# Let y=y(x) be the solution of the differential equation

Question:

Let $y=y(x)$ be the solution of the differential equation $\mathrm{e}^{\mathrm{x}} \sqrt{1-\mathrm{y}^{2}} \mathrm{~d} \mathrm{x}+\left(\frac{\mathrm{y}}{\mathrm{x}}\right) \mathrm{dy}=0, \mathrm{y}(1)=-1$.

Then the value of $(\mathrm{y}(3))^{2}$ is equal to:

1. $1-4 \mathrm{e}^{3}$

2. $1-4 e^{6}$

3. $1+4 \mathrm{e}^{3}$

4. $1+4 \mathrm{e}^{6}$

Correct Option: , 2

Solution:

$\mathrm{e}^{\mathrm{x}} \sqrt{1-\mathrm{y}^{2}} \mathrm{~d} \mathrm{x}+\frac{\mathrm{y}}{\mathrm{x}} \mathrm{dy}=0$

$\Rightarrow \mathrm{e}^{\mathrm{x}} \sqrt{1-\mathrm{y}^{2}} \mathrm{~d} \mathrm{x}+\frac{-\mathrm{y}}{\mathrm{x}} \mathrm{dy}$

$\Rightarrow \int \frac{-\mathrm{y}}{\sqrt{1-\mathrm{y}^{2}}} \mathrm{dy}=\int_{\mathrm{II}}^{\mathrm{e}} \quad 1$

$\Rightarrow \sqrt{1-\mathrm{y}^{2}}=\mathrm{e}^{\mathrm{x}}(\mathrm{x}-1)+\mathrm{c}$

Given : At $x=1, y=-1$

$\Rightarrow 0=0+\mathrm{c} \Rightarrow \mathrm{c}=0$

$\therefore \sqrt{1-y^{2}}=e^{x}(x-1)$

At $x=3 \quad 1-y^{2}=\left(e^{3} 2\right)^{2} \Rightarrow y^{2}=1-4 e^{6}$