Question

If Y is the solution of $(1+t)\frac{dy}{dt}-ty=1$ and Y(0) = -1 then y(1) 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)

$\frac{-1}{2}$

$(1+t)\frac{dy}{dt}-ty=1$
$\frac{dy}{dt}\frac{t}{1+t}=\frac{1}{1+t}$
If $e^{-}i\frac{t}{1+t}dt=e^{-}t+/n(t+1)=(t+1)e^{-}t$
$y(t+1)e^{-}t=|\frac{1}{t+1}(t+1)e^{-}tdt+C$
$Y(t+1)e^{-}t=-e^{-}t+C$
at$t=0,Y=-1C=0$
$Y=-\frac{1}{t+1}\Rightarrow Y\left(1\right)=\frac{-1}{2}$