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:
- 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 $y=-1$.
