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Chapter 6: Problem 52

Finding a General Solution In Exercises \(41-52,\) use integration to find a general solution of the differential equation. $$ \frac{d y}{d x}=5 e^{-x / 2} $$

Short Answer

Expert verified
The general solution to the given differential equation is \( y = -10 e^{-x / 2} + C \).

Step by step solution

01

Recognize the type of the differential equation

The given differential equation \( \frac{d y}{d x}=5 e^{-x / 2} \) is a non-homogeneous first order differential equation. To solve it, we need to isolate the variables and perform direct integration.
02

Perform integration

Now, integrate both sides of the equation with respect to x. Doing so looks like \( \int dy = \int 5 e^{-x / 2} dx \). The left side integral is simply y, while the right side involves a simple substitution to simplify the integral. Let \( u = -x / 2 \), then \( du = -dx / 2 \) or \( dx = -2du \). Substituting these values in, we get \( y = -10 \int e^u du \). The integral of \( e^u \) is simply \( e^u \) itself, giving us \( y = -10e^u + C \).
03

Substituting back and Simplifying

Substitute back for \( u \) by replacing \( u \) with \( -x / 2 \) in the last equation: \( y = -10 e^{-x / 2} + C \)

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Key Concepts

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

First Order Differential Equations
First order differential equations are equations that involve the derivative of a function and the function itself, without any higher derivatives involved.
In these equations, the highest order of derivative is one, hence the name "first order". They come in various forms and can model a wide range of phenomena in physics, biology, economics, and many other fields.
For example, if we have \( \frac{dy}{dx} = f(x, y) \), this represents a first order differential equation. Here, \( y \) is the function of \( x \) that we need to find, and \( \frac{dy}{dx} \) is the derivative of \( y \) with respect to \( x \).
This type of equation often requires methods like separation of variables or integrating factors to solve, depending on its structure.
Integration Techniques
Integration is a crucial tool used to solve differential equations. When dealing with first order differential equations, particularly linear ones, integration helps find the function from its derivative.
In our exercise, the equation \( \frac{dy}{dx} = 5e^{-x/2} \) requires integrating the right-hand side. Simple substitution is a commonly used technique to simplify complex integrals, such as setting \( u = -x/2 \).
  • Perform substitution: Transform the variable to simplify the function with respect to \( u \).
  • Evaluate the integral: Integrate the function of \( u \) - in this case, \( e^u \).
  • Back-substitute: Replace \( u \) with the original variables to express the solution back in terms of \( x \).
Correct use of integration techniques can solve many ordinary first order differential equations efficiently.
Separation of Variables
Separation of variables is a specific method used to solve first order differential equations when they can be rewritten so that each variable appears on a different side of the equation.
One of the key ideas in this technique is simplifying the equation into a product of two functions, each solely in terms of one variable.
Here's how it works: Suppose we have an equation \( \frac{dy}{dx} = g(x)h(y) \). To separate variables, you rewrite it as \( \frac{1}{h(y)}dy = g(x)dx \), then integrate both sides.
  • Separate terms: Distribute the terms so that all \( y \) terms are on one side and \( x \) terms are on the other.
  • Integrate: Calculate the integral of both sides independently.
  • Solve for \( y \): Often, you'll need further manipulation to make \( y \) the subject.
This method works well for equations that can be easily split and is typically one of the first strategies taught in dealing with such problems.

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Most popular questions from this chapter

Weight Gain \(A\) calf that weighs 60 pounds at birth gains weight at the rate $$\frac{d w}{d t}=k(1200-w)$$ where \(w\) is weight in pounds and \(t\) is time in years. (a) Solve the differential equation. (b) Use a graphing utility to graph the particular solutions for \(k=0.8,0.9,\) and \(1 .\) (c) The animal is sold when its weight reaches 800 pounds. Find the time of sale for each of the models in part (b). (d) What is the maximum weight of the animal for each of the models in part (b)?

Mixture In Exercises \(35-38\) , consider a tank that at time \(t=0\) contains \(v_{0}\) gallons of a solution of which, by weight, \(q_{0}\) pounds is soluble concentrate. Another solution containing \(q_{1}\) pounds of the concentrate per gallon is running into the tank at the rate of \(r_{1}\) gallons per minute. The solution in the tank is kept well stirred and is withdrawn at the rate of \(r_{2}\) gallons per minute. Let \(Q\) be the amount of concentrate in the solution at any time \(t .\) Write the differential equation for the rate of change of \(Q\) with respect to \(t\) when \(r_{1}=r_{2}=r .\)

Mixture In Exercises \(35-38\) , consider a tank that at time \(t=0\) contains \(v_{0}\) gallons of a solution of which, by weight, \(q_{0}\) pounds is soluble concentrate. Another solution containing \(q_{1}\) pounds of the concentrate per gallon is running into the tank at the rate of \(r_{1}\) gallons per minute. The solution in the tank is kept well stirred and is withdrawn at the rate of \(r_{2}\) gallons per minute. A 200 -gallon tank is full of a solution containing 25 pounds of concentrate. Starting at time \(t=0\) , distilled water is admitted to the tank at a rate of 10 gallons per minute, and the well-stirred solution is withdrawn at the same rate. (a) Find the amount of concentrate \(Q\) in the solution as a function of \(t .\) (b) Find the time at which the amount of concentrate in the tank reaches 15 pounds. (c) Find the quantity of the concentrate in the solution as \(t \rightarrow \infty\) .

Solving a Bernoulli Differential Equation In Exercises \(59-66,\) solve the Bernoulli differential equation. The Berchoulli equation is a well-known nonlinear equation of the form $$ y^{\prime}+P(x) y=Q(x) y^{n} $$ that can be reduced to a linear form by a substitution. The general solution of a Bernoulli equation is $$ y^{1-n} e^{\int(1-n) P(x) d x}=\int(1-n) Q(x) e^{f(1-n) P(x) d x} d x+C $$ $$ x y^{\prime}+y=x y^{3} $$

Electric Circuits In Exercises 33 and \(34,\) use the differential equation for electric circuits given by \(L \frac{d I}{d t}+R I=E\) . In this equation, \(I\) is the current, \(R\) is the resistance, \(L\) is the inductance, and \(E\) is the electromotive force (voltage). Solve the differential equation for the current given a constant voltage \(E_{0} .\)
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