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Magazine Chapter 1: Problem 36
\(d y / d x=x / \sqrt{x^{2}-16}\)
Short Answer
Expert verified
The solution is \( y = \sqrt{x^2 - 16} + C \).
Step by step solution
01
Simplify the Given Differential Equation
Look at the given differential equation: \[ \frac{dy}{dx} = \frac{x}{\sqrt{x^2 - 16}} \]This is a separable differential equation, which means we can separate the variables involving \(x\) and \(y\).
02
Separate the Variables
Rewrite the equation to isolate the differentials:\[ dy = \frac{x}{\sqrt{x^2 - 16}} dx \]
03
Integrate Both Sides
Integrate both sides to solve for \(y\) and remove the differentials:\[ \int dy = \int \frac{x}{\sqrt{x^2 - 16}} dx \]The integral on the left side is simple:\[ y = \int \frac{x}{\sqrt{x^2 - 16}} dx \]
04
Simplify the Integral
To solve the integral on the right, use the substitution method. Let \(u = x^2 - 16 \). Then, \(du = 2x dx\). Rewrite the integral in terms of \(u\):\[ \int \frac{x}{\sqrt{x^2 - 16}} dx = \int \frac{1}{2} \frac{du}{\sqrt{u}} \]This simplifies to:\[ \frac{1}{2} \int u^{-1/2} du \]
05
Perform the Integration
Next, perform the integration:\[ \frac{1}{2} \int u^{-1/2} du = \frac{1}{2} \left( 2u^{1/2} \right) + C \]\[ = \sqrt{u} + C \]
06
Substitute Back to Original Variable
Finally, substitute \(u = x^2 - 16\) back into the equation:\[ y = \sqrt{x^2 - 16} + C \]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Integration
Integration is a fundamental concept in calculus that is essentially the reverse process of differentiation. It is used to find the accumulated area under a curve or to solve differential equations.
To successfully integrate a function, one can use various techniques like substitution, integration by parts, partial fractions, and trigonometric integration.
In the given exercise, we use integration to solve the differential equation:
Understanding these integration methods allows us to find the function y, showing how x and y vary together.
To successfully integrate a function, one can use various techniques like substitution, integration by parts, partial fractions, and trigonometric integration.
In the given exercise, we use integration to solve the differential equation:
- We separate the variables to isolate the integrals involving differentials.
- We then solve the integral on the left side, which is simple.
- Next, we solve the more complex integral on the right side.
Understanding these integration methods allows us to find the function y, showing how x and y vary together.
Substitution Method
The substitution method is a powerful integration technique used to simplify the integration of complex functions.
It involves substituting a part of the integral with a new variable to make the integration easier.
In the given exercise, the substitution method was key to simplifying the integral:
It involves substituting a part of the integral with a new variable to make the integration easier.
In the given exercise, the substitution method was key to simplifying the integral:
- First, we let \( u = x^2 - 16 \).
- Then, we calculate the differential: \( du = 2x dx \).
- This transforms the integral into a simpler form, making it easier to solve.
Differential Equations
Differential equations involve functions and their derivatives and are used to describe various phenomena like motion, growth, and decay.
A separable differential equation is one where the variables can be separated on different sides of the equation for ease of integration.
In this exercise, the given differential equation is separable:
A separable differential equation is one where the variables can be separated on different sides of the equation for ease of integration.
In this exercise, the given differential equation is separable:
- We separate the variables involving \(x\) and \(y\).
- Next, we integrate both sides to find \(y\).
- Using substitution, we simplify and solve the right side integral.
