Find the "general solution" (that is, a solution containing...

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 $x=\frac{1}{2}$.

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 $x=\frac{1}{2}$.

Key Concepts

Separable Differential Equation

A separable differential equation is one where the variables can be rearranged so that all terms involving one variable (and its differential) are on one side of the equation and all terms involving the other variable are on the opposite side. This property makes it possible to integrate both sides independently with respect to their own variable, effectively reducing a complex differential equation into simpler integrals.

Separation of Variables Method

This is a common technique used to solve ordinary differential equations that are separable. After rearranging the equation to isolate the different variables on opposite sides, the method involves integrating both sides of the equation separately. The result is an equation that typically involves an arbitrary constant, representing the general solution of the differential equation.

General Solution

The general solution of a differential equation is a family of solutions expressed in terms of an arbitrary constant. This constant arises from the integration process and indicates that there are infinitely many solutions to the differential equation. It encapsulates all possible solutions satisfying the equation before any specific initial or boundary conditions are applied.

Particular Solution

A particular solution is obtained from the general solution by applying given boundary or initial conditions. These conditions enable the determination of the arbitrary constant, thereby yielding one unique solution that not only satisfies the differential equation but also adheres to the prescribed conditions.

Integration Process

Integration plays a crucial role in the separation of variables method. Once the differential equation is split into parts involving only one variable each, integration is used to find the antiderivatives. This step often introduces an arbitrary constant, which reflects the general nature of the solution prior to the application of specific conditions.

Transcript

00:01 In this question we have a differential equation x into under root of 1 minus y square d x plus y into under root of 1 minus x square d y is equal to 0.

00:18 We have y is equals to 1 upon 2 when x is equals to 1 upon 2.

00:27 We are required to evaluate the general solution for this differential equation.

00:31 So let's see how to solve this question.

00:35 The given differential equation can also be written as x into under root of 1 minus y square, dx is equal to minus y into under root of 1 minus x square, d .y.

00:52 Now let's cross multiply this, so we will have x upon under root of 1 minus x square, dx is equal.

01:03 Minus y upon under root of 1 minus y square d y and now let's integrate both of the sides we can observe that both of the integrals are identical the only difference is here we have x and here we have the variable y so the method to integrate these integrals will remain same so now let's integrate x upon under root 1 minus x squared d x so consider 1 minus x square is equal to u so by the differentiation we can write minus 2 x d x is equal to d u therefore x d x will be equal to minus d u upon 2 now substitute all these values in the above integral so we will have integration x d x upon under root 1 minus x square d x is equals to minus 1 upon 2 integration d u upon under root u and this can also be written as minus 1 upon 2 integration u to the power minus 1 upon 2 d u we know that the integration of x to the power and d x is equals to f to the power n plus 1 upon n plus 1.

02:50 So by using this formula, this integral will be equals to minus 1 upon 2, u to the power minus 1 upon 2 plus 1 divided by minus 1 upon 2 plus 1...

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