/*! 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 22 \(\frac{d y}{d x}=\frac{\ln x}{y... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 9: Problem 22

\(\frac{d y}{d x}=\frac{\ln x}{y} ; y=100\) when \(x=1\)

Short Answer

Expert verified
\(\frac{d y}{d x} = 0\) when \(x = 1\) and \(y = 100\).

Step by step solution

01

Identify Given Information

Given the differential equation \(\frac{d y}{d x} = \frac{\text{ln} x}{y}\). We need to find \(\frac{dy}{dx}\) at \(x = 1\) and \(y = 100\).
02

Substitute the Known Values

Substitute \(x = 1\) and \(y = 100\) into the equation \( \frac{d y}{d x} = \frac{\text{ln} x}{y} \). This gives us: \( \frac{d y}{d x} = \frac{\text{ln} 1}{100} \).
03

Calculate \(\text{ln} 1\)

Recall that \( \text{ln} 1 = 0 \). So the equation simplifies to: \( \frac{d y}{d x} = \frac{0}{100} \).
04

Simplify the Result

Simplify \( \frac{0}{100} \) to get \( \frac{d y}{d x} = 0 \).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Differentiation
Differentiation is a fundamental concept in calculus. It allows us to determine the rate at which a function is changing at any given point.
In the context of the exercise, we use differentiation to find how the variable 'y' changes with respect to 'x'. The differential equation provided is \(\frac{dy}{dx} = \frac{\ln x}{y}\).
Here, \(\frac{dy}{dx}\) represents the rate of change of 'y' with respect to 'x'.
By substituting the given values of 'x' and 'y' into the equation, we are able to explore how 'y' behaves at that specific point.
Logarithms
Logarithms are a way to solve for the exponent in equations of the form \(a^b = c\). The notation \(\ln x\) specifically refers to the natural logarithm, which uses the base 'e'.
In the given problem, the natural logarithm appears as \(\ln x\).
When we substitute the value of 'x' as 1 into \(\ln x\), we use the known property: \(\ln 1 = 0\).
This crucial step simplifies the differential equation and helps us solve it efficiently.
  • Logarithms help to transform multiplicative relationships into additive ones.
  • The natural logarithm function is especially common in continuous growth models.
  • Understanding logarithms is key for solving complex algebraic equations.
Initial Conditions
Initial conditions are specific values provided for the variables in a differential equation. They allow us to find a unique solution to the equation.
In our exercise, the initial conditions are given as \(x = 1\) and \y = 100\.
By substituting these values into the differential equation, we evaluate the equation at a specific point.
This approach helps us determine the behavior or value of the solution at that particular point.
  • Initial conditions provide a starting point for solving differential equations.
  • They ensure the solution fits a specific scenario or physical situation.
  • Using initial conditions can simplify the integration process in solving the equation.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

\(\frac{d y}{d x}=y e^{-2 x}\)

\(y_{n}-y_{n-1}=3 ; y_{0}=1\)

\(\frac{d y}{d x}=5 x^{4}-3 x^{2}-2 ; y=4\) when \(x=1\)

\(\frac{d y}{d x}=\frac{\ln x}{x y}\)

\(\frac{d^{2} y}{d x^{2}}=2 ; y=5\) and \(\frac{d y}{d x}=3\) when \(x=0\)
See all solutions

Recommended explanations on Math Textbooks

Decision Maths

Read Explanation

Geometry

Read Explanation

Probability and Statistics

Read Explanation

Pure Maths

Read Explanation

Mechanics Maths

Read Explanation

Theoretical and Mathematical Physics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.