If y(t) is the solution of the differential equation (1+t)dydt−yt=1 and y(0)=−1 then y(1) is

dydt−tt+1y=1t+1I⋅F=e−∫tt+1dt=e−∫1-1t+1dt=e-t+ln(t+1) =e−t⋅(1+t)the equation is ye−t⋅(1+t)=∫e−t (1+t)11+tdt+c ye−t⋅(1+t)=∫e−tdt ye−t⋅(1+t)=-e−t+c⇒put t=0 y(0)=−1 (-1)e0(1+0)= -e0+c ⇒ c=0⇒put t=1 y(1)e-1·2=-e-1y(1)=−12

If y(t) is the solution of the differential equation (1+t)dydt−yt=1 and y(0)=−1 then y(1) is

dydt−tt+1y=1t+1I⋅F=e−∫tt+1dt=e−∫1-1t+1dt=e-t+ln(t+1) =e−t⋅(1+t)the equation is ye−t⋅(1+t)=∫e−t (1+t)11+tdt+c ye−t⋅(1+t)=∫e−tdt ye−t⋅(1+t)=-e−t+c⇒put t=0 y(0)=−1 (-1)e0(1+0)= -e0+c ⇒ c=0⇒put t=1 y(1)e-1·2=-e-1y(1)=−12

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

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$- \frac{1}{2}$

$e + \frac{1}{2}$

$e- \frac{1}{2}$

$\frac{1}{2}$

answer is A.

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Detailed Solution

$\begin{array}{l} \frac{dy}{dt} - \frac{t}{t + 1} y = \frac{1}{t + 1} \\ I \cdot F = e^{- \int \frac{t}{t + 1} dt} = e^{- \int (1 - \frac{1}{t + 1}) dt} \\ = e^{- t + \ln (t + 1)} = e^{- t} \cdot (1 + t) \\ t h e e q u a t i o n i s {ye}^{- t} \cdot (1 + t) = \int e^{- t} (1 + t) \frac{1}{1 + t} dt + c \end{array}$

${ye}^{- t} \cdot (1 + t) = \int e^{- t} dt$

${ye}^{- t} \cdot (1 + t) = - e^{- t} + c$

$\begin{array}{l} \Rightarrow put t = 0 y (0) = - 1 \\ (- 1) e^{0} (1 + 0) = - e^{0} + c \Rightarrow c = 0 \\ \Rightarrow p u t t = 1 \\ y (1) e^{- 1} \cdot 2 = - e^{- 1} \\ y (1) = - \frac{1}{2} \end{array}$