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  • Find the equation of the curve through the point (1, 0) if the slope of the tangent to the curve any point (x, y) is \frac{y-1}{x^{2}+x}

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

Answers (1)

Given: Slope of the tangent is \frac{y-1}{x^{2}+x}$
Slope of tangent of a curve \mathrm{y}=\mathrm{f}(\mathrm{x})$ is given by $\mathrm{dy} / \mathrm{dx}$

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

\Rightarrow \int \frac{\mathrm{dy}}{\mathrm{y}-1}=\int \frac{\mathrm{dx}}{\mathrm{x}(\mathrm{x}+1)} \ldots(\mathrm{a})$ $\frac{1}{x(x+1)}$
Use partial fraction for
\\\Rightarrow \frac{1}{x(x+1)}=\frac{A}{x}+\frac{B}{x+1}$ \\$\Rightarrow \frac{1}{x(x+1)}=\frac{A(x+1)+B x}{x(x+1)}$
Equate the numerator
A(x+1)+B x=1$
Put x=0
A=1

Put x=-1

B=-1

Hence \Rightarrow \frac{1}{x(x+1)}=\frac{1}{x}+\frac{-1}{x+1}$
Hence equation (a) becomes,
$$ \\ \Rightarrow \int \frac{d y}{y-1}=\int\left(\frac{1}{x}-\frac{1}{x+1}\right) \mathrm{dx} \\ \Rightarrow \int \frac{d y}{y-1}=\int \frac{1}{x} d x-\int \frac{1}{x+1} d x $$
\log (y-1)=\log x-\log (x+1)+c \ldots(b)$
Now it is given that the curve is passing through (1,0)
Hence (1,0) will satisfy the equation (b)
Put  x=1  and y=0  in b
When we put  y=0 in equation b the result is \log (-1)$ which is undefined
hence, we must simplify equation (b) further
$$ \log (y-1)-\log x=-\log (x+1)+c $$
using loga-logb=loga/b

$$ \Rightarrow \log \left(\frac{y-1}{x}\right)(x+1)=c $$
Constant c  must be taken as log c to eliminate undefined elements in the
equation.(log cand not any other terms because taking logc completely
eliminates the log terms and we don't have to worry about undefined terms
in the equation)
$$ \Rightarrow \log \left(\frac{y-1}{x}\right)(x+1)=\log c $$
Eliminate log
$$ \Rightarrow\left(\frac{y-1}{x}\right)(x+1)=c \ldots(c) $$
Substitute  x=1 and y=0 
$$ \Rightarrow\left(\frac{0-1}{1}\right)(1+1)=c $$
 c=-2 
put back  c=-2  in  (c) 

\begin{aligned} &\Rightarrow\left(\frac{y-1}{x}\right)(x+1)=-2\\ &\text { Hence the equation of the curve is }(y-1)(x+1)=-2 x \end{aligned}

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