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

a
−12
b
e+12
c
e-12
d
12

detailed solution

Correct option is A

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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