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Chapter 11: Problem 20

use separation of variables to find the solution to the differential equation subject to the initial condition. $$\frac{d y}{d x}=2 y-4, \quad \text { through }(2,5)$$

Short Answer

Expert verified
The solution to the differential equation is \( y = \frac{3}{e^{4}} e^{2x} + 2 \).

Step by step solution

01

Write the differential equation in separable form

We start by rewriting the given differential equation \(\frac{d y}{d x} = 2y - 4\) to a separable form. Notice that the right-hand side can be rewritten as \(2(y - 2)\). This gives us: \[ \frac{d y}{d x} = 2(y - 2) \]
02

Separate variables

Separate the variables by rearranging terms to isolate \(y\) and \(x\). Divide both sides by \(y - 2\) and multiply both sides by \(dx\):\[ \frac{1}{y - 2} dy = 2 dx \]
03

Integrate both sides

Integrate both sides of the new equation. The left side integrates to the natural logarithm, and the right side integrates to a simple linear expression:\[ \int \frac{1}{y - 2} dy = \int 2 dx \]which gives:\[ \ln |y - 2| = 2x + C \]where \(C\) is the integration constant.
04

Solve for \(y\)

Exponentiate both sides to solve for \(y\):\[ y - 2 = e^{2x + C} \] \[ y = e^{2x + C} + 2 \]Denote \(e^C\) by \(D\) (a new constant), so we have:\[ y = De^{2x} + 2 \]
05

Apply the initial condition

Use the initial condition \((2, 5)\) to solve for \(D\). Substituting \(x = 2\) and \(y = 5\) into the solution, we have:\[ 5 = De^{4} + 2 \]\[ 3 = De^{4} \]\[ D = \frac{3}{e^{4}} \]
06

Write the specific solution

Substitute \(D\) back into the equation to get the specific solution:\[ y = \frac{3}{e^{4}} e^{2x} + 2 \]

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Key Concepts

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

Understanding Initial Conditions in Differential Equations
When solving differential equations, initial conditions play a crucial role. They help in determining a specific solution from an infinite number of possible solutions. Without initial conditions, the solution is generally a family of functions that differ by a constant parameter.
In the original exercise, the initial condition given is (2,5), which means that when the variable \(x\) is 2, the function \(y\) should be 5. This condition allows us to find the constant of integration C after we solve the differential equation.
Therefore, use initial conditions to pinpoint the exact path or curve out of many that fulfills the differential equation given. It's like finding the key to open a particular door amongst many in a hallway. Each initial condition corresponds to one clear path or solution.
Separation of Variables: Breaking Down the Process
Separation of variables is a method used to solve simple types of differential equations. It involves separating the variables \(y\) and \(x\) onto different sides of the equation. Hence, the name separation of variables.
Start with the given differential equation \(\frac{dy}{dx} = 2y - 4\). Transform it into a form where each variable appears only once. By rewriting the equation as \(\frac{1}{y-2} dy = 2 dx\), each side has a single variable for integration purposes.
Use this technique when you can manipulate the equation easily so that the variables separate without complexity. After separation, each side can be integrated individually, providing a path to solve the equation. It's an elegant method for tackling otherwise complicated problems and points towards integration as the next step.
Integration Techniques in Solving Differential Equations
Integration techniques are the mathematical tools we use to solve the separated equation in the separation of variables method. Once the differential equation is in an integrable form with each side containing a single variable, integrate each side accordingly. In our exercise, the left side \(\int \frac{1}{y - 2} dy \) turns into the natural logarithm \(\ln |y - 2|\), and the right side \(\int 2 dx\) results in \(2x + C\), where \(C\) is an integration constant.
Remember, integration involves finding the function whose derivative is the given function. Use constants of integration to ensure the method is precise. The newfound expression allows us to solve for \(y\) and include the initial condition to pinpoint a specific solution. The overarching principle here is to reverse the process of differentiation effectively.

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Most popular questions from this chapter

Find the general solution to the given differential equation. $$z^{\prime \prime}+2 z^{\prime}=0$$

Solve the boundary value problem. $$y^{\prime \prime}+5 y^{\prime}+6 y=0, \quad y(-2)=0, y(2)=3$$

We analyze world oil production. \(^{25}\) When annual world oil production peaks and starts to decline. major economic restructuring will be needed. We investigate when this slowdown is projected to occur. We define \(P\) to be the total oil production worldwide since 1859 in billions of barrels. In \(1993,\) annual world oil production was 22.0 billion barrels and the total production was \(P=724\) billion barrels. In 2013 , annual production was 27.8 billion barrels and the total production was \(P=1235\) billion barrels. Let \(t\) be time in years since 1993 (a) Estimate the rate of production, \(d P / d t,\) for 1993 and 2013 (b) Estimate the relative growth rate, \((1 / P)(d P / d \imath)\) for 1993 and 2013 (c) Find an equation for the relative growth rate. \((1 / P)(d P / d t),\) as a function of \(P,\) assuming that the function is linear. (d) Assuming that \(P\) increases logistically and that all oil in the ground will ultimately be extracted, estimate the world oil reserves in 1859 to the nearest billion barrels. (e) Write and solve the logistic differential equation modeling \(P\)

Two companies share the market for a new technology. They have no competition except each other. Let \(A(t)\) be the net worth of one company and \(B(t)\) the net worth of the other at time \(t .\) Suppose that net worth cannot be negative and that \(A\) and \(B\) satisfy the differential equations $$\begin{aligned}&A^{\prime}=2 A-A B\\\&B^{\prime}=B-A B\end{aligned}$$ (a) What do these equations predict about the net worth of each company if the other were not present? What effect do the companies have on each other? (b) Are there any equilibrium points? If so, what are they? (c) Sketch a slope field for these equations (using a computer or calculator), and hence describe the different possible long-run behaviors.

Give an example of: A differential equation all of whose solutions have their critical points on the parabola \(y=x^{2}.\)
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