If y(t) is the solution of the differential equation (1+t)dydt−yt=1 and y(0)=−1 then y(1) is
−12
e+12
e-12
12
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