Question:
If a curve passes through the origin and the slope of the tangent to it at any point $(x, y)$ is
$\frac{x^{2}-4 x+y+8}{x-2}$, then this curve also passes
through the point:
Correct Option: , 4
Solution:
Given
$y(0)=0$
$\& \frac{d y}{d x}=\frac{(x-2)^{2}+y+4}{x-2}$
$\Rightarrow \frac{d y}{d x}-\frac{y}{x-2}=(x-2)+\frac{4}{x-2}$
$\Rightarrow$ I.F. $=\mathrm{e}^{-\int \frac{1}{x-2} d x}=\frac{1}{x-2}$
Solution of L.D.E.
$\Rightarrow \mathrm{y} \frac{1}{\mathrm{x}-2}=\int \frac{1}{\mathrm{x}-2}\left((\mathrm{x}-2)+\frac{4}{\mathrm{x}-2}\right) \cdot \mathrm{dx}$
$\Rightarrow \frac{y}{x-2}=x-\frac{4}{x-2}+C$
Now, at $x=0, y=0 \Rightarrow C=-2$
$y=x(x-2)-4-2(x-2)$
$\Rightarrow y=x^{2}-4 x$
This curve passes through $(5,5)$