Question

If $y\left(x\right)$ is a solution of $\frac{d y}{d x}-\frac{x y}{1 \, + \, x}=\frac{1}{1 \, + \, x}$ and $y\left(0\right)=-1$ , then the value of $y\left(2\right)$ is

• A. $-\frac{1}{2}$
• B. $-\frac{1}{3}$
• C. $-\frac{1}{4}$
• D. $-\frac{1}{5}$

Answer 1

Answer: (B)

I.F. $=e^{- \int \frac{x}{1 + x} d x}=e^{- \int \frac{x + 1 - 1}{x + 1} d x}=e^{- x + ln \left|\left(x + 1\right)\right|}$
$= \, e^{- x}\cdot \left|\right.\left(\right.x+1\left.\right)\left|\right.$
$y\cdot \left|x + 1\right|e^{- x}= \int \frac{\left|x + 1\right|}{x + 1}e^{- x}dx+c$
$y\cdot \left|\right.\left(\right.x+1\left.\right)\left|\right.e^{- x}=-\frac{\left|\right. \, x + 1 \, \left|\right.}{x + 1}e^{- x}+c$
$\Rightarrow \, y=-\frac{1}{x + 1}+\frac{c e^{x}}{\left|\right. \, x + 1 \, \left|\right.}$
$x=0, \, y=-1\Rightarrow \, -1=-1+c\Rightarrow \, c=0$
$\Rightarrow \, \, \, y=-\frac{1}{1 + x}$
$y\left(2\right)=-\frac{1}{3}$