Question

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

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

Answer 1

Answer: (A)

$-\frac{1}{2}$

Given, $(1+t)\frac{dy}{dt}-ty=1$

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