Problem 13

Question

From $\frac{e^{y}}{\left(e^{y}+1\right)^{2}} d y=\frac{-e^{x}}{\left(e^{x}+1\right)^{3}} d x$ we obtain $-\left(e^{y}+1\right)^{-1}=\frac{1}{2}\left(e^{x}+1\right)^{-2}+c$.

Step-by-Step Solution

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Answer

Integrate both sides separately to find the expression. Combine integral constants to match the given form.

1Step 1: Understanding the Problem

We start with the given differential equation $ \frac{e^{y}}{(e^{y}+1)^{2}} \, dy = \frac{-e^{x}}{(e^{x}+1)^{3}} \, dx $. The task is to integrate both sides to express $-\frac{1}{e^{y}+1}$ in terms of $\frac{1}{2(e^{x}+1)^{2}}$ and a constant of integration $c$.

2Step 2: Integration of the Left Side

The left side of the equation is $ \frac{e^{y}}{(e^{y}+1)^{2}} \, dy$. We perform the integration with respect to $y$. We use the substitution $u = e^{y} + 1$, where $du = e^{y} \cdot dy$. This converts the integral into $\int \frac{1}{u^2} \, du$, which evaluates to $-\frac{1}{u} + C$. Substituting back, we have $-\frac{1}{e^{y}+1} + C$.

3Step 3: Integration of the Right Side

The right side of the equation is $-\frac{e^{x}}{(e^{x}+1)^{3}} \, dx$. We integrate with respect to $x$ using substitution or a polynomial long division, setting $v = e^{x} + 1$, $dv = e^{x} \, dx$. This gives $-\int \frac{1}{v^{3}} \, dv$, which evaluates to $\frac{1}{2v^{2}} + D$. Substituting back, we have $\frac{1}{2(e^{x}+1)^{2}} + D$.

4Step 4: Equating Solutions and Solving for Constants

Since both sides of the equation come from the original equivalence, we set $-\frac{1}{e^{y}+1} + C = \frac{1}{2(e^{x}+1)^{2}} + D$. Recognizing that the constant of integration $C$ includes any constant term from both integrals, we combine $C - D = c$, where $c$ is the constant of integration used in the exercise.

5Step 5: Final Expression

The resulting equation from both sides is $-\frac{1}{e^{y}+1} = \frac{1}{2(e^{x}+1)^{2}} + c$, where $c$ is the new combined constant. This matches the form of the equation to be proven.

Key Concepts

Integration TechniquesSubstitution MethodConstants of Integration

Integration Techniques

Integration is a mathematical method used to find the area under a curve, or to solve differential equations like the one in our problem. Techniques of integration vary and are chosen based on the form of the function we are integrating. In this case, the function was split into two integrals, each requiring a specific method for integration.

The left side of the equation involved the function $ \frac{e^{y}}{(e^{y}+1)^{2}} $. This is not a simple polynomial, so we turned to substitution.
The chosen technique transformed the variable $ y $ into a more manageable form using substitution, specifically $ u = e^{y} + 1 $.
By substituting, the integral became $ \int \frac{1}{u^2} \, du $, a known form that is easier to integrate, resulting in a solution involving inverse powers.
The right side followed a similar substitution technique, changing the variable from $ x $ to $ v $, making the integral simpler, $ -\int \frac{1}{v^{3}} \, dv $.

With integration techniques like substitution, complex problems become approachable by converting them to easier integral forms.

Substitution Method

The substitution method is an integral technique where we replace one variable with another to simplify the function we are dealing with. This helps in converting complex expressions into integrals that are more straightforward to evaluate. In this exercise, substitution played a critical role.We saw it applied twice:

In the left side of the equation: The substitution $ u = e^{y} + 1 $ converted $ dy $ into terms of $ du $, making the integral simpler.
Similarly, on the right side, $ v = e^{x} + 1 $ was used, facilitating the transition from $ dx $ to $ dv $.

After substitution, both integrals appeared in forms that are elementary in integration, $ \int \frac{1}{u^2} \, du $ and $ -\int \frac{1}{v^{3}} \, dv $. This method effectively reduces the complexity of the process, solving the integrals with ease and leading us to standard solutions.

Constants of Integration

In calculus, constants of integration appear every time we integrate a function. When we solve indefinite integrals, an arbitrary constant, usually noted as $ C $ or $ D $, must be included in the solution.Let's break down their role in this exercise:

For the integral of the left side of the equation, the constant of integration was signified as $ C $.
The right side's integral produced a different constant, $ D $.
Since both sides of our differential equation were integrals of the same original expression, these constants needed to be equated to ensure consistency. This was achieved by combining $ C - D = c $.

The constant $ c $ thus represents the equivalence of integration results from both sides, uniting them into a single equation. Constants of integration are crucial in ensuring that different integrated expressions precisely fit the conditions or match each other's domain, as was seen in this problem.

Problem 12

Problem 13

Other exercises in this chapter

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Problem 13

Letting $y=u x$ we have $$\begin{aligned} \left(x+u x e^{u}\right) d x-x e^{u}(u d x+x d u) &=0 \\ d x-x e^{u} d u &=0 \\ \frac{d x}{x}-e^{u} d u &=0 \\ \ln |

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Problem 13

For $y^{\prime}+\left(1+\frac{2}{x}\right) y=\frac{e^{x}}{x^{2}}$ an integrating factor is $e^{\int[1+(2 / x)] d x}=x^{2} e^{x}$ so that \(\frac{d}{d x}\lef

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