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Q.
A curve passes through $(2,0)$ and slope at point $P(x, y)$ is $\frac{(x+1)^2+(y-3)}{(x+1)}$. The area between curve and $x$-axis in 4th quadrant is
Differential Equations
Solution:
Here, slope of tangent
$ \frac{d y}{d x}=\frac{(x+1)^2+y-3}{(x+1)}$
$ \Rightarrow \frac{d y}{d x}=(x+1)+\frac{(y-3)}{(x+1)} $,
put $ x+1=X $ and $ y-3=Y\left(\text { here } \frac{d y}{d x}=\frac{d Y}{d X}\right)$
$ \therefore \frac{ dY }{ dX }= X +\frac{ Y }{ X } \Rightarrow \frac{ dY }{ dX }-\frac{1}{ X } Y = X $
where integrating factor
$=e^{\int-\frac{1}{x} d X}=e^{-\ell n X}=\frac{1}{X}$
$ \therefore $ Solution is, $ Y \cdot \frac{1}{X}=\int X \cdot \frac{1}{X} d X+c \Rightarrow \frac{Y}{X}=X+c $
$ y-3=(x+1)^2+c(x+1) $, which passes through $(2,0)$
$-3=9+3 c$
$\Rightarrow c=-4$
$\therefore $ Required curve
$\therefore y=(x+1)^2-4(x+1)+3$
$\Rightarrow y=x^2-2 x$
Drawing curve
Thus, required area
$=\left|\int\limits_0^2\left(x^2-2 x\right) d x\right|=\left|\left(\frac{x^3}{3}-x^2\right)_0^2\right|=\frac{4}{3}$ sq. units
