Problem 14
Question
Determine whether the given differential equation is exact. If it is exact, solve it. $$ \left(1-\frac{3}{y}+x\right) \frac{d y}{d x}+y=\frac{3}{x}-1 $$
Step-by-Step Solution
Verified Answer
The differential equation is not exact.
1Step 1: Rewrite the equation in standard form
Given the differential equation: \( \left(1-\frac{3}{y}+x\right) \frac{dy}{dx} + y = \frac{3}{x} - 1 \). Begin by rewriting it in the form \( M(x,y)dx + N(x,y)dy = 0 \). Start by making the equation homogeneous by ensuring all terms are on one side. Multiply the entire equation by \( dx \): \(\left(1-\frac{3}{y}+x\right)dy + y\,dx = \left(\frac{3}{x} - 1\right)dx\). Reorder to get: \(y\,dx - \left(1 - \frac{3}{y} + x\right)dy = \left(\frac{3}{x} - 1\right)dx\). This can be simplified to \( M(x,y)dx + N(x,y)dy = 0 \) where \( M(x,y) = y - \left(\frac{3}{x} - 1\right) \) and \( N(x,y) = -\left(1 - \frac{3}{y} + x\right) \).
2Step 2: Determine if the equation is exact
For the equation to be exact, we need the condition: \( \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} \). Calculate \( \frac{\partial M}{\partial y} \): Start with \( M(x,y) = y - \frac{3}{x} + 1 \). Here, \( \frac{\partial M}{\partial y} = 1 \).Calculate \( \frac{\partial N}{\partial x} \): Start with \( N(x,y) = -\left(1 - \frac{3}{y} + x\right) \), so \( \frac{\partial N}{\partial x} = -1 \).Since \( \frac{\partial M}{\partial y} = 1 eq \frac{\partial N}{\partial x} = -1 \), the differential equation is not exact.
Key Concepts
Partial DerivativesHomogeneous Differential EquationsExactness Condition
Partial Derivatives
Understanding partial derivatives is crucial when dealing with differential equations like the one given in this exercise. A partial derivative represents how a function changes as one of its variables changes, while the other variables are held constant. This is different from a regular derivative, which considers changes in two or more variables simultaneously.
For example, if we have a function \( f(x,y) \), the partial derivative with respect to \( x \) is represented as \( \frac{\partial f}{\partial x} \). This notation means that we're looking at how \( f \) changes as \( x \) varies, keeping \( y \) unchanged.
The partial derivatives are crucial in determining exactness in differential equations. In the current example, we need to compute the partial derivatives of \( M(x,y) \) with respect to \( y \), and \( N(x,y) \) with respect to \( x \).
For example, if we have a function \( f(x,y) \), the partial derivative with respect to \( x \) is represented as \( \frac{\partial f}{\partial x} \). This notation means that we're looking at how \( f \) changes as \( x \) varies, keeping \( y \) unchanged.
The partial derivatives are crucial in determining exactness in differential equations. In the current example, we need to compute the partial derivatives of \( M(x,y) \) with respect to \( y \), and \( N(x,y) \) with respect to \( x \).
- \( \frac{\partial M}{\partial y} \) tells us the rate of change of \( M \) as \( y \) changes.
- \( \frac{\partial N}{\partial x} \) indicates how \( N \) changes with \( x \).
Homogeneous Differential Equations
A homogeneous differential equation is one where all terms are of the same degree when considering the variables involved. This type of equation allows for specific methods of solution, such as substitution, which can simplify the problem.
In our initial problem, the given equation needs some manipulation to fit this form. Bringing all terms to one side results in something that resembles the structure \( M(x,y)dx + N(x,y)dy = 0 \). This is a standard approach in rearranging linear forms into something more workable.
To classify a differential equation as homogeneous, every term in the equation must be proportional to the same power of the variables:
In our initial problem, the given equation needs some manipulation to fit this form. Bringing all terms to one side results in something that resembles the structure \( M(x,y)dx + N(x,y)dy = 0 \). This is a standard approach in rearranging linear forms into something more workable.
To classify a differential equation as homogeneous, every term in the equation must be proportional to the same power of the variables:
- For a simple linear equation of the form \( ay = bx \), both terms are first degree.
- The notion can be extended where terms such as \( xy \) or \( x^2 + y^2 \) retain homogeneity of degree two.
Exactness Condition
To determine if a differential equation like \( M(x,y)dx + N(x,y)dy = 0 \) is exact, we use the exactness condition. Simply put, this condition requires that the mixed partial derivatives of the functions \( M \) and \( N \) must be equal. This means:
\[ \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} \]
If this condition holds, then the differential is exact, and it implies the existence of a potential function \( \Psi(x,y) \) such that \( d\Psi = Mdx + Ndy \).
\[ \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} \]
If this condition holds, then the differential is exact, and it implies the existence of a potential function \( \Psi(x,y) \) such that \( d\Psi = Mdx + Ndy \).
- If \( \frac{\partial M}{\partial y} = 1 \)
- And \( \frac{\partial N}{\partial x} = -1 \)
- Here, these derivatives are not equal, indicating that the equation is not exact.
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