Differential Equations 2 Question 19

20. If $x d y=y(d x+y d y), y(1)=1$ and $y(x)>0$. Then, $y(-3)$ is equal to

(a) 3

(b) 2

(c) 1

(d) 0

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Answer:

Correct Answer: 20. (a)

Solution:

  1. Given, $x d y=y(d x+y d y), y>0$

$$ \begin{aligned} & \Rightarrow \quad x d y-y d x=y^{2} d y \\ & \Rightarrow \quad \frac{x d y-y d x}{y^{2}}=d y \Rightarrow \quad d \frac{x}{y}=-d y \end{aligned} $$

On integrating both sides, we get

$$ \frac{x}{y}=-y+c $$

Since,

$$ y(1)=1 \quad \Rightarrow x=1, y=1 $$

$\therefore \quad c=2$

Now, Eq. (i) becomes, $\frac{x}{y}+y=2$

Again, for $x=-3$

$$ \begin{array}{ll} \Rightarrow & -3+y^{2}=2 y \\ \Rightarrow & y^{2}-2 y-3=0 \\ \Rightarrow & (y+1)(y-3)=0 \end{array} $$

As $y>0$, take $y=3$, neglecting $y=-1$.



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