Question

If $y(t)$ is solution of $(t+1) \frac{d y}{d t}-t y=1, y(0)=-1$, then $y(1)=$

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

Answer 1

Answer: (C)

$\frac{d y}{d t}-\frac{t}{t+1} y=\frac{1}{t+1}$
$I . F=e^{-\int \frac{t+1-1}{t+1} d t}=e^{-t+t(t+1)}=(t+1) e^{-t}$
solution is $(t+1) e^{-t} y=-e^{-t}+c $
put $t=0$ and $y=-1 \Rightarrow c=0$
$\therefore 2 e^{-1} y=-e^{-1} $ put $t=1$
$ y=-\frac{1}{2}$