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Chapter 9: Problem 2

Solve the differential equation. $$\frac{d y}{d x}=\frac{\sqrt{x}}{e^{y}}$$

Short Answer

Expert verified
\( y = \ln\left(\frac{2}{3}x^{3/2} + C_1\right) \) is the solution.

Step by step solution

01

Separate the Variables

The given differential equation is \( \frac{dy}{dx} = \frac{\sqrt{x}}{e^y} \). We aim to separate the variables so that all terms involving \( y \) are on one side and all terms involving \( x \) are on the other. Rewriting, we get:\[ e^y \, dy = \sqrt{x} \, dx \].
02

Integrate Both Sides

Now that the variables are separated, we can integrate both sides of the equation:\[ \int e^y \, dy = \int \sqrt{x} \, dx \].For the left-side integral, the result is \( e^y + C \), where \( C \) is the constant of integration. For the right-side, \( \int \sqrt{x} \, dx = \frac{2}{3}x^{3/2} + C \).
03

Simplify and Solve for y

Equating the results from the integration, we have:\[ e^y = \frac{2}{3}x^{3/2} + C_1 \].Solve for \( y \) by taking the natural logarithm of both sides:\[ y = \ln\left(\frac{2}{3}x^{3/2} + C_1\right) \].
04

Final Answer

The solution of the original differential equation is expressed as a function of \( x \):\[ y = \ln\left(\frac{2}{3}x^{3/2} + C_1\right) \], where \( C_1 \) is an arbitrary constant from integration.

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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 used to solve certain types of differential equations. The main goal is to rearrange the equation so that each variable and its derivative are situated on opposite sides of the equation. This allows us to integrate both sides independently.
In the given exercise, the differential equation is \( \frac{dy}{dx} = \frac{\sqrt{x}}{e^y} \).
We separate the variables by multiplying both sides by \( e^y \, dy \) and \( \sqrt{x} \, dx \), resulting in \( e^y \, dy = \sqrt{x} \, dx \).
This ensures that all terms involving \( y \) are isolated on one side, and all terms involving \( x \) are on the other.
  • This technique greatly simplifies the equation, making it feasible to move on to integration, the next step in solving the problem.
  • The separation of variables is particularly useful for equations where direct integration of the original form is not straightforward.
Integration
Integration is the process of finding the antiderivative of a function, essentially reversing the operation of differentiation. After separating the variables in a differential equation, the next step is integrating each side individually.
For this exercise, we integrated both sides after separation: \( \int e^y \, dy = \int \sqrt{x} \, dx \). Integrating the left side (\( e^y \)) gives us \( e^y + C \), where \( C \) is the constant of integration.
For the right side, integrating \( \sqrt{x} \) results in \( \frac{2}{3}x^{3/2} + C \).
  • Integration is crucial as it allows us to find the general solution to the differential equation.
  • Remember, each indefinite integral will result in a constant of integration, often symbolized as \( C \), important for defining the general solution.
Natural Logarithm
The natural logarithm is used extensively in mathematics to solve equations where an exponential function is present. The natural logarithm, denoted as \( \ln \), is the inverse function of exponentiation to the base \( e \) (where \( e \approx 2.718 \)).
After integrating both sides of our differential equation, we arrived at \( e^y = \frac{2}{3}x^{3/2} + C_1 \). To solve for \( y \), we take the natural logarithm of both sides, resulting in \( y = \ln\left(\frac{2}{3}x^{3/2} + C_1\right) \).
  • This step is crucial as it allows us to express the solution for \( y \) explicitly.
  • Using \( \ln \) to solve for \( y \) aligns with our need to "invert" the exponential form present from the integration phase.
  • The natural logarithm efficiently handles calculations involving growth and decay processes often modeled by differential equations.

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Most popular questions from this chapter

According to Newton's Law of Universal Gravitation, the gravitational force on an object of mass \(m\) that has been projected vertically upward from the earth's surface is $$F=\frac{m g R^{2}}{(x+R)^{2}}$$ where \(x=x(t)\) is the object's distance above the surface at time \(t, R\) is the earth's radius, and \(g\) is the acceleration due to gravity. Also, by Newton's Second Law, \(F=m a=m(d v / d t)\) and so $$m \frac{d v}{d t}=-\frac{m g R^{2}}{(x+R)^{2}}$$ (a) Suppose a rocket is fired vertically upward with an initial velocity \(v_{0 .}\) Let \(h\) be the maximum height above the surface reached by the object. Show that $$v_{0}=\sqrt{\frac{2 g R h}{R+h}}$$ [Hint: By the Chain Rule, \(m(d v / d t)=m v(d v / d x) . ]\) (b) Calculate \(v_{c}=\lim _{h \rightarrow \infty} v_{0} .\) This limit is called the escape velocity for the earth. (c) Use \(R=3960 \mathrm{mi}\) and \(g=32 \mathrm{ft} / \mathrm{s}^{2}\) to calculate \(v_{e}\) in feet per second and in miles per second.

Find the solution of the differential equation that satisfies the given initial condition. $$\frac{d P}{d t}=\sqrt{P t}, \quad P(1)=2$$

A direction field for the differential equation \(y^{\prime}=y\left(1-\frac{1}{4} y^{2}\right)\) is shown. (a) Sketch the graphs of the solutions that satisfy the given initial conditions. $$\begin{array}{ll}{\text { (i) } y(0)=1} & {\text { (ii) } y(0)=-1} \\\ {\text { (iii) } y(0)=-3} & {\text { (iv) } y(0)=3}\end{array}$$ (b) Find all the equilibrium solutions.

When a raindrop falls, it increases in size and so its mass at time \(t\) is a function of \(t, m(t) .\) 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 \(k .\)

\(5-14\) Solve the differential equation. $$x y^{\prime}+y=\sqrt{x}$$
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