Question

Solve the given differential equation. $$ \frac{d y}{d x}=\frac{x-e^{-x}}{y+e^{y}} $$

Short answer

Question: Given the differential equation $$ \frac{dy}{dx}=\frac{x-e^{-x}}{y+e^{y}}, $$ find the implicit solution for y(x). Answer: The implicit solution for y(x) is given by $$ \frac{1}{2}y^2 + e^{y} = \frac{1}{2}x^2 + e^{-x} + C. $$

Step-by-step solution

Separate Variables

To separate the variables, we want to get all the y's on one side and all the x's on the other side of the equation. To do this, we can multiply both sides by the denominator of the right side and by dx: $$ (y+e^{y})dy = (x-e^{-x})dx $$ Now we have successfully separated the x's and y's into two separate expressions.

Integrate Both Sides

Now, we will integrate both sides of the equation with respect to their respective variables: $$ \int (y+e^{y})dy = \int (x-e^{-x})dx $$ First, integrate the left side with respect to y: $$ \int y dy + \int e^{y} dy = \frac{1}{2}y^2 + e^y + C_1 $$ Next, integrate the right side with respect to x: $$ \int x dx - \int e^{-x} dx = \frac{1}{2}x^2 + e^{-x} + C_2 $$

Combine Constants and Solve for y

Now that both sides have been integrated, we will combine the constants and rewrite the equation. We can do this by setting C = C_2 - C_1, and the equation becomes: $$ \frac{1}{2}y^2 + e^y = \frac{1}{2}x^2 + e^{-x} + C $$ To solve for y, we can express y as a function of x. However, due to the complexity, this can't be done explicitly. So the solution to the differential equation can be written implicitly as: $$ \frac{1}{2}y^2 + e^{y} = \frac{1}{2}x^2 + e^{-x} + C $$ The student should note the final solution represents a family of solutions, where C is the constant of integration and can represent any value, which is typical for first-order differential equations.

Key concepts

Separation of Variables

Solving differential equations can often be a challenging task, but one widely used method to simplify the process is the separation of variables. Separation of variables is a technique where you rearrange a differential equation, which involves a function and its derivative, in such a way that all terms involving one variable are on one side of the equation, and all terms involving the other variable are on the other side.

Imagine you have a dance where participants are initially mixed; separation of variables is like asking all dancers in red to move to one side of the room and all dancers in blue to the other side. In mathematical terms, you manipulate the equation to have one variable, let's say 'y', and its derivatives 'dy' on one side, and the other variable, 'x', and its derivatives 'dx' on the other side. By doing this, you essentially set the stage to solve the equation more straightforwardly through integration.

In the provided exercise, the separation of variables step effectively paves the way to find the solution by isolating y-terms with 'dy' and x-terms with 'dx'. You can view this as sorting like terms into their respective 'camps' to deal with them individually.

Integration of Functions

Once we've successfully separated our variables, the next step in solving the differential equation is to perform integration on both sides. Integration is the process of finding the function that originally produced our derivatives. If differentiation is the process of cutting the birthday cake (our function) into smaller pieces (derivatives), then integration is putting those pieces back together to recreate the whole cake.

In our example, after separation, we have two integrals: one with respect to y and the other with respect to x. By integrating both sides, we're basically summing up the small 'slices' of change outlined by the differential equation to find the original 'whole' functions. This process brings us a step closer to the solution, as it provides us with an equation that reflects the cumulative effect of those changes over the function's entire range.

Calculating the Integrals

Integrating the function on the left side with respect to y and the function on the right side with respect to x yields two separate antiderivatives. Don't forget to add the constant of integration on both sides, since indefinite integrals always include an arbitrary constant, representing the infinite possible positions where our original function could have started its journey.

Implicit Solutions

In the realm of differential equations, not every problem will allow for a neatly packaged explicit solution where the dependent variable is isolated on one side of the equation. This is where implicit solutions come in handy. An implicit solution is a way of defining a function indirectly through an equation in which the function and its derivatives are intermingled.

Think of it as being handed a treasure map with 'X' marking the spot, but the path to the treasure isn't drawn for you. You have to understand the landscape and the clues given to find the treasure. In our exercise, we end up with an equation where 'y' cannot be easily isolated on one side due to the complexity of the relationship between 'y' and 'x'. We solve as much as we can and then represent our answer with both variables interrelated.

The implicit solution is frequently the stopping point for many first-order differential equations. It acknowledges the relationship between x and y without forcibly separating them, allowing for a broader understanding of their connection. In the case of our example, the implicit solution looks like a statement of balance or equivalence between y's and x's behavior, held together by the integration constant 'C', which represents the various potential starting points of our function.