### Differential Equations Question 1

#### Question 1 - 24 January - Shift 1

Let $y=y(x)$ be the solution of the differential equation $x^{3} d y+(x y-1) d x=0, x>0$, $y(\frac{1}{2})=3-e$. Then $y(1)$ is equal to

(1) 1

(2) e

(3) 2-e

(4) 3

## Show Answer

#### Answer: (1)

#### Solution:

#### Formula: Linear differential equations of first order

$\frac{d y}{d x}=\frac{1-x y}{x^{3}}=\frac{1}{x^{3}}-\frac{y}{x^{2}}$

$\frac{d y}{d x}+\frac{y}{x^{2}}=\frac{1}{x^{3}}$

IF $=e^{\int \frac{1}{x^{2}} d x}=e^{-\frac{1}{x}}$

$y \cdot e^{-\frac{1}{x}}=\int e^{-\frac{1}{x}} \cdot \frac{1}{x^{3}} d x(.$ put $.-\frac{1}{x}=t)$

$y . e^{-\frac{1}{x}}=-\int e^{t} \cdot t d t$

$y=\frac{1}{x}+1+Ce^{\frac{1}{x}}$

Where $C$ is constant

Put $x=\frac{1}{2}$

$3-e=2+1+Ce^{2}$

$C=-\frac{1}{e}$

$y(1)=1$