Differential Equations 2 Question 15
15. If a curve $y=f(x)$ passes through the point $(1,-1)$ and satisfies the differential equation, $y(1+x y) d x=x d y$, then $f-\frac{1}{2}$ is equal to
(a) $-\frac{2}{5}$
(b) $-\frac{4}{5}$
(c) $\frac{2}{5}$
(d) $\frac{4}{5}$
(2016 Main)
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Answer:
Correct Answer: 15. (d)
Solution:
- Given differential equation is
$$ \begin{array}{rlrl} & & y(1+x y) d x & =x d y \\ \Rightarrow & y d x+x y^{2} d x & =x d y \\ \Rightarrow & \frac{x d y-y d x}{y^{2}} & =x d x \\ \Rightarrow & & -\frac{(y d x-x d y)}{y^{2}} & =x d x \Rightarrow-d \quad \frac{x}{y}=x d x \end{array} $$
On integrating both sides, we get
$$ -\frac{x}{y}=\frac{x^{2}}{2}+C $$
$\because$ It passes through $(1,-1)$.
$$ \therefore \quad 1=\frac{1}{2}+C \Rightarrow C=\frac{1}{2} $$
Now, from Eq. (i) $-\frac{x}{y}=\frac{x^{2}}{2}+\frac{1}{2}$
$$ \begin{aligned} \Rightarrow & x^{2}+1 & =-\frac{2 x}{y} \\ \Rightarrow & y & =-\frac{2 x}{x^{2}+1} \\ \therefore & f-\frac{1}{2} & =\frac{4}{5} \end{aligned} $$
