91Ó°ÊÓ

The method of variable separation is a powerful tool for solving differential equations. The central idea is to transform the equation in such a way that all terms containing one variable are on one side, and all terms with the other variable are on the opposite side. This way, the differential equation can be expressed in the form:

• \[ N(y) rac{dy}{dx} = M(x) \]
• and further into \[ N(y) \, dy = M(x) \, dx \]

This transformation allows you to handle the equation more easily because it separates the roles of each variable. For our specific problem, separating variables was straightforward as the equation was set up to make this possible. We moved \( (1+y)^3 \) with all terms involving \( y \) to one side and \( dx \) related to \( x \) to the other side, allowing us to integrate both sides independently.

Integration is a fundamental operation used to solve differential equations after separating variables. Once you've transformed the differential equation using variable separation, the next step is to integrate both sides. This helps in finding the function \( y \) in terms of \( x \). For example:

• The left-hand side, involving \( dy \), requires you to find the antiderivative with respect to \( y \).
• The right-hand side, involving \( dx \), is integrated with respect to \( x \).

In our example, for the left side \( \int \frac{1}{(1+y)^3} \, dy \), the integration requires substitution which will be explained in detail later. The right side is straightforward, giving \( \int dx = x + C_2 \), where \( C_2 \) is an integration constant.

The substitution method is a strategic approach that simplifies complex integrals, making them more manageable. In our differential equation, we used substitution to address the integral \( \int \frac{1}{(1+y)^3} \, dy \). Here's how substitution works:

• Identify a part of the expression to substitute, which simplifies the integral.
• In this case, set \( u = 1 + y \), making \( du = dy \).

This transforms the integral into \( \int u^{-3} \, du \). This is easier to solve because it's a simple power of \( u \), leading to the solution \( -\frac{1}{2} u^{-2} + C_1 \). By reverting to the original variables \( y \), we find the integrated form \( -\frac{1}{2(1+y)^2} + C_1 \). Substitution aids in greatly reducing the complexity of the integral.

Once integration is complete, the task is to express the function \( y \) as clearly as possible. This involves combining the results of the integrations from both sides. We substitute back, set the integrations equal, and solve for \( y \). In our example:

• Combine: \( -\frac{1}{2(1+y)^2} = x + C \)
• where \( C = C_2 - C_1 \).

After some rearrangements, you solve for \( y \) to obtain:

• \[ (1+y)^2 = -\frac{1}{2(x + C)} \]
• Thus, \( y = \sqrt{-\frac{1}{2(x + C)}} - 1 \).

This process of solving differential equations provides a powerful way to understand how different variables relate and change. Remember that careful algebraic manipulation is crucial to maintain the meaning and validity of the solution.