Differential Equations Ques 50
- Given that the slope of the tangent to a curve $y=y(x)$ at any point $(x, y)$ is $\frac{2 y}{x^{2}}$. If the curve passes through the centre of the circle $x^{2}+y^{2}-2 x-2 y=0$, then its equation is
(2019 Main, 8 April II)
(a) $x^{2} \log _e|y|=-2(x-1)$
(b) $x \log _e|y|=x-1$
(c) $x \log _e|y|=2(x-1)$
(d) $x \log _e|y|=-2(x-1)$
Show Answer
Answer:
Correct Answer: 50.(c)
Solution:
- Given, $\frac{d y}{d x}=\frac{2 y}{x^{2}}$
$\Rightarrow \quad \int \frac{d y}{y}=\int \frac{2}{x^{2}} d x \quad$ [integrating both sides]
$\Rightarrow \quad \log _e|y|=-\frac{2}{x}+C$
Since, curve (i) passes through centre $(1,1)$ of the circle
$$ \begin{gathered} x^{2}+y^{2}-2 x-2 y=0 \\ \therefore \quad \log _e(1)=-\frac{2}{1}+C \Rightarrow C=2 \end{gathered} $$
$\therefore$ Equation required curve is
$$ \begin{array}{rlrl} & \quad \log _e|y| & =-\frac{2}{x}+2 \quad \text { [put } C=2 \text { in Eq. (i)] } \\ \Rightarrow \quad & x \log _e|y|=2(x-1) \end{array} $$
BETA
