If y(t) is solution of (t+1) (d y/d t)-t y=1, y(0)=-1, then y(1)=

Question

If y(t) is solution of (t+1) (d y/d t)-t y=1, y(0)=-1, then y(1)= (A) (1/4) (B) -2 (C)  -(1/2) (D) (1/2)

Tardigrade Admin · Accepted Answer

(d y/d t)-(t/t+1) y=(1/t+1) I . F=e-∫ (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 ⇒ c=0 ∴ 2 e-1 y=-e-1 put t=1 y=-(1/2)

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

$\frac{1}{4}$

$- 2$

$- \frac{1}{2}$

$\frac{1}{2}$

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

Q. If y(t) is solution of (t+1)dtdy​−ty=1,y(0)=−1, then y(1)=

41​

−2

−21​

21​