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Home  >>  CBSE XII  >>  Math  >>  Differential Equations
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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 $\Large \frac{y-1}{x^2+x}.$

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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.
answered May 14, 2013 by meena.p
 

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