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Q.
If $y (t)$ is a solution of $(1+t) \frac {dy}{dt}-ty=1 $ and $ y(0)=-1, $ then $y(1)$ is equal to
IIT JEEIIT JEE 2003Differential Equations
Solution:
Given, $\frac{d y}{d t}-\left(\frac{t}{1+t}\right) y=\frac{1}{(1+t)}$ and $y(0)=-1$
Which represents linear differential equation of first order.
$\therefore $ IF $=e^{\int-\left(\frac{t}{1+t}\right) d t}=e^{-t+\log (1+t)}=e^{-t} \cdot(1+t)$
Required solution is,
$y e^{-t}(1+t)=\int \frac{1}{1+t} \cdot e^{-t}(1+t) d t+c=\int e^{-t} d t+c$
$\Rightarrow y e^{-t}(1+t)=-e^{-t}+c$
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}$