Question

Find the particular solution to the differential equation \(y^{\prime}\left(1-x^{2}\right)=1+y\) that passes through \((0,-2)\), given that \(y=C \frac{\sqrt{x+1}}{\sqrt{1-x}}-1\) is a general solution.

Short answer

The particular solution is \( y = -\frac{\sqrt{x+1}}{\sqrt{1-x}} - 1 \).

Step-by-step solution

Understand the General Solution

The general solution given is \(y = C \frac{\sqrt{x+1}}{\sqrt{1-x}} - 1\). It involves an arbitrary constant \(C\). Our task is to find the particular solution that satisfies the initial condition \((0, -2)\).

Apply the Initial Condition

Substitute \(x = 0\) and \(y = -2\) into the general solution: \(-2 = C \frac{\sqrt{0+1}}{\sqrt{1-0}} - 1\). This simplifies to \(-2 = C \cdot 1 - 1\).

Solve for the Constant C

From \(-2 = C - 1\), we add 1 to both sides to get \(-1 = C\). Thus, the constant \(C\) is equal to \(-1\).

Form the Particular Solution

Substitute \(C = -1\) back into the general solution: \(y = -1 \times \frac{\sqrt{x+1}}{\sqrt{1-x}} - 1\). This simplifies to \(y = -\frac{\sqrt{x+1}}{\sqrt{1-x}} - 1\).

Verify the Solution

Check to see if the particular solution satisfies the initial condition \((0, -2)\). Substitute \(x = 0\), resulting in \(y = -\frac{\sqrt{0+1}}{\sqrt{1-0}} - 1 = -1 - 1 = -2\). The initial condition is satisfied, confirming that this is the correct particular solution.

Key concepts

Differential Equations

Differential equations are equations that relate a function with its derivatives. These equations can often describe how a quantity changes in relation to another. For example, the differential equation \(y' (1-x^2) = 1+y\) represents a relationship between the function \(y\) and its derivative \(y'\), considering additional components like \((1-x^2)\). Differential equations are fundamental in mathematics because they can model real-world phenomena, such as

• population growth,
• heat diffusion,
• or motion of planets.

They exist in two main flavors: ordinary differential equations, which deal with single independent variables, and partial differential equations, applicable for multiple independent variables.Solving differential equations is about finding a function or a family of functions that satisfies the equation. This brings us to the next concept: the general solution.

General Solution

The general solution of a differential equation provides a broad family of solutions, showcasing a function that encompasses arbitrary constants. These arbitrary constants arise from the integration process, reflecting the infinite number of possible solutions that can satisfy an equation. In our exercise, the general solution is given as \(y = C \frac{\sqrt{x+1}}{\sqrt{1-x}} - 1\). Here, \(C\) is the arbitrary constant, representing an infinite number of curves, each differing by this constant. The goal, frequently, is to use additional information—often termed initial conditions—to determine this constant and thus narrow down to a single, specific solution known as the particular solution. Let's explore how initial conditions come into play.

Initial Condition

An initial condition is a specific piece of information that allows us to find the particular solution from a general solution. It can be thought of as a starting point that a function must pass through, providing us with the means to solve for the arbitrary constants. In this exercise, the initial condition provided is the point \((0, -2)\). This tells us that when \(x = 0\), the value of \(y\) is \(-2\). Plugging the initial condition into the general solution \(y = C \frac{\sqrt{x+1}}{\sqrt{1-x}} - 1\) helps us resolve the value of \(C\). More generally, initial conditions are crucial as they ensure that the solution fits a real-world scenario or the context of a specific problem.

Constant of Integration

The constant of integration, represented as \(C\) in our solution, is an integral part of the solution to differential equations. It emerges from indefinite integration, where the derivative does not provide enough information to define a single function completely. In the context of the solved exercise, \(C\) is the unknown constant in the equation \(y = C \frac{\sqrt{x+1}}{\sqrt{1-x}} - 1\). The determination of this constant is critical to converting the general solution into a particular solution. By applying the initial condition \((0, -2)\), we solved for \(C = -1\). The constant of integration bridges the abstract world of calculus with practical applications, ensuring solutions are tailored to specific, real-world conditions.