91Ó°ÊÓ

Differential equations can often be solved using a method known as separation of variables. This technique is especially useful when you can express the equation in such a way that all terms involving one variable (say, \( y \)) are on one side, while the terms involving the other variable (\( x \)) are on the opposite side.

In our problem, we start with the equation:

• \( \frac{\sqrt{1+x^{2}}}{1+y} \frac{d y}{d x} = -x \)

To separate the variables, we multiply both sides by \((1+y) \, dx\), resulting in:

• \( \frac{dy}{1+y} = -x \frac{dx}{\sqrt{1+x^2}} \)

In this form, each side of the equation involves only one of the variables and its respective differential. This allows us to integrate each side with respect to its own variable. Separation of variables simplifies the integration process and makes it possible to solve the differential equation step by step.

Once we've separated the variables, the next logical step is integration. In calculus, integration provides the antiderivative, or the reverse, of the differentiation process. Let's take a closer look at how this is done in our problem.

After separation of variables, our equation is:

• \( \int \frac{dy}{1+y} = \int -x \frac{dx}{\sqrt{1+x^2}} \)

The left-hand side involves a straightforward integral of a natural logarithmic function, resulting in:

• \( \ln|1+y| + C_1 \)

For the right-hand side, we apply substitution because it simplifies complex expressions. Setting \( u = 1+x^2 \), then \( du = 2x \, dx \), transforms the integral into a simpler form. The integration of \( -\frac{1}{2} \int \frac{du}{\sqrt{u}} \) leads to:

• \(-\sqrt{u} + C_2 = -\sqrt{1+x^2} \)

Combining these results, we merge the constants into one, often denoted by \( C \), to represent the general solution before moving to the final goal of finding an explicit function when possible.

Expressing solutions as explicit functions is a goal in solving differential equations, as it provides a clear relationship between dependent and independent variables. In this scenario, our aim is to express \( y \) as a function of \( x \).

From our integration results, we have:

• \( \ln|1+y| = -\sqrt{1+x^2} + C \)

To solve for \( y \), we need to eliminate the natural logarithm by exponentiating:

• \( |1+y| = e^{-\sqrt{1+x^2} + C} \)
• Letting \( e^C = C' \), gives: \( |1+y| = C' e^{-\sqrt{1+x^2}} \)

Consequently, the explicit function becomes:

• \( y = C' e^{-\sqrt{1+x^2}} - 1 \)

This explicit expression clearly showcases how \( y \) is dependent on \( x \), with \( C' \) representing the constant that can adjust based on initial conditions or particular solutions. This step is crucial because it offers a clear and direct form, making it easier to compute values and further analyze the trend and behavior of the solution in various contexts.