Differential Equations 2 Question 8
8. If a curve passes through the point $(1,-2)$ and has slope of the tangent at any point $(x, y)$ on it as $\frac{x^{2}-2 y}{x}$, then the curve also passes through the point
(a) $(\sqrt{3}, 0)$
(b) $(-1,2)$
(c) $(-\sqrt{2}, 1)$
(d) $(3,0)$
(2019 Main, 12 Jan II)
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Answer:
Correct Answer: 8. (a)
Solution:
- We know that, slope of the tangent at any point $(x, y)$ on the curve is
$$ \begin{aligned} \frac{d y}{d x} & =\frac{x^{2}-2 y}{x} \\ \Rightarrow \quad \frac{d y}{d x}+\frac{2}{x} y & =x \end{aligned} $$
which is a linear differential equation of the form $\frac{d y}{d x}+P(x) \cdot y=Q(x)$
where
$$ P(x)=\frac{2}{x} \text { and } Q(x)=x $$
Now, integrating factor
$$ \begin{aligned} \text { (IF) } & =e^{\int P(x) d x}=e^{\int \frac{2}{x} d x}=e^{2 \log _e x} & \\ & =e^{\log _e x^{2}} & {\left[\because m \log a=\log a^{m}\right] } \\ & =x^{2} & {\left[\because e^{\log _e f(x)}=f(x)\right] } \end{aligned} $$
and the solution of differential Eq. (i) is
$$ \begin{array}{cc} & y(IF)=\int Q(x)(IF) d x+C \Rightarrow y\left(x^{2}\right)=\int x \cdot x^{2} d x+C \\ \Rightarrow \quad & y x^{2}=\frac{x^{4}}{4}+C \end{array} $$
$\because$ The curve (ii) passes through the point $(1,-2)$, therefore
$$ -2=\frac{1}{4}+C \Rightarrow C=-\frac{9}{4} $$
$\therefore$ Equation of required curve is $4 y x^{2}=x^{4}-9$.
Now, checking all the option, we get only $(\sqrt{3}, 0)$ satisfy the above equation.
