Question

Solve the differential equation $(1+y^2)(1+logx)dx+xdy=0$, given that when $x=1$ then $y=1$.

Answer 1

$(1+y^2)(1+\log x)dx=xdy$

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

Integrate

${\tan }^{-1}y=-\int \frac{1+\log x}{x}dx$

Let $1+\log x=t$

$\frac{dx}{x}=dt$

${\tan }^{-1}y=\frac{-1}{2}(1+\log x{)}^2+C$

Put $x=1,y=1$

$\frac{\pi }{4}=\frac{-1}{2}+C$

$C=\frac{\pi }{4}+\frac{1}{2}$

$C=\frac{\pi +2}{4}$

So ${\tan }^{-1}y=\frac{-1}{2}(1+\log x{)}^2+\frac{\pi +2}{4}$