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

Question:

Let $y(x)$ be the solution of the differential equation $2 x^{2} d y+\left(e^{y}-2 x\right) d x=0, x>0$. If $y(e)=1$, then $\mathrm{y}(1)$ is equal to :

1. 0

2. 2

3. $\log _{e} 2$

4. $\log _{\mathrm {e}}(2 \mathrm{e})$

Correct Option: , 3

Solution:

$2 x^{2} d y+\left(e^{y}-2 x\right) d x=0$

$\frac{d y}{d x}+\frac{e^{y}-2 x}{2 x^{2}}=0 \Rightarrow \frac{d y}{d x}+\frac{e^{y}}{2 x^{2}}-\frac{1}{x}=0$

$e^{-y} \frac{d y}{d x}-\frac{e^{-y}}{x}=-\frac{1}{2 x^{2}} \Rightarrow$ Put $e^{-y}=z$

$\frac{-d z}{d x}-\frac{z}{x}=-\frac{1}{2 x^{2}} \Rightarrow x d z+z d x=\frac{d x}{2 x}$

$\mathrm{d}(\mathrm{xz})=\frac{\mathrm{dx}}{2 \mathrm{x}} \Rightarrow \mathrm{xz}=\frac{1}{2} \log _{\mathrm{e}} \mathrm{x}+\mathrm{c}$

$\mathrm{xe}^{-\mathrm{y}}=\frac{1}{2} \log _{\mathrm{e}} \mathrm{x}+\mathrm{c}$, passes through $(\mathrm{e}, 1)$

$\Rightarrow \mathrm{C}=\frac{1}{2}$

$x e^{-y}=\frac{\log _{e} e x}{2}$

$\mathrm{e}^{-\mathrm{y}}=\frac{1}{2} \Rightarrow \mathrm{y}=\log _{\mathrm{e}} 2$