Question

Solve \(\left(1+\mathrm{e}^{\frac{x}{y}}\right) \mathrm{d} \mathrm{x}+\mathrm{e}^{\frac{x}{y}}\left(1-\frac{\mathrm{x}}{\mathrm{y}}\right) \mathrm{dy}=0\)

Short answer

Based on the solution above, provide a short answer to the given problem by solving the first-order differential equation and presenting its implicit solution. The implicit solution of the given first-order differential equation is found to be: \[x(y-y^2)(1+y)=C.\]

Step-by-step solution

Rewrite the given equation

\[(1+e^{\frac{x}{y}})\mathrm{d}x + e^{\frac{x}{y}}\left(1-\frac{x}{y}\right)\mathrm{d}y=0.\] Now, let's check the exactness condition, ∂M/∂y = ∂N/∂x, where M(x, y) = 1 + e^(x/y) and N(x, y) = e^(x/y)(1-x/y):

Calculate the partial derivatives of M and N

\[\frac{\partial M}{\partial y}=-\frac{x}{y^2}e^{\frac{x}{y}},\qquad \frac{\partial N}{\partial x}=\frac{1}{y}e^{\frac{x}{y}}-\frac{1}{y^2}e^{\frac{x}{y}}.\] Comparing the partial derivatives, we can see ∂M/∂y ≠ ∂N/∂x. Thus the given equation is not exact. Now let's find an integrating factor μ(x, y) such that the equation becomes exact. A common approach is trying to find an integrating factor that depends on x only or y only. To do this, let's try μ = μ(y) and rewrite the exactness condition accordingly:

Rewrite the exactness condition involving μ(y)

\[\frac{\partial}{\partial y}(\mu(y)(1+e^{\frac{x}{y}}))=\frac{\partial }{\partial x}(\mu(y)(e^{\frac{x}{y}}(1-\frac{x}{y}))).\] Now, we will calculate the partial derivatives of M(y) = μ(y)(1 + e^(x/y)) and N(y) = μ(y)e^(x/y)(1-x/y), and divide one equation by the other to solve for μ(y):

Solve for μ(y)

\[\frac{\mu'(y)(1+e^{\frac{x}{y}})-\mu(y)\frac{x}{y^2}e^{\frac{x}{y}}}{\mu(y)\frac{1}{y}e^{\frac{x}{y}}-\mu(y)\frac{1}{y^2}e^{\frac{x}{y}}}=1.\] Let's simplify the left side of the equation and observing that the e^(x/y) term cancels out, we get: \[1-\frac{\mu'(y)}{\mu(y)}+\frac{\mu'(y)}{\mu(y)y}=\frac{\mu'(y)(1-y)}{\mu(y)y}.\] Now, we will integrate both sides with respect to y and solve for μ(y):

Integrate the equation for μ(y)

\[\int\frac{\mu'(y)(1-y)}{\mu(y)y}\mathrm{d}y=\int\mathrm{d}y\Rightarrow \ln|\mu(y)|=\ln|y-y^2|+c\Rightarrow \mu(y)=y-y^2.\] Now, we will multiply M and N with the integrating factor μ(y) and check if the exactness condition is met:

Multiply M and N with the integrating factor and check the exactness condition

\[M'(x,y)=(y-y^2)(1+e^{\frac{x}{y}}),\qquad N'(x,y)=(y-y^2)e^{\frac{x}{y}}(1-\frac{x}{y}).\] Calculating the partial derivatives, we get:

Calculate the partial derivatives of M' and N'

\[\frac{\partial M'}{\partial y}=\frac{x}{y^2}(y-y^2)e^{\frac{x}{y}},\qquad \frac{\partial N'}{\partial x}=\frac{x}{y^2}(y-y^2)e^{\frac{x}{y}}.\] Now, as ∂M'/∂y = ∂N'/∂x, we have an exact equation:

Write the exact equation using M' and N'

\[(y-y^2)(1 + e^{\frac{x}{y}})\mathrm{d}x + (y-y^2)e^{\frac{x}{y}}(1-\frac{x}{y})\mathrm{d}y = 0.\] From the exact equation, we have dF(x, y) = M'dx + N'dy, where F'(x, y) = C is the potential function. Integrate M' with respect to x and N' with respect to y and equate the results to obtain the potential function, F(x, y) = C:

Calculate the potential function F(x, y)

\[\begin{aligned} \int M'\, dx &=\int (y-y^2)(1 + e^{\frac{x}{y}})\, dx = xy-xy^2+x(y-y^2),\\ \int N'\, dy &=\int (y-y^2)e^{\frac{x}{y}}(1-\frac{x}{y})\, dy=xy-xy^2+x(y-y^2). \end{aligned}\] Thus, the potential function is F(x, y) = xy - xy^2 + x(y - y^2) = C. Rearranging the terms for the potential function gives us the implicit solution of the differential equation:

Implicit solution of the differential equation

\[x(y-y^2)(1+y)=C.\]

Key concepts

Integrating Factor

The concept of an integrating factor is a powerful tool used to solve non-exact differential equations. It’s essentially a function, denoted as μ(x, y), that, when multiplied with the original equation, makes it exact. This means that after finding the right integrating factor, the new equation satisfies the condition where the partial derivative of M with respect to y equals the partial derivative of N with respect to x.

In our step by step solution, we see that the original equation isn't exact, but by strategically choosing an integrating factor that depends only on y, the equation is transformed into an exact one. It’s like finding the right key to unlock a door — once we have it, we can open the door to the solution. The technique involves a trial and error process and sometimes integrating to finally arrive at the right integrating factor.

Partial Derivatives

Understanding partial derivatives is critical when working with exact differential equations. A partial derivative represents the rate at which a function changes as one variable shifts, while keeping other variables constant. In the context of our problem, we're looking at the functions M(x, y) and N(x, y) that are part of the differential equation. Checking whether an equation is exact involves comparing the partial derivatives ∂M/∂y and ∂N/∂x.

As shown in the exercise, when these two derivatives aren't equal, the equation is not exact. However, this comparison guides us toward finding the correct integrating factor to adjust the equation so that this condition of exactness is satisfied, making the problem solvable.

Potential Function

A potential function, often denoted as F(x, y), is related to conservative vector fields and is a crucial concept in solving exact differential equations. It's like a hidden treasure map that outlines how two variables interact within a system. The potential function satisfies the property that its partial derivatives with respect to x and y give us back the functions M(x, y) and N(x, y) in the differential equation.

In the solution provided, we found F(x, y) by integrating M and N after applying the integrating factor. The process of uncovering F(x, y) is akin to tracing back to the source of a river — once uncovered, it gives a comprehensive view of how the variables x and y interplay in the equation.

Implicit Solution

An implicit solution of a differential equation represents the relationship between the dependent and independent variables without explicitly solving for one in terms of the other. It contrasts with an explicit solution, where the dependent variable is directly expressed as a function of the independent variable.

In the final step of our problem, we arrange the terms obtained from the potential function to present the solution in an implicit form, which is a common outcome in solving exact differential equations. This form of the solution often encapsulates the relationship between the variables in a compact and mathematically rich way, even though it may not give an immediate explicit expression for either variable.