Question

A curve passes through (2,0) and the slope of tangent at a point P(x,y) is equal to $(x+1{)}^2+y-3/(x+1).$ Then equation of the curve is: [IIT 2004]

• A. $y=x^2+2x$
• B. $y=x^2-2x$
• C. $y=2x^2-x$
• D. None of these

Answer 1

The correct option is B $y=x^2-2x$

$\frac{dy}{dx}=\frac{(x+1{)}^2+y-3}{x+1}=(x+1)+\frac{y-3}{x+1}$

Putting x+1=X, y-3=Y, $\frac{dy}{dx}=\frac{dY}{dX},$ the equation becomes $\frac{dY}{dX}=X+\frac{Y}{X}or\frac{dY}{dX}-\frac{1}{X}.$Y = X [L.D.E]

I.F. = $e^{\int (-1/X)dX}=e^{-logX}=X^{-1}=\frac{1}{X}$

∴ the solution is $Y.\left(\frac{1}{X}\right)=c+\int X.\left(\frac{1}{X}\right)dx=c+Xor\frac{(y-3)}{(x+1)}=c+x+1$

x = 2, y = 0 $\Rightarrow \frac{0-3}{2+1}c+2+1c=-4.$

∴ the equation of the curve is $\frac{y-3}{x+1}=x-3ory=x^2-2x.$