Question

If $y\left(x\right)$ is a solution of $\frac{dy}{dx}-\frac{xy}{1+x}=\frac{1}{1+x}$ 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. $=e^{-\int \frac{x}{1+x}dx}=e^{-\int \frac{x+1-1}{x+1}dx}=e^{-x+ln|\left(x+1\right)|}$
$=e^{-x}\cdot |(x+1)|$
$y\cdot \left|x+1\right|e^{-x}=\int \frac{|x+1|}{x+1}{e}^{-x}dx+c$
$y\cdot |(x+1)|e^{-x}=-\frac{|x+1|}{x+1}{e}^{-x}+c$
$\Rightarrow y=-\frac{1}{x+1}+\frac{c{e}^x}{|x+1|}$
$x=0,y=-1\Rightarrow -1=-1+c\Rightarrow c=0$
$\Rightarrow y=-\frac{1}{1+x}$
$y\left(2\right)=-\frac{1}{3}$