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:

  1. 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) $-\frac{x}{y}=\frac{x^{2}}{2}+\frac{1}{2}$

$$ \begin{aligned} \Rightarrow & x^{2}+1 & =-\frac{2 x}{y} \\ \Rightarrow & y & =-\frac{2 x}{x^{2}+1} \\ \therefore & f-\frac{1}{2} & =\frac{4}{5} \end{aligned} $$



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