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Chapter 9: Problem 15

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

Short Answer

Expert verified
Integrating and applying the initial condition yields the particular solution for \( y \) in terms of \( x \).

Step by step solution

01

Analyze the Given Equation

The differential equation given is \( x \ln x = y(1+\sqrt{3+y^2}) y' \). We need to solve this first-order differential equation with the initial condition \( y(1) = 1 \).
02

Separate Variables

To solve this, let's try separating the variables. Rearrange the equation by moving all terms involving \( y \) and \( y' \) to one side and \( x \) to the other side: \[ \frac{y'}{y(1+\sqrt{3 + y^2})} = \frac{1}{x \ln x}. \]
03

Integrate Both Sides

Integrate both sides separately. The left side becomes \( \int \frac{1}{y(1+\sqrt{3+y^2})} \, dy \) and the right side becomes \( \int \frac{1}{x \ln x} \, dx \).
04

Solve the Integral

Solving the left integral is complex and may require substitution. For now, assume integration yields \( f(y) \) while the right integral of \( \frac{1}{x \ln x} \) results in \( h(x) \).
05

Apply the Initial Condition

Using the initial condition \( y(1) = 1 \), apply this to find the constant of integration, \( C \).
06

Solve for the Function y(x)

Using the results from the integrals and the constant found using the initial condition, solve for \( y \) as a function of \( x \). This gives us our particular solution to the differential equation.

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Key Concepts

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

Initial Value Problem
An initial value problem in differential equations involves finding a specific solution that passes through a given point on the curve. In this context, the initial condition provided is \( y(1) = 1 \). This condition tells us the value of the function \( y \) when \( x = 1 \). It is crucial for determining the particular solution of the differential equation from a family of potential solutions derived through integration.
  • You first solve the differential equation as if you are finding a general solution.
  • Then, incorporate the initial condition to find the constant of integration.
  • Finally, substitute this constant back to get the specific solution that fits the initial condition.
This tailor-fit approach ensures that the solution is unique to the needs of the problem, satisfying the equation as well as the initial condition.
Separation of Variables
Separation of variables is a method used to solve first-order differential equations where you manipulate the equation to isolate variables on different sides. This means putting all terms involving \( y \) on one side of the equation and all terms involving \( x \) on the other side. In the given problem, the rearrangement leads us to:
\[ \frac{y'}{y(1+\sqrt{3+y^2})} = \frac{1}{x \ln x}. \]
This form makes it possible to integrate each side independently. Key steps include:
  • Carefully move variables without dropping terms or incorrect manipulation.
  • Ensure the differential \( dx \) and \( dy \) are appropriately placed, making it separable.
  • Once separated, each side can be integrated separately to solve the differential equation.
This method is powerful for solving differential equations where a straightforward integration is possible once variables are separated.
Integration of Functions
Integration is a major step in solving differential equations after separating the variables. It involves finding the antiderivative of each side of the equation. For the exercise at hand:
  • The left side involves integrating \( \int \frac{1}{y(1+\sqrt{3+y^2})} \, dy \).
  • The right side involves \( \int \frac{1}{x \ln x} \, dx \).
These integrals might require sophisticated methods such as substitution or partial fraction decomposition depending on the complexity of the terms. Integration essentially reverses differentiation, allowing us to construct the original function from its derivative.
After obtaining these integrals, they often include a constant of integration \( C \). This constant is determined using the initial value given, creating a specific solution rather than a family of solutions.
First-order Differential Equation
A first-order differential equation involves derivatives of the first degree with respect to the unknown function. The given equation, \( x \ln x = y(1+\sqrt{3+y^{2}}) y' \), exemplifies a first-order equation as it includes the first derivative \( y' \).
Key characteristics:
  • It includes the function \( y \) and its first derivative \( y' \), but no higher derivatives.
  • Such equations often model growth or decay processes, physical phenomena, and other dynamic changes.
  • Solving the equation requires techniques like separation of variables, exact equations, or integrating factors.
First-order differential equations are foundational in calculus and applied mathematics, providing insight into various real-world situations where change is constant.

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Most popular questions from this chapter

Sketch a direction field for the differential equation. Then use it to sketch three solution curves. $$ y^{\prime}=x-y+1 $$

The population of the world was about 6.1 billion in 2000 . Birth rates around that time ranged from 35 to 40 million per year and death rates ranged from 15 to 20 million per year. Let's assume that the carrying capacity for world population is 20 billion. (a) Write the logistic differential equation for these data. (Because the initial population is small compared to the carrying capacity, you can take \(k\) to be an estimate of the initial relative growth rate.) (b) Use the logistic model to estimate the world population in the year 2010 and compare with the actual population of 6.9 billion. (c) Use the logistic model to predict the world population in the years 2100 and \(2500 .\)

There is considerable evidence to support the theory that for some species there is a minimum population \(m\) such that the species will become extinct if the size of the population falls below \(m .\) This condition can be incorporated into the logistic equation by introducing the factor \((1-m / P) .\) Thus the modified logistic model is given by the differential equation $$ \frac{d P}{d t}=k P\left(1-\frac{P}{M}\right)\left(1-\frac{m}{P}\right) $$ (a) Use the differential equation to show that any solution is increasing if \(m

Solve the initial-value problem. $$ x y^{\prime}+y=x \ln x, \quad y(1)=0 $$

Solve the initial-value problem. $$ x y^{\prime}=y+x^{2} \sin x, \quad y(\pi)=0 $$
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