Question

Find the "general solution" (that is, a solution containing an arbitrary constant) of each of the following differential equations, by separation of variables. Then find a particular solution of each equation satisfying the given boundary conditions. \(x \sqrt{1-y^{2}} d x+y \sqrt{1-x^{2}} d y=0, \quad y=\frac{1}{2}\) when \(x=\frac{1}{2}\).

Short answer

General solution: \( \sqrt{1-x^{2}} + \sqrt{1-y^{2}} = C \); Particular solution: \( \sqrt{1-x^{2}} + \sqrt{1-y^{2}} = \sqrt{3} \)

Step-by-step solution

- Rewrite the differential equation

Rearrange the given differential equation to separate variables. The given differential equation is: \[ x \sqrt{1-y^{2}} \ d x + y \sqrt{1-x^{2}} \ d y = 0 \]Rewriting, we get: \[ x \sqrt{1-y^{2}} \ d x = - y \sqrt{1-x^{2}} \ d y \]Next, we split the terms to isolate the variables on each side.

- Separate the variables

Divide both sides to achieve separation of variables: \[ \frac{x}{\sqrt{1-x^{2}}} dx = - \frac{y}{\sqrt{1-y^{2}}} dy \]Integrate both sides to find the general solution.

- Integrate both sides

Integrate both sides: \[ \int \frac{x}{\sqrt{1-x^{2}}} dx = - \int \frac{y}{\sqrt{1-y^{2}}} dy \]Using substitution \( u = 1-x^2 \) and \( v = 1-y^2 \) integrates to:\[ -\sqrt{1-x^{2}} = -\sqrt{1-y^{2}} + C \]where \(C\) is the constant of integration.

- Simplify the general solution

Simplify to get: \[ \sqrt{1-x^{2}} + \sqrt{1-y^{2}} = C \]where \(C\) is the arbitrary constant.

- Apply the boundary condition

Use the given boundary condition to find the particular solution. Substitute \(x = \frac{1}{2}\) and \( y = \frac{1}{2} \) into the general solution: \[ \sqrt{1-\left(\frac{1}{2}\right)^2} + \sqrt{1-\left(\frac{1}{2}\right)^2} = C \]Calculate the values:\[ \sqrt{1-\frac{1}{4}} + \sqrt{1-\frac{1}{4}} = C \]\[ \sqrt{\frac{3}{4}} + \sqrt{\frac{3}{4}} = C \]\[ 2 \cdot \frac{\sqrt{3}}{2} = C \]\[ \sqrt{3} = C \]

- Write the particular solution

Substitute the value of \(C\) back into the general solution to get the particular solution: \[ \sqrt{1-x^{2}} + \sqrt{1-y^{2}} = \sqrt{3} \]

Key concepts

Separation of Variables

In solving differential equations, separation of variables is a valuable technique. This method involves rearranging the equation so that each variable and its differential are isolated on opposite sides. Starting with the given equation:
\[ x \sqrt{1-y^{2}} \, dx + y \sqrt{1-x^{2}} \, dy = 0 \]
Our goal is to separate the terms involving x on one side and those involving y on the other. By moving terms around, we arrive at:
\[ x \sqrt{1-y^{2}} \, dx = - y \sqrt{1-x^{2}} \, dy \]
Now, each variable is on a different side of the equation, making it easier to integrate.

Integration

Integrating both sides of the separated differential equation helps find the general solution. For our exercise, we start with:
\[ \int \frac{x}{\sqrt{1-x^{2}}} \, dx = - \int \frac{y}{\sqrt{1-y^{2}}} \, dy \]
Perform the integrations using substitutions:

• Let \( u = 1-x^2 \), so \( du = -2x \, dx \)
• Let \( v = 1-y^2 \), so \( dv = -2y \, dy \)

The integrals simplify to:
\[ - \sqrt{1-x^2} = - \sqrt{1-y^2} + C \]
where C is the constant of integration. This equation describes a family of solutions, forming the general solution.

Boundary Conditions

Boundary conditions are specific values given for the variables that allow us to find a particular solution. In this exercise, the boundary conditions are x = 1/2 and y = 1/2. These conditions help us determine the constant C. Substitute the boundary conditions into the general solution:
\[ \sqrt{1-\left( \frac{1}{2} \right)^2 } + \sqrt{1- \left( \frac{1}{2} \right)^2 } = C \]
Calculate each term:
\[ \sqrt{1-\frac{1}{4}} + \sqrt{1-\frac{1}{4}} = C \]
Which results in:
\[ \sqrt{\frac{3}{4}} + \sqrt{\frac{3}{4}} = C \]
Thus, C = √3.

General Solution

The general solution contains an arbitrary constant, representing all possible solutions to the differential equation. From our integration, we know that:
\[ \sqrt{1-x^2} + \sqrt{1-y^2} = C \]
This describes a family of curves depending on the value of C. By substituting different constants, we get different solutions, outlining the behavior of the function for various initial conditions. The general solution is useful as it encapsulates all potential scenarios under the given equation.

Particular Solution

A particular solution is derived from the general solution by applying the boundary conditions. Once we found the constant C with given boundary conditions x = 1/2 and y = 1/2:
\[ \sqrt{1-x^2} + \sqrt{1-y^2} = \sqrt{3} \]
This particular solution uniquely fits the original differential equation with the given boundary conditions. Such solutions are incredibly important in real-world applications where specific initial or boundary conditions are known and influential in determining the behavior of systems modeled by differential equations.