Question

Equation of the curve passing through (1, 0) whose slope is $\frac{y-1}{x^2+x}$

• A. $(y-1)(x+1)+2x=0$
• B. $2x(y-1)+x+1=0$
• C. $x(y-1)(x+1)+2=0$
• D. $(x+1)(y+1)+2x=0$

Answer 1

Answer: (A)

$(y-1)(x+1)+2x=0$

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

$\Rightarrow \frac{dy}{y-1}=\frac{dx}{x(x+1)}=(\frac{1}{x}-\frac{1}{x+1})dx$

$\Rightarrow \int \frac{dy}{y-1}=\int (\frac{1}{x}-\frac{1}{x+1})dx$

$\Rightarrow \log (y-1)=\log x-\log (x+1)+\log c$

$\Rightarrow \log (y-1)=\log \left(\frac{cx}{x+1}\right)$

$\Rightarrow (y-1)=\frac{cx}{x+1}$

Put $(x,y)=(1,0)$

$\Rightarrow -1=\frac{c}{2}$

$\Rightarrow c=-2$

$\Rightarrow (y-1)(x+1)+2x=0$