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Chapter 9: Problem 1

What is a separable first-order differential equation?

Short Answer

Expert verified
Answer: y = Ce^((1/2)x^2)

Step by step solution

01

Definition of a separable first-order differential equation

A separable first-order differential equation is a differential equation that can be written in the form: dy/dx = g(x) * h(y) where g(x) is a function of x only, and h(y) is a function of y only.
02

Separate the variables

To solve a separable first-order differential equation, we begin by separating the variables. This means that we rewrite the equation so that all expressions involving the independent variable (x) are on one side, and all expressions involving the dependent variable (y) are on the other side. For example: 1/h(y) * dy/dx = g(x)
03

Integrate both sides

Now that we have separated the variables, we need to integrate both sides of the equation with respect to the appropriate variable. For example: ∫(1/h(y)) dy = ∫g(x) dx
04

Solve for the dependent variable (y)

After integrating both sides, we may need to solve for the dependent variable (y) or, in some cases, express the result as an implicit equation relating x and y. It is essential to consider any integration constant that could represent the solution's family. The result will be the general solution of the separable first-order differential equation.
05

Example: Simple separable first-order differential equation

Let's consider a very simple example: dy/dx = x * y First, separate the variables: 1/y * dy/dx = x Now, integrate both sides with respect to the appropriate variables: ∫(1/y) dy = ∫x dx The integrals are: ln|y| = (1/2)x^2 + C We can solve for y: y = e^((1/2)x^2 + C) Simplifying, we get the general solution: y = Ce^((1/2)x^2) In conclusion, a separable first-order differential equation is a type of differential equation that can be written as a product of two functions that depend only on their respective variables. The steps to solve it involve separating the variables, integrating both sides, and solving for the dependent variable.

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Key Concepts

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

Differential Equations
Differential equations are equations that involve an unknown function and its derivatives. They play a crucial role in various fields of science and engineering because they describe how quantities change. These equations can vary in complexity and type. A **first-order differential equation** involves the first derivative of the unknown function. An important class of these equations is the **separable differential equation**. This specific type allows us to break down a complex equation into simpler parts, simplifying the integration process.
Separable differential equations are particularly useful because they allow for solving by separation of variables, making them accessible to solve with basic calculus techniques.
Calculus Integration
In solving separated differential equations, **calculus integration** is a critical step. Once variables are separated, the next phase is to integrate both sides of the equation. Integration is essentially the reverse process of differentiation, providing a way to recover a function from its derivative.
  • On the left side, you'll integrate with respect to the dependent variable, often labeled as \( y \).
  • On the right side, you'll integrate with respect to the independent variable, typically \( x \).
Care is needed during integration to include the integration constant, which is vital since it represents all possible solutions that satisfy the initial differential equation. This constant is usually represented by \( C \). Without this constant, the general solution would not be fully complete.
General Solution
Once the integration is complete, the resulting expression represents a solution for the differential equation. This solution is known as the **general solution**, as it includes an arbitrary constant, which makes it adaptable to a range of initial conditions defined by additional parameters.
  • The general solution provides a formula expressing the dependent variable in terms of the independent variable, possibly including the constant \( C \).
  • Through the constant \( C \), the solution encompasses a family of functions, each of which could potentially satisfy a specific initial condition applied to the differential equation.
After determining the general solution, it may be necessary to solve the expression for the dependent variable explicitly. If not easily solvable, the expression might remain as an implicit function involving both \( x \) and \( y \).
Variable Separation
**Variable separation** is the foundational concept behind solving separable first-order differential equations. This technique involves rearranging the equation so that each side depends on a single variable: one side on \( x \) and the other on \( y \). By rewriting the equation in this way:
  • The differential equation \( \frac{dy}{dx} = g(x) \cdot h(y) \) is transformed to \( \frac{1}{h(y)} dy = g(x) dx \),
  • We create two separate integrals involving only one variable each, which simplifies the solution process by enabling straightforward integration.
Variable separation is uniquely powerful because it transforms an intertwined problem into a more manageable form, allowing us to solve it using basic integration methods. This method can effectively tackle many real-world problems modeled by separable differential equations, leading to a clear path to finding the general solution.

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Most popular questions from this chapter

Convergence of Euler's method Suppose Euler's method is applied to the initial value problem \(y^{\prime}(t)=a y, y(0)=1,\) which has the exact solution \(y(t)=e^{a t} .\) For this exercise, let \(h\) denote the time step (rather than \(\Delta t\) ). The grid points are then given by \(t_{k}=k h .\) We let \(u_{k}\) be the Euler approximation to the exact solution \(y\left(t_{k}\right),\) for \(k=0,1,2, \ldots\). a. Show that Euler's method applied to this problem can be written \(u_{0}=1, u_{k+1}=(1+a h) u_{k},\) for \(k=0,1,2, \ldots\). b. Show by substitution that \(u_{k}=(1+a h)^{k}\) is a solution of the equations in part (a), for \(k=0,1,2, \ldots\). c. Recall from Section 4.7 that \(\lim _{h \rightarrow 0}(1+a h)^{1 / h}=e^{a} .\) Use this fact to show that as the time step goes to zero \((h \rightarrow 0,\) with \(\left.t_{k}=k h \text { fixed }\right),\) the approximations given by Euler's method approach the exact solution of the initial value problem; that is, \(\lim _{h \rightarrow 0} u_{k}=\lim _{h \rightarrow 0}(1+a h)^{k}=y\left(t_{k}\right)=e^{a t_{k}}\).

For each of the following stirred tank reactions, carry out the following analysis. a. Write an initial value problem for the mass of the substance. b. Solve the initial value problem. A \(2000-\) Leank is initially filled with a sugar solution with a concentration of \(40 \mathrm{g} / \mathrm{L} .\) A sugar solution with a concentration of \(10 \mathrm{g} / \mathrm{L}\) flows into the tank at a rate of \(10 \mathrm{L} / \mathrm{min}\). The thoroughly mixed solution is drained from the tank at a rate of \(10 \mathrm{L} / \mathrm{min}\).

Consider the general first-order linear equation \(y^{\prime}(t)+a(t) y(t)=f(t) .\) This equation can be solved, in principle, by defining the integrating factor \(p(t)=\exp \left(\int a(t) d t\right) .\) Here is how the integrating factor works. Multiply both sides of the equation by \(p\) (which is always positive) and show that the left side becomes an exact derivative. Therefore, the equation becomes $$p(t)\left(y^{\prime}(t)+a(t) y(t)\right)=\frac{d}{d t}(p(t) y(t))=p(t) f(t)$$ Now integrate both sides of the equation with respect to t to obtain the solution. Use this method to solve the following initial value problems. Begin by computing the required integrating factor. $$y^{\prime}(t)+\frac{3}{t} y(t)=1-2 t, y(2)=0$$

A special class of first-order linear equations have the form \(a(t) y^{\prime}(t)+a^{\prime}(t) y(t)=f(t),\) where \(a\) and \(f\) are given functions of t. Notice that the left side of this equation can be written as the derivative of a product, so the equation has the form $$a(t) y^{\prime}(t)+a^{\prime}(t) y(t)=\frac{d}{d t}(a(t) y(t))=f(t)$$ Therefore, the equation can be solved by integrating both sides with respect to \(t .\) Use this idea to solve the following initial value problems. $$t y^{\prime}(t)+y=1+t, y(1)=4$$

An endowment is an investment account in which the balance ideally remains constant and withdrawals are made on the interest earned by the account. Such an account may be modeled by the initial value problem \(B^{\prime}(t)=r B-m,\) for \(t \geq 0,\) with \(B(0)=B_{0} .\) The constant \(r>0\) reflects the annual interest rate, \(m>0\) is the annual rate of withdrawal, \(B_{0}\) is the initial balance in the account, and \(t\) is measured in years. a. Solve the initial value problem with \(r=0.05, m=\$ 1000 /\) year, and \(B_{0}=\$ 15,000 .\) Does the balance in the account increase or decrease? b. If \(r=0.05\) and \(B_{0}=\$ 50,000,\) what is the annual withdrawal rate \(m\) that ensures a constant balance in the account? What is the constant balance?
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