91Ó°ÊÓ

Separation of variables is a technique used to solve differential equations. It involves rearranging an equation to isolate all terms involving the variable y on one side and all terms involving the variable x on the other side. In our example, starting with the differential equation \( y^{\prime} + x e^{y} = 0 \), we first rearrange it to \( y^{\prime} = -x e^{y} \).

We can then manipulate it further so that each side of the equation contains a single variable. By dividing both sides by \( e^y \) and multiplying by \( dx \), we obtain \( e^{-y} dy = -x dx \). Now, the equation is perfectly set for integration, with variables separated, making this technique particularly useful for easily solvable problems.

Integration is the process of finding the integral of an equation, which is essentially the reverse of differentiation. It allows us to determine the function that results in a given derivative.

After separating variables, we have: \( \int e^{-y} dy \) on the left side and \( \int -x dx \) on the right side. Let's explore these integrations separately:

• The integral of \( e^{-y} dy \) is \(-e^{-y} + C_1\), where \( C_1 \) is a constant of integration.
• The integral of \(-x dx\) is \(-\frac{x^2}{2} + C_2\), where \( C_2 \) is another constant of integration.

Combining these results gives us \(-e^{-y} = -\frac{x^2}{2} + C\), where \( C = C_2 - C_1 \). Integration is a powerful tool that converts our derivative equation into a more familiar algebraic form.

Exponential functions involve the variable as an exponent, and they appear frequently in solving differential equations. In our equation, we deal with \( e^y \), which means our solution involves understanding and manipulating exponential terms.

This is evident in our integration, yielding \(- e^{-y} = -\frac{x^2}{2} + C \). To solve for \( y \), one must be comfortable with the properties of exponential functions:

• Taking reciprocals: From \( e^{-y} \), one finds the reciprocal by rewriting it as \( \frac{1}{e^y} \).
• Natural logarithms: Logarithms allow us to "undo" the exponentiation, which is crucial in solving for \( y \).

Understanding how exponential and logarithmic functions work is essential for manipulating and solving the solutions derived from differential equations.

After integrating and rewriting our equation, the next step is to solve explicitly for \( y \). We use algebraic manipulations and properties of logarithms to isolate \( y \). The rearranged equation from the integration is \( e^{-y} = \frac{x^2}{2} - C \).

By taking the reciprocal on both sides, we convert \( e^{-y} \) into \( e^y \), leading to the expression \( y = -\ln\left( \frac{2}{x^2 - 2C} \right) \). This involves using natural logarithms to solve the equation, which is a common practice when dealing with exponential solutions.

This final step transforms the originally given differential equation into a solution that directly expresses \( y \) as a function of \( x \) and the constant \( C \). Remember to check the domain restrictions placed by the expression within the logarithm, ensuring meaningful and valid solutions.