Question

Find the solution of $\frac{dy}{dx}+\frac{y}{x}=\frac{1}{{(1+\log x+\log y)}^2}$

• A. $xy[1-{\left(\log xy\right)}^2]=-\frac{x^2}{2}+C$
• B. $xy[1+{\left(\log xy\right)}^2]=\frac{x^2}{2}+C$
• C. $xy[1+{\left(\log xy\right)}^2]=-\frac{x^2}{2}+C$
• D. $xy[1-{\left(\log xy\right)}^2]=\frac{x^2}{2}+C$

Answer 1

The correct option is B $xy[1+{\left(\log xy\right)}^2]=\frac{x^2}{2}+C$

$\frac{dy}{dx}+\frac{y}{x}=\frac{1}{{(1+\log x+\log y)}^2}$
$xdy+ydx=\frac{xdy}{{(1+\log xy)}^2}=dxy$
or ${[1+\log \left(xy\right)]}^{2}d\left(xy\right)=xdy$
Put $xy=t$ and integrating
$\int 1{(1+\log t)}^{2}dt=\int xdx$
or $t{(1+\log t)}^2-\int t\frac{2(1+\log t)}{t}dt=\frac{x^2}{2}+c$
Now $\int (1+\log t)dt=t+[t\log t-t]=t\log t$
Thus solution is,
$t{(1+\log t)}^2-2\left(t\log t\right)=\frac{x^2}{2}+C$
or $t[1+{\left(\log t\right)}^2+2\log t-2\log t]=\frac{x^2}{2}+C$
or $xy[1+{\left(\log xy\right)}^2]=\frac{x^2}{2}+C$