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Chapter 2: Problem 18

Show that any separable equation, $$ M(x)+N(y) y^{\prime}=0 $$ is also exact.

Short Answer

Expert verified
#Conclusion# Any separable equation, which can be written in the form $M(x) + N(y) y' = 0$, is also an exact equation. This is because the partial derivatives of $M(x)$ with respect to $y$ and $N(y)$ with respect to $x$ are both zero, satisfying the condition for an exact equation: $\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}$.

Step by step solution

01

Write the given equation

We are given the separable equation: $$ M(x) + N(y) y' = 0 $$
02

Consider it as an exact equation

Now let's consider the given equation as an exact equation. We have: $$ M(x,y) = M(x) \quad \text{and} \quad N(x,y) = N(y) $$
03

Find the partial derivatives

To show that the separable equation is an exact equation, we need to find the partial derivatives and check if the condition is satisfied: $$ \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} $$ Let's find these partial derivatives: $$ \frac{\partial M}{\partial y} = \frac{\partial M(x)}{\partial y} = 0 \quad \text{as M(x) is independent of y} $$ $$ \frac{\partial N}{\partial x} = \frac{\partial N(y)}{\partial x} = 0 \quad \text{as N(y) is independent of x} $$
04

Check if the condition is satisfied

Now let's check if the condition is satisfied: $$ \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} \Rightarrow 0 = 0 $$ Since the condition is satisfied, we can conclude that any separable equation is also an exact equation.

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Key Concepts

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

Separable Equations
Separable equations are a class of differential equations that can be rewritten so that each variable appears on opposite sides of the equation. This type of equation typically takes the form \( M(x) + N(y) y' = 0 \), where \( M \) and \( N \) are functions of \( x \) and \( y \), respectively. These equations allow us to manipulate and integrate each side independently.
Here鈥檚 how it works:
  • Separate the variables: \( M(x) + N(y) y' = 0 \) can be rearranged to \( M(x)dx = -N(y)dy \).
  • Integrate both sides: The separated equation can often be integrated on both sides to find a general solution.
This process essentially transforms the original differential equation into two independent integrals, simplifying the process of solving it. Separable equations are particularly useful because of their simplicity, allowing students to solve them with basic calculus techniques.
Exact Equations
Exact equations are a specific type of differential equations, which have a very strict yet fascinating condition. An equation \( M(x,y)dx + N(x,y)dy = 0 \) is considered exact if there exists a potential function \( \Psi(x, y) \) such that \( \frac{\partial \Psi}{\partial x} = M \) and \( \frac{\partial \Psi}{\partial y} = N \).
To determine its exactness:
  • Find \( \frac{\partial M}{\partial y} \).
  • Find \( \frac{\partial N}{\partial x} \).
  • If \( \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} \), the equation is exact.
Exact equations allow students to find solutions using potential functions, which streamline solving the equation since it pivots on matching partial derivatives of the component functions.
Partial Derivatives
Partial derivatives are fundamental in understanding how a function changes with respect to one variable while keeping other variables constant. When dealing with functions of multiple variables, partial derivatives express the rate of change of the function as the input variables change incrementally.
Here's how they are used:
  • Determine \( \frac{\partial M}{\partial y} \), treating \( x \) as a constant.
  • Calculate \( \frac{\partial N}{\partial x} \), viewing \( y \) as a constant.
  • Compare results to verify exactness condition.
By finding partial derivatives, you acquire essential relationships between multiple variables in the context of exactness, crucial for determining solutions to certain types of differential equations.
Mathematical Proof
Mathematical proof is a logical argument that demonstrates the truth of a statement using a sequence of logical steps. In solving the problem of whether a separable equation is exact, proof involves verifying that specific mathematical conditions hold true. Here鈥檚 a breakdown:
  • Analyze the initial equation, \( M(x) + N(y) y' = 0 \).
  • Transform and verify necessary conditions for exactness, such as comparing partial derivatives.
  • Conclude with a logical affirmation, like showing \( 0 = 0 \), which demonstrates the exactness condition is met.
Proofs like this build a strong foundation in mathematical reasoning by providing clear and logical explanations, crucial for strengthening understanding and problem-solving skills in differential equations.

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Most popular questions from this chapter

let \(\phi_{0}(t)=0\) and use the method of successive approximations to solve the given initial value problem. (a) Determine \(\phi_{n}(t)\) for an arbitrary value of \(n .\) (b) Plot \(\phi_{n}(t)\) for \(n=1, \ldots, 4\). Observe whether the iterates appear to be converging. $$y^{\prime}=t y+1, \quad y(0)=0$$

transform the given initial value problem into an equivalent problem with the initial point at the origin. $$ d y / d t=t^{2}+y^{2}, \quad y(1)=2 $$

draw a direction field and plot (or sketch) several solutions of the given differential equation. Describe how solutions appear to behave as \(t\) increases, and how their behavior depends on the initial value \(y_{0}\) when \(t=0\). $$ y^{\prime}=t y(3-y) $$

Suppose that a rocket is launched straight up from the surface of the earth with initial velocity \(v_{0}=\sqrt{2 g R}\), where \(R\) is the radius of the earth. Neglect air resistance. (a) Find an expression for the velocity \(v\) in terms of the distance \(x\) from the surface of the earth. (b) Find the time required for the rocket to go \(240,000\) miles (the approximate distance from the earth to the moon). Assume that \(R=4000\) miles.

Find the escape velocity for a body projected upward with an initial velocity \(v_{0}\) from a point \(x_{0}=\xi R\) above the surface of the earth, where \(R\) is the radius of the earth and \(\xi\) is a constant. Neglect air resistance. Find the initial altitude from which the body must be launched in order to reduce the escape velocity to \(85 \%\) of its value at the earth's surface.
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