Question

The integrating factor of the differential equation is
$x^2(x^2-1)\frac{dy}{dx}+x(x^2+1)y=x^2-1$

• A. $\frac{x^2-1}{x}$
• B. $\frac{x^2+1}{x(x^2-1)}$
• C. $\log \frac{x^2-1}{x}$
• D. None of these

Answer 1

The correct option is A $\frac{x^2-1}{x}$

Given, Differential equation as $x^2(x^2-1)\frac{dy}{dx}+x(x^2+1)y=(x^2-1)$
$\Rightarrow \frac{dy}{dx}+\frac{(x^2+1)y}{x(x^2-1)}=\frac{1}{x^2}$
This is in the form of linear differential equation as $\frac{dy}{dx}+p\left(X\right)y=q\left(X\right)$.
Then the solution of the equation is $y\ast u\left(x\right)=(\int (u\left(x\right)\ast q\left(x\right)dx)+C$
where $u\left(x\right)=e^{\int \left(p\right(x\left)dx\right)}$ which is the integration factor of the equation.
$\Rightarrow$ On comparing the given equation with general equation.$P\left(x\right)=\frac{(x^2+1)}{x(x^2-1)},q\left(x\right)=\frac{1}{x^2}$
$\Rightarrow Integratingfactor,u\left(x\right)=e^{\int (\frac{(x^2+1)}{x(x^2-1)}dx)}$
$=e^{\int (\frac{1}{(x-1)}+\frac{1}{(1+x)}-\frac{1}{x})dx)}=e^{\left(ln\right(x-1)+ln(x+1)-lnx)}$
$=e^{ln\left(\frac{(x^2-1)}{x}\right)}$