Question

The equation of the curve through the point $(1,0)$, whose slope is $\frac{y-1}{x^2+x}$, is.-

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

Answer 1

The correct option is A $(y-1)(x+1)+2x=0$

Given that,

Slope of the curve$=\frac{y-1}{x^2+x}$

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

We can rewrite this as

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

Integrate on both sides

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

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

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

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

Note that this integral is in the form of $\int \frac{1}{a}da=\log a$

Applying this formula, we get

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

We know that $\Rightarrow \log m+\log n=\log mn$ and

$\Rightarrow \log m-\log n=\log \left(\frac{m}{n}\right)$

Using these formulae in the above expressions we get,

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

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

Applying Anti-log on both sides we get,

$\Rightarrow y-1=\frac{Cx}{x+1}$ ....$eq\left(1\right)$

It is given that the curve passes through $(1,0)$

Substituting these values of $x,y$ in the above equation we get,

$\Rightarrow 0-1=\frac{C\times 1}{1+1}$

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

$\Rightarrow C=-2$

Substituting this value of $C$ in $eq\left(1\right)$

$\Rightarrow y-1=\frac{-2x}{x+1}$

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

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

Hence, the required equation of the curve is $\Rightarrow (y-1)(x+1)+2x=0$