Question

$x(x^2+1)\frac{dy}{dx}+(x^2-1)y=x^{3}lnx$

• A. $4(x^2-1)y-x^3(1-2lny)=cx$
• B. $4(x^2+1)y+x^3(1-2lnx)=cx$
• C. $4(x^2-1)y+x^3(1+2lnx)=cx$
• D. $4(x^2-1)y-x^3(1+2lny)=cx$

Answer 1

The correct option is B $4(x^2+1)y+x^3(1-2lnx)=cx$

$x(x^2+1)\frac{dy}{dx}+(x^2-1)y=x^3\ln x$
$\frac{dy}{dx}+\frac{(x^2-1)y}{x(x^2+1)}=\frac{x^2\ln x}{x^2+1}$
I.F. $=e^{\int \frac{x^2-1}{x(x^2+1)}dx}=\frac{x^2+1}{x}$
$\therefore \left(\frac{x^2+1}{x}\right)y=\int \frac{x^2+1}{x}\times \frac{x^2}{x^2+1}\ln xdx$
$\Rightarrow \left(\frac{x^2+1}{x}\right)y=\int x\ln xdx$
$\Rightarrow \left(\frac{x^2+1}{x}\right)y=\ln x\frac{x^2}{2}-\frac{x^2}{4}+c$
$\Rightarrow 4(x^2+1)y+x^3(1-2\ln x)=cx$