Question

If a curve passes through the origin and the slope of the tangent to it at any point $(x,y)$ is $\frac{x^2-4x+y+8}{x-2}$, then this curve also passes through the point:

• A. (5,4)
• B. (4,5)
• C. (4,4)
• D. (5,5)

Answer 1

Answer: (D)

Given
$y(0)=0$
$\&\frac{dy}{dx}=\frac{(x-2{)}^2+y+4}{x-2}$
$\Rightarrow \frac{dy}{dx}-\frac{y}{x-2}=(x-2)+\frac{4}{x-2}$
$\Rightarrow I.F.=e^{-\int \frac{1}{x-2}dx}=\frac{1}{x-2}$
Solution of $L.D.E$
$\Rightarrow y\frac{1}{x-2}=\int \frac{1}{x-2}\left((x-2)+\frac{4}{x-2}\right)\cdot dx$
$\Rightarrow \frac{y}{x-2}=x-\frac{4}{x-2}+C$
Now, at $x=0,y=0\Rightarrow C=-2$
$y=x(x-2)-4-2(x-2)$
$\Rightarrow y=x^2-4x$
This curve passes through (5,5)