Question

The integrating factor of the differential equation is $${x^2}({x^2} - 1)\dfrac{{dy}}{{dx}} + {x^2}({x^2} + 1)y = {x^2} - 1$$
A.$$\dfrac{{{x^2} - 1}}{x}$$
B.$$\dfrac{{{x^2} + 1}}{{{x^2}({x^2} - 1)}}$$
C.$$\log \dfrac{{{x^2} - 1}}{x}$$
D.None of this.

Answer 1

Hint: Make given equation in form of linear differential equation as
$$\dfrac{{dx}}{{dy}} + p(x)y = q(x)$$.solution of this equation is $$y \times u\left( x \right) = (\smallint \left( {u\left( x \right) \times q\left( x \right)dx} \right) + C$$
Where $$u(x) = {e^{\int {(p(x)dx)} }}$$

Complete step-by-step answer:
Given, Differential equation as $${x^2}({x^2} - 1)\dfrac{{dy}}{{dx}} + {x^2}({x^2} + 1)y = {x^2} - 1$$
 $$\dfrac{{dy}}{{dx}} + \dfrac{{({x^2} + 1)y}}{{x({x^2} - 1)}} = \dfrac{1}{{{x^2}}}$$
 This is in the form of linear differential equation as
$$\dfrac{{dx}}{{dy}} + p(x)y = q(x)$$ .Then the solution of the equation is $$y \times u\left( x \right) = (\smallint \left( {u\left( x \right) \times q\left( x \right)dx} \right) + C$$
Where $$u(x) = {e^{\int {(p(x)dx)} }}$$ which is the integration factor of the equation.
⇒ On comparing the given equation with the general equation. $$p(x) = \dfrac{{({x^2} + 1)}}{{x({x^2} - 1)}},q(x) = \dfrac{1}{{{x^2}}}$$
Integrating factor,
$$\eqalign{
  & u(x) = {e^{\int {(\dfrac{{({x^2} + 1)}}{{x({x^2} - 1)}}dx)} }} \cr
  & = {e^{\int {(\dfrac{1}{{(x - 1)}} + \dfrac{1}{{(1 + x)}} - \dfrac{1}{x})dx} }} \cr
  & = {e^{(\ln (x - 1) + \ln (x + 1) - \ln x)}} \cr
  & = {e^{\ln (\dfrac{{({x^2} - 1)}}{x})}} \cr} $$
So the correct option is a. $$\dfrac{{{x^2} - 1}}{x}$$

Note: It is necessary to form a given equation into $$\dfrac{{dx}}{{dy}} + p(x)y = q(x)$$. Solution of this equation is $$y \times u\left( x \right) = (\smallint \left( {u\left( x \right) \times q\left( x \right)dx} \right) + C$$
Where $$u(x) = {e^{\int {(p(x)dx)} }}$$