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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

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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.

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’s 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’s 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

Consider a cylindrical water tank of constant cross section \(A .\) Water is pumped into the tank at a constant rate \(k\) and leaks out through a small hole of area \(a\) in the bottom of the tank. From Torricelli's theorem in hydrodynamics it follows that the rate at which water flows through the hole is \(\alpha a \sqrt{2 g h},\) where \(h\) is the current depth of water in the tank, \(g\) is the acceleration due to gravity, and \(\alpha\) is a contraction coefficient that satisfies \(0.5 \leq \alpha \leq 1.0 .\) (a) Show that the depth of water in the tank at any time satisfies the equation \(-\) $$ d h / d t=(k-\alpha a \sqrt{2 g h}) / A . $$ (b) Determine the equilibrium depth \(h_{e}\) of water and show that it it is asymptotically stable. Observe that \(h_{e}\) does not depend on \(A .\)

Use the technique discussed in Problem 20 to show that the approximation obtained by the Euler method converges to the exact solution at any fixed point as \(h \rightarrow 0 .\) $$ y^{\prime}=\frac{1}{2}-t+2 y, \quad y(0)=1 \quad \text { Hint: } y_{1}=(1+2 h)+t_{1} / 2 $$

Determine whether or not each of the equations is exact. If it is exact, find the solution. $$ \left(3 x^{2}-2 x y+2\right) d x+\left(6 y^{2}-x^{2}+3\right) d y=0 $$

Show that if \(\left(N_{x}-M_{y}\right) /(x M-y N)=R,\) where \(R\) depends on the quantity \(x y\) only, then the differential equation $$ M+N y^{\prime}=0 $$ has an integrating factor of the form \(\mu(x y)\). Find a general formula for this integrating factor.

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

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