Problem 31
Question
Show that the given relation defines an implicit solution to the given differential equation, where \(c\) is an arbitrary constant. $$e^{y / x}+x y^{2}-x=c, \quad y^{\prime}=\frac{x^{2}\left(1-y^{2}\right)+y e^{y / x}}{x\left(e^{y / x}+2 x^{2} y\right)}$$.
Step-by-Step Solution
Verified Answer
To show that the given relation defines an implicit solution to the given differential equation, we perform implicit differentiation on the relation \(e^{y / x}+x y^{2}-x=c\) with respect to \(x\). We find that after reorganizing the equation and solving for \(y'\), we get \(y' = \frac{x^2\left(1-y^{2}\right)+y e^{y / x}}{x\left(e^{y / x}+2 x^{2} y\right)}\), which is the same as the given differential equation. Therefore, the relation is an implicit solution to the differential equation.
1Step 1: Implicit differentiation of the given relation
Perform the implicit differentiation of the relation: \(e^{y / x}+x y^{2}-x=c\) with respect to \(x\). Remember to apply the chain rule when necessary.
Case 1: \(e^{y / x}\)
The derivative is: \(\frac{d}{dx}e^{y / x} = e^{y / x} \cdot \frac{d}{dx}(y / x)\)
Now, find \(\frac{d}{dx}(y / x)\):
\[\frac{d}{dx}(y / x) = \frac{1}{x^2}\left(x\frac{dy}{dx} - y\right)\]
So, \(\frac{d}{dx}e^{y / x} = e^{y / x} \cdot \frac{1}{x^2}\left(x\frac{dy}{dx} - y\right) = \frac{e^{y / x}}{x^2}(x \cdot y' - y)\)
Case 2: \(x y^{2}\)
The derivative is: \[\frac{d}{dx}(x y^{2}) = y^2 + 2x y y'\]
Case 3: \(- x\)
The derivative is: \(\frac{d}{dx}(-x) = -1\)
Combining all cases, we have:
\[\frac{d}{dx}(e^{y / x}+x y^{2}-x) = \frac{e^{y / x}}{x^2}(x \cdot y' - y) + y^2 + 2x y y' - 1\]
Now, since the LHS of the given equation does not depend on x, the derivative with respect to x is 0:
\[0 = \frac{e^{y / x}}{x^2}(x \cdot y' - y) + y^2 + 2x y y' -1\]
2Step 2: Simplify and solve for \(y'\)
Now, solve the equation from step 1 for \(y'\):
\[0 = \frac{e^{y / x}}{x^2}(x \cdot y' - y) + y^2 + 2x y y' -1\]
Reorganize the equation to isolate terms containing \(y'\):
\[1 - y^2 = \frac{e^{y / x}}{x^2}(x \cdot y' - y) + 2x y y'\]
Solve for \(y'\):
\[y' = \frac{x^2\left(1-y^{2}\right)+y e^{y / x}}{x\left(e^{y / x}+2 x^{2} y\right)}\]
The result is the same as the given differential equation. Therefore, we have shown that the given relation defines an implicit solution to the given differential equation.
Key Concepts
Implicit DifferentiationChain Rule in DifferentiationSolving Differential Equations
Implicit Differentiation
Implicit differentiation is a technique used to differentiate equations where the variables are not separated, meaning that the equation is not solved for one variable in terms of the other. Instead of directly finding the derivative of a single function, implicit differentiation allows us to differentiate both sides of an equation with respect to a given variable. This is particularly useful for relations like
\(e^{y / x}+x y^{2}-x=c\) where it's not practical or possible to solve for \(y\) explicitly in terms of \(x\).
In our case, to show that the given relation is an implicit solution to a differential equation, we differentiate each term with respect to \(x\), treating \(y\) as a function of \(x\) (\(y(x)\)). The use of implicit differentiation in the solution process applies the chain rule frequently, where each derivative of a term involving \(y\) also includes \(y'\), which is the derivative of \(y\) with respect to \(x\). This is because \(y\) is itself a function of \(x\), and any change in \(x\) could potentially affect \(y\).
This implicit differentiation step results in a new equation that includes the term \(y'\), which can then be solved to find the derivative in terms of \(x\) and \(y\). This process is crucial to find the slope of the curve at any point without explicitly defining the function \(y\).
\(e^{y / x}+x y^{2}-x=c\) where it's not practical or possible to solve for \(y\) explicitly in terms of \(x\).
In our case, to show that the given relation is an implicit solution to a differential equation, we differentiate each term with respect to \(x\), treating \(y\) as a function of \(x\) (\(y(x)\)). The use of implicit differentiation in the solution process applies the chain rule frequently, where each derivative of a term involving \(y\) also includes \(y'\), which is the derivative of \(y\) with respect to \(x\). This is because \(y\) is itself a function of \(x\), and any change in \(x\) could potentially affect \(y\).
This implicit differentiation step results in a new equation that includes the term \(y'\), which can then be solved to find the derivative in terms of \(x\) and \(y\). This process is crucial to find the slope of the curve at any point without explicitly defining the function \(y\).
Chain Rule in Differentiation
The chain rule is a fundamental principle in calculus used to calculate the derivative of the composition of two or more functions. When one function is nested within another, the chain rule allows us to differentiate the entire expression cleanly and systematically. This rule essentially states that the derivative of a composed function is the derivative of the outer function evaluated at the inner function, multiplied by the derivative of the inner function.
In the context of implicit differentiation, the chain rule is applied when differentiating terms like \(e^{y / x}\). Since \(y\) is itself a function of \(x\), we treat the expression \(y / x\) as the inner function and \(e^{u}\) (where \(u = y / x\)) as the outer function. We must first differentiate the outer function, leaving the inner function unchanged, and then multiply by the derivative of the inner function.
For instance, in the term \(e^{y / x}\), the derivative of the outer function \(e^u\) with respect to \(u\) is simply \(e^u\). The inner function \(y / x\) has its own derivative with respect to \(x\), which we compute as \(\frac{dy}{dx}/x - y/x^2\). The application of the chain rule seamlessly integrates these steps, yielding the derivative of the compound function with respect to \(x\). The ability to differentiate such composite functions correctly is crucial for solving many implicit differential equations.
In the context of implicit differentiation, the chain rule is applied when differentiating terms like \(e^{y / x}\). Since \(y\) is itself a function of \(x\), we treat the expression \(y / x\) as the inner function and \(e^{u}\) (where \(u = y / x\)) as the outer function. We must first differentiate the outer function, leaving the inner function unchanged, and then multiply by the derivative of the inner function.
For instance, in the term \(e^{y / x}\), the derivative of the outer function \(e^u\) with respect to \(u\) is simply \(e^u\). The inner function \(y / x\) has its own derivative with respect to \(x\), which we compute as \(\frac{dy}{dx}/x - y/x^2\). The application of the chain rule seamlessly integrates these steps, yielding the derivative of the compound function with respect to \(x\). The ability to differentiate such composite functions correctly is crucial for solving many implicit differential equations.
Solving Differential Equations
Solving differential equations is a core activity in understanding dynamic systems in mathematics. A differential equation is an equation that relates a function with its derivatives. Solutions to these equations describe how a particular quantity changes over time or another variable, and they play a significant role in physics, engineering, economics, and other sciences.
In the context of our exercise, the differential equation is given in an implicit form involving both \(x\) and \(y\), needing implicit differentiation to handle. The solution process involves manipulating the differentiated equation to isolate \(y'\), which represents the derivative of \(y\) with respect to \(x\).
After applying implicit differentiation and the chain rule, the next step is to collect like terms and to solve for \(y'\), as seen in the step-by-step solutions. Simplifying such equations may involve algebraic manipulation, which includes expanding, factoring, and canceling terms. The goal is to express \(y'\) in terms of \(x\) and \(y\), demonstrating that the original relation satisfies the differential equation. By successfully finding a valid expression for \(y'\), we confirm that we have an implicit solution to the given differential equation, thus solving the problem at hand.
In the context of our exercise, the differential equation is given in an implicit form involving both \(x\) and \(y\), needing implicit differentiation to handle. The solution process involves manipulating the differentiated equation to isolate \(y'\), which represents the derivative of \(y\) with respect to \(x\).
After applying implicit differentiation and the chain rule, the next step is to collect like terms and to solve for \(y'\), as seen in the step-by-step solutions. Simplifying such equations may involve algebraic manipulation, which includes expanding, factoring, and canceling terms. The goal is to express \(y'\) in terms of \(x\) and \(y\), demonstrating that the original relation satisfies the differential equation. By successfully finding a valid expression for \(y'\), we confirm that we have an implicit solution to the given differential equation, thus solving the problem at hand.
Other exercises in this chapter
Problem 31
Use the technique derived in the previous problem to solve the given differential equation. $$y^{\prime}+y=e^{-2 x}$$
View solution Problem 31
In a certain chemical reaction \(9 \mathrm{g}\) of \(\mathrm{C}\) are formed when \(6 \mathrm{g}\) of A combine with \(3 \mathrm{g}\) of \(\mathrm{B}\). Initial
View solution Problem 31
Determine the slope field and some representative solution curves for the given differential equation. $$\diamond y^{\prime}=-2 x y$$
View solution Problem 32
Determine which of the five types of differential equations we have studied the given equation falls into (see Table \(1.12 .1),\) and use an appropriate techni
View solution