91Ó°ÊÓ

When solving differential equations, initial conditions play a crucial role. An initial condition is a value that allows us to find a specific solution among the many potential solutions of a differential equation. In this exercise, the initial condition is given as \( y(0) = 2 \).
This condition helps us determine the constant \( C \) in the solution. It narrows down the general solution to one that satisfies \( y(0) = 2 \).
By substituting \( x = 0 \) and \( y = 2 \) into the solution form \( y = -\frac{1}{x^2 + C} \), we solve for \( C \). This gives the specific constant that makes the solution unique to the initial condition.

Integration is the process of finding an antiderivative, and it is used to solve differential equations by reversing differentiation. In this exercise, once we have separated the variables, the integration process allows us to find a solution function. We need to integrate both sides of the equation \( \frac{dy}{y^2} = 2x \, dx \).

• The left side integrates to \( -\frac{1}{y} \), which is the antiderivative of \( \frac{1}{y^2} \).
• The right side integrates to \( x^2 + C \), where \( C \) is the constant of integration, reflecting the antiderivative of \( 2x \).

Combining these results is key to formulating a general solution, which we can customize by applying the initial condition.

Variable separation is a technique used in solving differential equations where we rewrite the equation so that each variable and its differential is on a separate side of the equation. In our equation, \( y^2 y' = 2x \), we manipulate it so that all \( y \)-related terms and differentials are on one side whereas \( x \)-related terms are on the other side.

• We leave \( \frac{dy}{y^2} = 2x \, dx \) with all \( y \) terms on the left and \( x \) terms on the right.
• This separation allows us to integrate each part with respect to its variables.

Variable separation is fundamental because it simplifies the ordinary differential equation into a format where direct integration is possible.

Solution verification involves checking that the solution derived indeed satisfies both the original differential equation and the initial condition. We do this through substitution. For our solution, \( y = \frac{1}{\frac{1}{2} - x^2} \), we need to verify its validity.
By substituting this expression back into the differential equation, compute the derivatives, and confirm that \( y^2 y' = 2x \) holds true. This step ensures that the substitution meets the requirements of the differential equation.
Additionally, verify the initial condition by plugging in \( x = 0 \). The solution \( y = 2 \) confirms that the initial condition is satisfied, thus validating the uniqueness and correctness of the solution.