# 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 $\Large \frac{y-1}{x^2+x}.$

Toolbox:
• The slope of the tangent to the curve is $\large\frac{dy}{dx}$
• If a proper rational expression is $\large\frac{1}{(x+a)(x+b)},$ then it can be resolved into its partial fraction as $\large\frac{A}{x+a}+\frac{B}{x+b}$
• A linear differential equation of the form $\large\frac{dy}{dx}$$=f(x) can be solved by seperating the variables and then integrating it . Given :The slope of the tangent to the curve at any point is \large\frac{y-1}{x^2+x} The slope of the tangent to the curve is \large\frac{dy}{dx} Therefore \large\frac{dy}{dx}=\frac{y-1}{x^2+x} Seperating the variables we get \large\frac{dy}{y-1}=\frac{dx}{x(x+1)} Now \large\frac{1}{x(x+1)} can be resolved into paritial fractions as \large\frac{A}{x}+\frac{B}{x+1} Hence 1=A(x+1)+B(x) Now let us equate the coefficient of like terms First let us equate the 'x' term 0=A+B-----(1) Equating the constant term 1=+A\qquad B=-1 Hence \large\frac{1}{x(x+1)}=\frac{1}{x}-\frac{1}{x+1} Therefore \large \frac{dy}{y-1}$$=+\large\frac{dx}{x}-\frac{dx}{x+1}$
Integrating on both sides we get,
$\int \large \frac{dy}{y-1}$$=+\int \large\frac{dx}{x}-\int \frac{dx}{x+1}$
=>$\log (y-1)=+\log x-\log (x+1)+\log c$
=>$\log (y-1)=\log \large\frac{c(x)}{(x+1)}$
=>$(y-1)=\Large\frac{cx}{x+1}$
=>$(y-1)(x+1)=cx$
To evaluate the value of c,let us now substitute the value of x and y with the given point (1,0)
$(0-1)(1+1)=c(1)$
=>$c=-2$
$(y-1)(x+1)=-2x$
$(y-1)(x+1)+2x=0$ is the required solution.