Question

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

Answer 1

We can convert the given differential equation into a linear ordinary differential equation. Then we can find the integrating factor using the equation $IF = {e^{\int {P\left( x \right)} dx}}$ . We can proceed the integration using the method of integration by partial fractions

Complete step by step answer:

We have the differential equation ${x^2}\left( {{x^2} - 1} \right)\dfrac{{dy}}{{dx}} + x\left( {{x^2} + 1} \right)y = {x^2} - 1$
To make it to a linear equation of the form $\dfrac{{dy}}{{dx}} + P\left( x \right)y = Q\left( x \right)$ , we can divide the equation throughout with ${x^2}\left( {{x^2} - 1} \right)$ . Thus the equation becomes,
$\Rightarrow \dfrac{{dy}}{{dx}} + \dfrac{{\left( {{x^2} + 1} \right)}}{{x\left( {{x^2} - 1} \right)}}y = \dfrac{1}{{{x^2}}}$
Comparing it with standard equation, we get , $P\left( x \right) = \dfrac{{\left( {{x^2} + 1} \right)}}{{x\left( {{x^2} - 1} \right)}}$
The integrating factor of a linear equation of the form ${x^2}\left( {{x^2} - 1} \right)\dfrac{{dy}}{{dx}} + x\left( {{x^2} + 1} \right)y = {x^2} - 1$ is given by $IF = {e^{\int {P\left( x \right)} dx}}$
We have $P\left( x \right) = \dfrac{{\left( {{x^2} + 1} \right)}}{{x\left( {{x^2} - 1} \right)}}$ . To integrate, we can use partial fraction,
$\dfrac{{\left( {{x^2} + 1} \right)}}{{x\left( {x - 1} \right)\left( {x + 1} \right)}} = \dfrac{A}{x} + \dfrac{B}{{\left( {x - 1} \right)}} + \dfrac{C}{{\left( {x + 1} \right)}}$ ……….(a)
Now we can make the denominator of the RHS equal to that of LHS by multiplying the numerators and denominators by required terms.
We can multiply and divide the 1st term with $\left( {x - 1} \right)\left( {x + 1} \right)$ , 2nd term with $\left( {x + 1} \right)x$ and 3rd term with $\left( {x - 1} \right)x$ to make the denominator same.
\Rightarrow $$$$\dfrac{{\left( {{x^2} + 1} \right)}}{{x\left( {x - 1} \right)\left( {x + 1} \right)}} = \dfrac{{A\left( {x - 1} \right)\left( {x + 1} \right)}}{{x\left( {x - 1} \right)\left( {x + 1} \right)}} + \dfrac{{B\left( {x + 1} \right)x}}{{x\left( {x - 1} \right)\left( {x + 1} \right)}} + \dfrac{{C\left( {x - 1} \right)x}}{{x\left( {x - 1} \right)\left( {x + 1} \right)}}
Comparing the numerators, we get,
\Rightarrow $$${x^2} + 1 = A{x^2} - 1A + B{x^2} + Bx + C{x^2} - Cx$ On taking like terms together we get, \Rightarrow $$${x^2} + 1 = \left( {A + B + C} \right){x^2} + \left( {B - C} \right)x - A$Comparing the constants, we get,$ - 1 = A$.. (1) On comparing the coefficients of$x$, we get,$B - C = 0$On rearranging, we get,$ \Rightarrow B = C$..(2) On comparing the coefficients of${x^2}$, we get,$A + B + C = 1$.. (3) Substituting (1) and (2) in (3), we get,$ - 1 + C + C = 1$On simplification we get,$ \Rightarrow 2C = 2$On dividing by 2, we get,$ \Rightarrow C = 1$As$B = C$, we get,$ \Rightarrow B = 1 Substituting the values of A B and C, in (a) we get, $$\dfrac{{\left( {{x^2} + 1} \right)}}{{x\left( {x - 1} \right)\left( {x + 1} \right)}} = - \dfrac{1}{x} + \dfrac{1}{{\left( {x - 1} \right)}} + \dfrac{1}{{\left( {x + 1} \right)}}$$ … (b) Now, we can integrate, \Rightarrow \int {P\left( x \right)} = \int {\dfrac{{\left( {{x^2} + 1} \right)}}{{x\left( {{x^2} - 1} \right)}}} Substituting equation (b), we get, $$ \Rightarrow \int {P\left( x \right)} = - \int {\dfrac{1}{x}} + \int {\dfrac{1}{{\left( {x - 1} \right)}}} + \int {\dfrac{1}{{\left( {x + 1} \right)}}} $$ We know that\int {\dfrac{1}{{x - a}} = \log \left( {x - a} \right)} $$ = - \log \left( x \right) + \log \left( {x - 1} \right) + \log \left( {x + 1} \right)$We know that$\log ab = \log a + \log b$and$\log \dfrac{a}{b} = \log a - \log b$$ = \log \dfrac{{\left( {x - 1} \right)\left( {x + 1} \right)}}{x}$We know that,$\left( {a - b} \right)\left( {a + b} \right) = {a^2} - {b^2}. So, we get, $$ \Rightarrow $$\int {P\left( x \right)} $$ = \log \dfrac{{{x^2} - 1}}{x}$. Now we can calculate integrating factor.$IF = {e^{\log \dfrac{{{x^2} - 1}}{x}}}$We know that${e^{\log a}} = a$$ \Rightarrow IF = \dfrac{{{x^2} - 1}}{x}$So the integrating factor is$\dfrac{{{x^2} - 1}}{x}$ .
Therefore the correct answer is option A.

Note: A linear ordinary differential equation is the equation of the form $\dfrac{{dy}}{{dx}} + P\left( x \right)y = Q\left( x \right)$ where P and Q are functions of x alone. We can find its solution by integrating Q multiplied with an integrating factor. We have the equation $y = \int {IF \times Q\left( x \right)dx}$ . The integrating factor is given by the equation $IF = {e^{\int {P\left( x \right)} dx}}$ . Here we used the method of partial fraction to find the integral. This method is used when the denominator is the product of more than one term. In this method we factorise the denominator and add them together and solve for their numerator.