Question

The solution of the differential equation $\frac{dy}{dx}+\frac{y}{x}=\frac{1}{(1+\ln x+\ln y{)}^2}$ is (where, $c$ is an arbitrary constant)

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

Answer 1

Answer: (A)

$\frac{dy}{dx}+\frac{y}{x}=\frac{1}{(1+\ln xy{)}^2}$
Let $xy=u$ so that $\frac{du}{dx}=\frac{x}{(1+\ln u{)}^2}$
$\therefore \int (1+\ln u{)}^{2}du=\int xdx+c$
$\Rightarrow u(1+\ln u{)}^2-\int \frac{2(1+\ln u)}{u}\cdot udu=\frac{x^2}{2}+c$
$\Rightarrow u\left(1+2\ln u+2u(\ln u{)}^2\right)-2u\ln u=\frac{x^2}{2}+c$
$\therefore xy\left(1+(\ln (xy){)}^2\right)=\frac{x^2}{2}+c$