Question

Find the equation of the curve through the point $(1,0)$. If the slope of the tangent to the curve at any point $(x,y)$ is $\frac{y-1}{x^2+x}$ ?

Answer 1

Given, the slope of the tangent to the curve at any point $(x,y)$ is $\frac{y-1}{x^2+x}$

Since, the slope of the tangent to the curve is $\frac{dy}{dx}$,

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

Separating the variables we get,

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

Now, resolving $\frac{1}{x(x+1)}$ into partial fractions as,

$\frac{A}{x}+\frac{B}{x+1}$

$\therefore 1=A(x+1)+B\left(x\right)$

Now, let us equate the $x$ term,

$0=A+B$

Equating the constant term,

$A=1,B=-1$

Hence, $\frac{1}{x(x+1)}=\frac{1}{x}-\frac{1}{x+1}$

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

Integrating on both sides we get,

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

Substituting the value of $x$ and $y$ with the given point $(1,0)$,

$(0-1)(1+1)=c\left(1\right)\Rightarrow c=-2\therefore (y-1)(x+1)=-2x\Rightarrow (y-1)(x+1)+2x=0$

Hence, it is the required equation of the curve through the point $(1,0).$