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Prove: If the equations \(M_{1} d x+N_{1} d y=0\) and \(M_{2} d x+N_{2} d y=0\) are exact on an open rectangle \(R,\) so is the equation $$ \left(M_{1}+M_{2}\right) d x+\left(N_{1}+N_{2}\right) d y=0 $$

Short Answer

Expert verified
Answer: Yes, the sum of two exact differential equations is also an exact differential equation on an open rectangle R. This is because the partial derivatives of the combined equation satisfy the condition for exactness.

Step by step solution

01

Write the combined equation

Add the given exact differential equations to obtain a single equation: \((M_1 + M_2)dx + (N_1 + N_2)dy = 0\)
02

Compute the partial derivatives

Find the partial derivatives of the combined equation with respect to \(x\) and \(y\): $$ \frac{\partial (M_1+ M_2)}{\partial y} = \frac{\partial M_1}{\partial y} + \frac{\partial M_2}{\partial y} $$ and $$ \frac{\partial (N_1+ N_2)}{\partial x} = \frac{\partial N_1}{\partial x} + \frac{\partial N_2}{\partial x} $$
03

Show that the combined equation is exact

We know that the given equations are exact, which means: $$ \frac{\partial M_1}{\partial y} = \frac{\partial N_1}{\partial x} $$ and $$ \frac{\partial M_2}{\partial y} = \frac{\partial N_2}{\partial x} $$ Thus, when we add the two equalities, we get: $$ \left(\frac{\partial M_1}{\partial y} + \frac{\partial M_2}{\partial y}\right) = \left(\frac{\partial N_1}{\partial x} + \frac{\partial N_2}{\partial x}\right) $$ From Step 2, we know that: $$ \frac{\partial (M_1+ M_2)}{\partial y} = \frac{\partial (N_1+ N_2)}{\partial x} $$ This implies that the combined equation \((M_1 + M_2)dx + (N_1 + N_2)dy = 0\) is also exact on the open rectangle \(R\), as its partial derivatives satisfy the condition for exactness.

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Key Concepts

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

Partial Derivatives
In mathematics, a partial derivative represents how a function changes as one of the variables changes, while all other variables are held constant. Suppose you have a function of multiple variables, such as \(f(x, y)\). To find the partial derivative with respect to \(x\), we will differentiate \(f\) treating \(y\) as a constant.
In the context of exact differential equations, understanding partial derivatives helps us describe the change in a function along different axes in multi-dimensional space. They form the basis for determining how variables in equations affect one another.
Consider two functions \(M(x, y)\) and \(N(x, y)\) which are parts of a differential equation \(M dx + N dy = 0\). Calculating the partial derivatives \(\frac{\partial M}{\partial y}\) and \(\frac{\partial N}{\partial x}\) plays a crucial role. These derivatives are central to checking if a differential equation is exact.
To sum up, partial derivatives offer a powerful tool to uncover the underlying relationships between variables in a function, a key step when dealing with exactness in differential equations.
Condition for Exactness
The condition for exactness in differential equations is a critical criterion to test whether a given differential equation represents a differential form that's integrable, or in simpler terms, if we can find a potential function \(\psi(x, y)\) such that \(d\psi = M dx + N dy = 0\).
  • To check for exactness, you need to compute the partial derivatives: \(\frac{\partial M}{\partial y}\) and \(\frac{\partial N}{\partial x}\).
  • The exactness condition states that these partial derivatives must be equal: \(\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}\).
This condition ensures that there is indeed a function \(\psi(x, y)\) satisfying \(d\psi = M dx + N dy\), which means the equation is integrable into such a function. This ability to transform a differential equation into a simpler function is invaluable as it allows us to solve or analyze the behavior of the system more easily.
Exactness on Open Rectangle
In calculus, the notion of open rectangles comes into play when discussing regions in a plane that do not include their boundary. These open rectangles are crucial in ensuring the function involved behaves nicely without breaches or sharp changes at the boundary, which can complicate solutions involving partial derivatives and integrability.
When a differential equation is described as exact on an open rectangle, it guarantees that the equation satisfies the exactness condition not just at isolated points, but throughout the rectangular region. This uniformity across the region ensures we can derive a potential function \(\psi\) over the entire rectangle.
Thus, proving that combined equations remain exact on open rectangles is significant. It suggests that new combinations of original equations still adhere to this smoothness and integrability property. With this assurance, solutions derived from these differential equations are valid and applicable across the entire open rectangle, providing a stable set of solutions without edge-case problems.

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Most popular questions from this chapter

Verify that the following functions are harmonic, and find all their harmonic conjugates. (See Exercise \(43 .)\) (a) \(x^{2}-y^{2}\) (b) \(e^{x} \cos y\) (c) \(x^{3}-3 x y^{2}\) (d) \(\cos x \cosh y\) (e) \(\sin x \cosh y\)

Solve the initial value problem. $$ x^{2} y^{\prime}=2 x^{2}+y^{2}+4 x y, \quad y(1)=1 $$

We've shown that if \(p\) and \(f\) are continuous on \((a, b)\) then every solution of $$ y^{\prime}+p(x) y=f(x) $$ on \((a, b)\) can be written as \(y=u y_{1},\) where \(y_{1}\) is a nontrivial solution of the complementary equation for \((\mathrm{A})\) and \(u^{\prime}=f / y_{1} .\) Now suppose \(f, f^{\prime}, \ldots, f^{(m)}\) and \(p, p^{\prime}, \ldots, p^{(m-1)}\) are continuous on \((a, b),\) where \(m\) is a positive integer, and define $$ \begin{array}{l} f_{0}=f \\ f_{j}=f_{j-1}^{\prime}+p f_{j-1}, \quad 1 \leq j \leq m \end{array} $$ Show that $$ u^{(j+1)}=\frac{f_{j}}{y_{1}}, \quad 0 \leq j \leq m $$

Find an integrating factor; that is a function of only one variable, and solve the given equation. $$ \left(x^{2} y+4 x y+2 y\right) d x+\left(x^{2}+x\right) d y=0 $$

According to Theorem \(2.1 .2,\) the general solution of the linear nonhomogeneous equation $$ y^{\prime}+p(x) y=f(x) $$ is $$ y=y_{1}(x)\left(c+\int f(x) / y_{1}(x) d x\right). $$ where \(y_{1}\) is any nontrivial solution of the complementary equation \(y^{\prime}+p(x) y=0 .\) In this exercise we obtain this conclusion in a different way. You may find it instructive to apply the method suggested here to solve some of the exercises in Section 2.1 . (a) Rewrite (A) as $$ [p(x) y-f(x)] d x+d y=0, $$ and show that \(\mu=\pm e^{\int p(x) d x}\) is an integrating factor for (C). (b) Multiply (A) through by \(\mu=\pm e^{\int p(x) d x}\) and verify that the resulting equation can be rewritten as $$ (\mu(x) y)^{\prime}=\mu(x) f(x) $$ Then integrate both sides of this equation and solve for \(y\) to show that the general solution of (A) is $$ y=\frac{1}{\mu(x)}\left(c+\int f(x) \mu(x) d x\right). $$ Why is this form of the general solution equivalent to (B)?

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