If y(t) is a solution of (1+t) d y d t -t y=1 and y(0)=-1 , then y(1) equal to

Step-by-step solution

Here, d y d t - ( t 1+t ) y= 1 1+t and y(0)=-1 which represents linear differential equation of first order. aligned IF & =e^ - ( t 1+t ) d t =e^ - (t+1-1) 1+t d t & =e^ -t+ (1+t) & =e^ -t (1+t) aligned Required solution is aligned y( IF ) & = [ Q ( IF ) d t ]+C y e^ -t (1+t) & = 1 1+t e^ -t (1+t) d t+C & = e^ -t d t+C=-e^ -t +C aligned At t=0, y=-1 aligned & & -1(1+0) & =-e^0+C & & C & =0 & & y e^ -t (1+t) & =-e^ -t aligned At t=1 aligned y e⁻¹(1+1) & =-e⁻¹ y & =- 1 2 aligned