A differential equation is defined by the sum of two mathematical expressions and the mathematical expressions are defined in terms of two variables $x$ and $y$ as follows.

$y\sqrt{1-x^2}\,dy$ $+$ $x\sqrt{1-y^2}\,dx$ $\,=\,$ $0$

This differential equation can be solved by using the separation of variables in three steps. Now, let’s learn how to solve this equation in the variable separable method from the understandable steps.

Let us try to separate the functions in terms of $x$ with the respective differential element $dx$ from the functions in terms of $y$ with the corresponding differential $dy$.

$\implies$ $y\sqrt{1-x^2}\,dy$ $\,=\,$ $-$ $x\sqrt{1-y^2}\,dx$

Now, shift the functions in terms of $y$ to the left hand side from the right hand side of the equation. Similarly, move the functions in terms of $x$ to the right hand side from the left hand side of the equation. Now, the functions in terms of both variables with their respective differential are separated successfully as follows.

$\implies$ $\dfrac{y}{\sqrt{1-y^2}}\,dy$ $\,=\,$ $-\dfrac{x}{\sqrt{1-x^2}}\,dx$

$\implies$ $\dfrac{y}{\sqrt{1-y^2}}\,dy$ $\,=\,$ $\dfrac{-x}{\sqrt{1-x^2}}\,dx$

The simplified differential equation has to be integrated to solve the given differential equation mathematically.

$\implies$ $\displaystyle \int{\dfrac{y}{\sqrt{1-y^2}}}\,dy$ $\,=\,$ $\displaystyle \int{\dfrac{-x}{\sqrt{1-x^2}}}\,dx$

Let’s take $u^2 \,=\, 1-y^2$ for finding the indefinite integration of the left hand side equation. Now, differentiate this equation with respect to $y$.

$\implies$ $\dfrac{d}{dy}{\,(u^2)}$ $\,=\,$ $\dfrac{d}{dy}{\,(1-y^2)}$

The difference rule of derivatives can be used to differentiate each term in the expression on the right hand side of the equation.

$\implies$ $\dfrac{d}{dy}{\,(u^2)}$ $\,=\,$ $\dfrac{d}{dy}{\,(1)}-\dfrac{d}{dy}{\,(y^2)}$

Now, use the power rule of the differentiation and also the derivative rule of a constant.

$\implies$ $2u^{\,2-1} \times \dfrac{du}{dy}$ $\,=\,$ $0-2y^{\,2-1}$

Now, simplify the above equation by using the mathematical fundamentals.

$\implies$ $2u^{\,1} \times \dfrac{du}{dy}$ $\,=\,$ $0-2y^{\,1}$

$\implies$ $2u \times \dfrac{du}{dy}$ $\,=\,$ $-2y$

$\implies$ $\cancel{2}u \times \dfrac{du}{dy}$ $\,=\,$ $-\cancel{2}y$

$\implies$ $u \times \dfrac{du}{dy}$ $\,=\,$ $-y$

$\implies$ $u \times du$ $\,=\,$ $-y \times dy$

$\implies$ $u\,du$ $\,=\,$ $-y\,dy$

$\implies$ $-y\,dy$ $\,=\,$ $u\,du$

$\,\,\,\therefore\,\,\,\,\,\,$ $y\,dy$ $\,=\,$ $-u\,du$

It is time to focus on the simplified integral equation.

$\displaystyle \int{\dfrac{y}{\sqrt{1-y^2}}}\,dy$ $\,=\,$ $\displaystyle \int{\dfrac{-x}{\sqrt{1-x^2}}}\,dx$

$\implies$ $\displaystyle \int{\dfrac{y \times dy}{\sqrt{1-y^2}}}$ $\,=\,$ $\displaystyle \int{\dfrac{-x}{\sqrt{1-x^2}}}\,dx$

$\implies$ $\displaystyle \int{\dfrac{y\,dy}{\sqrt{1-y^2}}}$ $\,=\,$ $\displaystyle \int{\dfrac{-x}{\sqrt{1-x^2}}}\,dx$

We have taken that $u^2 \,=\, 1-y^2$ and have obtained a mathematical relation $y\,dy$ $\,=\,$ $-u\,du$ by the differentiation. Now, express the left hand side integral expression in terms of $u$ by using these two rules.

$\implies$ $\displaystyle \int{\dfrac{-u\,du}{\sqrt{(u^2)}}}$ $\,=\,$ $\displaystyle \int{\dfrac{-x}{\sqrt{1-x^2}}}\,dx$

Now, simplify the above integral equation by the fundamentals of mathematics.

$\implies$ $\displaystyle \int{\dfrac{-u\,du}{u}}$ $\,=\,$ $\displaystyle \int{\dfrac{-x}{\sqrt{1-x^2}}}\,dx$

$\implies$ $\displaystyle \int{\dfrac{-\cancel{u}\,du}{\cancel{u}}}$ $\,=\,$ $\displaystyle \int{\dfrac{-x}{\sqrt{1-x^2}}}\,dx$

$\implies$ $\displaystyle \int{\,-du}$ $\,=\,$ $\displaystyle \int{\dfrac{-x}{\sqrt{1-x^2}}}\,dx$

$\implies$ $\displaystyle \int{\,(-1) \times du}$ $\,=\,$ $\displaystyle \int{\dfrac{(-1) \times x}{\sqrt{1-x^2}}}\,dx$

$\implies$ $\displaystyle \int{(-1) \times \,du}$ $\,=\,$ $\displaystyle \int{\bigg((-1) \times \dfrac{x}{\sqrt{1-x^2}}\bigg)}\,dx$

The constants that multiply the functions can be taken out from the integration by the constant multiple rule of the integration.

$\implies$ $(-1) \times \displaystyle \int{\,du}$ $\,=\,$ $(-1) \times \displaystyle \int{\dfrac{x}{\sqrt{1-x^2}}}\,dx$

$\implies$ $\cancel{(-1)} \times \displaystyle \int{\,du}$ $\,=\,$ $\cancel{(-1)} \times \displaystyle \int{\dfrac{x}{\sqrt{1-x^2}}}\,dx$

$\implies$ $\displaystyle \int{\,du}$ $\,=\,$ $\displaystyle \int{\dfrac{x}{\sqrt{1-x^2}}}\,dx$

$\implies$ $\displaystyle \int{\,du}$ $\,=\,$ $\displaystyle \int{\dfrac{x \times dx}{\sqrt{1-x^2}}}$

$\implies$ $\displaystyle \int{\,du}$ $\,=\,$ $\displaystyle \int{\dfrac{x\,dx}{\sqrt{1-x^2}}}$

The procedure that we have used above is used one more time to find the indefinite integration of the right hand side expression of the equation.

Take $v^2 \,=\, 1-x^2$ and differentiate it with respect to $x$.

$\implies$ $\dfrac{d}{dx}{\,(v^2)}$ $\,=\,$ $\dfrac{d}{dx}{\,(1-x^2)}$

$\implies$ $\dfrac{d}{dx}{\,(v^2)}$ $\,=\,$ $\dfrac{d}{dx}{\,(1)}-\dfrac{d}{dx}{\,(x^2)}$

$\implies$ $2v^{\,2-1} \times \dfrac{dv}{dx}$ $\,=\,$ $0-2x^{\,2-1}$

$\implies$ $2v^{\,1} \times \dfrac{dv}{dx}$ $\,=\,$ $0-2x^{\,1}$

$\implies$ $2v \times \dfrac{dv}{dx}$ $\,=\,$ $-2x$

$\implies$ $\cancel{2}v \times \dfrac{dv}{dx}$ $\,=\,$ $-\cancel{2}x$

$\implies$ $v \times \dfrac{dv}{dx}$ $\,=\,$ $-x$

$\implies$ $v \times dv$ $\,=\,$ $-x \times dx$

$\implies$ $v\,dv$ $\,=\,$ $-x\,dx$

$\implies$ $-x\,dx$ $\,=\,$ $v\,dv$

$\,\,\,\therefore\,\,\,\,\,\,$ $x\,dx$ $\,=\,$ $-v\,dv$

Now, focus on the simplified integral equation for eliminating the functions in terms of $x$.

$\implies$ $\displaystyle \int{\,du}$ $\,=\,$ $\displaystyle \int{\dfrac{x\,dx}{\sqrt{1-x^2}}}$

In this problem, we have taken $v^2 = 1-x^2$ and this equation is differentiated with respect to $x$. Thus, we have obtained that $x\,dx$ $\,=\,$ $-v\,dv$. Now, we can eliminate the functions in terms of $x$ by expressing it in terms of $v$.

$\implies$ $\displaystyle \int{\,du}$ $\,=\,$ $\displaystyle \int{\dfrac{-v\,dv}{\sqrt{v^2}}}$

$\implies$ $\displaystyle \int{\,du}$ $\,=\,$ $\displaystyle \int{\dfrac{-v\,dv}{v}}$

$\implies$ $\displaystyle \int{\,du}$ $\,=\,$ $\displaystyle \int{\dfrac{-\cancel{v}\,dv}{\cancel{v}}}$

$\implies$ $\displaystyle \int{\,du}$ $\,=\,$ $\displaystyle \int{\,-dv}$

$\implies$ $\displaystyle \int{\,du}$ $\,=\,$ $\displaystyle \int{(-1) \times \,dv}$

$\implies$ $\displaystyle \int{\,du}$ $\,=\,$ $(-1) \times \displaystyle \int{\,dv}$

$\implies$ $\displaystyle \int{\,du}$ $\,=\,$ $-\displaystyle \int{\,dv}$

The integration on both sides of the equation can be performed by the integral one rule.

$\implies$ $u+c_1$ $\,=\,$ $-(v+c_2)$

Here, $c_1$ and $c_2$ are the integral constants.

$\implies$ $u+c_1$ $\,=\,$ $-v-c_2$

$\implies$ $u$ $\,=\,$ $-v-c_1-c_2$

$\implies$ $u$ $\,=\,$ $-v-(c_1+c_2)$

$\implies$ $u+v$ $\,=\,$ $-(c_1+c_2)$

In the above equation, the expression on the right hand side of the equation represents a constant. Hence, it can be simply denoted by $c$.

$\implies$ $u+v \,=\, c$

The solution of the given differential equation is obtained in terms of $u$ and $v$ in the previous step but the given differential equation is expressed in terms of $x$ and $y$. So, we have to express the above equation in terms of the variables $x$ and $y$.

Actually, we have taken above that $u^2 \,=\, 1-y^2$ and $v^2 \,=\, 1-x^2$. Therefore, $u \,=\, \sqrt{1-y^2}$ and $v \,=\, \sqrt{1-x^2}$. Now, substitute the values of $u$ and $v$ in the solution of the given differential equation.

$\,\,\,\therefore\,\,\,\,\,\,$ $\sqrt{1-y^2}+\sqrt{1-x^2}$ $\,=\,$ $c$

Latest Math Topics

Mar 21, 2023

Feb 25, 2023

Feb 17, 2023

Feb 10, 2023

Latest Math Problems

Mar 03, 2023

Mar 01, 2023

Feb 27, 2023

A best free mathematics education website for students, teachers and researchers.

Learn each topic of the mathematics easily with understandable proofs and visual animation graphics.

Learn how to solve the math problems in different methods with understandable steps and worksheets on every concept for your practice.

Copyright © 2012 - 2022 Math Doubts, All Rights Reserved