Question

Integrating factor of $\left( 1-{{x}^{2}} \right)\dfrac{dy}{dx}-xy=1$ is:
A. $1-{{x}^{2}}$
B. $\sqrt{1-{{x}^{2}}}$
C. $\dfrac{1}{1-{{x}^{2}}}$
D. $\dfrac{1}{\sqrt{1-{{x}^{2}}}}$

Answer 1

Hint : A differential equation of the from $\dfrac{dy}{dx}+P\times y=Q$ , where P and Q are constants or functions of x only, is known as a first order linear differential equation.
Rearrange the terms of the given equation to reflect the standard form and identify the expression of the function P.
Find the Integrating Factor (F) by using the formula: $F={{e}^{\int{P}\;dx}}$ .
Make appropriate substitution to simplify the integral. Recall that $\int{\dfrac{1}{x}dx}=\log x+C$ .

Complete step-by-step answer :
The given first order ordinary differential equation is $\left( 1-{{x}^{2}} \right)\dfrac{dy}{dx}-xy=1$ .
We can observe that if we divide both sides by $\left( 1-{{x}^{2}} \right)$ , we will be able to get it in the standard form $\dfrac{dy}{dx}+P\times y=Q$ , where P and Q are constants or functions of x only.
⇒ $\dfrac{dy}{dx}+\left( \dfrac{-x}{1-{{x}^{2}}} \right)y=\dfrac{1}{1-{{x}^{2}}}$
∴ $P=\dfrac{-x}{1-{{x}^{2}}}$ .
Let's calculate $\int{Pdx}$ .
$\int{Pdx}=\int{\dfrac{-x}{1-{{x}^{2}}}dx}$
Substituting $1-{{x}^{2}}=t$ , so that $-2xdx=dt$ , we get:
$\int{Pdx}=\dfrac{1}{2}\int{\dfrac{1}{t}dt}$
Using $\int{\dfrac{1}{x}dx}=\log x+C$ and ignoring the constant C, we get:
⇒ $\int{Pdx}=\dfrac{1}{2}\log t$
Back substituting $1-{{x}^{2}}=t$ , we get:
⇒ $\int{Pdx}=\dfrac{1}{2}\log \left( 1-{{x}^{2}} \right)$
Using $a\log x=\log {{x}^{a}}$ , we can write it as:
⇒ $\int{Pdx}=\log \sqrt{1-{{x}^{2}}}$
Now, the integrating factor will be:
$F={{e}^{\int{P}\;dx}}$
⇒ $F={{e}^{\log \sqrt{1-{{x}^{2}}}}}$
Knowing that ${{e}^{\log a}}=a$ , it can be written as:
⇒ $F=\sqrt{1-{{x}^{2}}}$ , which is the required integrating factor.
So, the correct answer is “ $F=\sqrt{1-{{x}^{2}}}$ ”.

Note : An ordinary differential equation is a differential equation containing one or more functions of one independent variable and the derivatives of those functions.
Steps to solve a First Order Ordinary Linear Differential Equation:
Convert into the standard form $\dfrac{dy}{dx}+P\times y=Q$ , where P and Q are constants or functions of x only.
Find the Integrating Factor (F) by using the formula: $F={{e}^{\int{P}\;dx}}$ .
Write the solution using the formula: $y\times F=\int{Q\times F\ dx}+C$ where C is the constant of integration.
Find the value of C by using the values of $y(0)$ , $y(1)$ etc.