Question

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

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

Answer 1

Answer: (A)

$\frac{d y}{d x}+\frac{y}{x}=\frac{1}{(1+\ln x y)^{2}}$
Let $x y=u$ so that $\frac{d u}{d x}=\frac{x}{(1+\ln u)^{2}}$
$\therefore \quad \int(1+\ln u)^{2} d u=\int x d x+c$
$\Rightarrow u(1+\ln u)^{2}-\int \frac{2(1+\ln u)}{u} \cdot u d u=\frac{x^{2}}{2}+c$
$\Rightarrow u\left(1+2 \ln u+2 u(\ln u)^{2}\right)-2 u \ln u=\frac{x^{2}}{2}+c$
$\therefore x y\left(1+(\ln (x y))^{2}\right)=\frac{x^{2}}{2}+c$