open up paths to precisely describe and predict real-world behaviors through mathematical models.

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Most popular questions from this chapter

Suppose that \(2 y d x+(x+y) d y=0\) has an integrating factor that is a function of \(y\) alone [i.e., \(\mu=\mu(y)]\). Find the integrating factor and use it to solve the differential equation.

An electric circuit, consisting of a capacitor, resistor, and an electromotive force can be modeled by the differential equation $$ R \frac{d q}{d t}+\frac{1}{C} q=E(t) $$ where \(R\) and \(C\) are constants (resistance and capacitance) and \(q=q(t)\) is the amount of charge on the capacitor at time \(t\). For simplicity in the following analysis, let \(R=C=1\), forming the differential equation \(d q / d t+q=E(t)\). In Exercises 17-20, an electromotive force is given in piecewise form, a favorite among engineers. Assume that the initial charge on the capacitor is zero \([q(0)=0]\). (i) Use a numerical solver to draw a graph of the charge on the capacitor during the time interval \([0,4]\). (ii) Find an explicit solution and use the formula to determine the charge on the capacitor at the end of the four-second time period. E(t)= \begin{cases}5, & \text { if } 0

A tank contains 100 gal of pure water. At time zero, a sugar-water solution containing \(0.2\) lb of sugar per gal enters the tank at a rate of 3 gal per minute. Simultaneously, a drain is opened at the bottom of the tank allowing the sugar solution to leave the tank at 3 gal per minute. Assume that the solution in the tank is kept perfectly mixed at all times. (a) What will be the sugar content in the tank after 20 minutes? (b) How long will it take the sugar content in the tank to reach \(15 \mathrm{lb}\) ? (c) What will be the eventual sugar content in the tank?

Suppose that \(\left(y^{2}+2 x y\right) d x-x^{2} d y=0\) has an integrating factor that is a function of \(y\) alone [i.e., \(\mu=\mu(y)\) ]. Find the integrating factor and use it to solve the differential equation.

A tank contains 500 gal of a salt-water solution containing \(0.05 \mathrm{lb}\) of salt per gallon of water. Pure water is poured into the tank and a drain at the bottom of the tank is adjusted so as to keep the volume of solution in the tank constant. At what rate (gal/min) should the water be poured into the tank to lower the salt concentration to \(0.01 \mathrm{lb} / \mathrm{gal}\) of water in under one hour?

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