Question

If $y(t)$ is a solution of $(1+t)\frac{dy}{dt}-ty=1$ and $y(0)=-1,$ then $y(1)$ is equal to

• A. $-1/2$
• B. $e+1/2$
• C. $e-1/2$
• D. $1/2$

Answer 1

Answer: (A)

Given, $\frac{dy}{dt}-\left(\frac{t}{1+t}\right)y=\frac{1}{(1+t)}$ and $y(0)=-1$
Which represents linear differential equation of first order.
$\therefore$ IF $=e^{\int -\left(\frac{t}{1+t}\right)dt}=e^{-t+\log (1+t)}=e^{-t}\cdot (1+t)$
Required solution is,
$y{e}^{-t}(1+t)=\int \frac{1}{1+t}\cdot e^{-t}(1+t)dt+c=\int e^{-t}dt+c$
$\Rightarrow y{e}^{-t}(1+t)=-e^{-t}+c$
Since, $y(0)=-1$
$\Rightarrow -1\cdot e^0(1+0)=-e^0+c$
$c=0$
$\therefore y=-\frac{1}{(1+t)}\Rightarrow y(1)=-\frac{1}{2}$