Find the particular solution of the differential equation ` x(dy)/(dx)

1.	Find the particular solution of the differential equation ` x(dy)/(dx) - y = (x + 1 ) e^(-x) ` given that `y = 0 ` when ` x = 1 `.
Answer» The given differential equation may be written as ` (dy)/(dx) - (1)/(x) * y = ((x + 1) e ^(-x))/( x )`...(i) This is of the form ` (dy)/(dx) + Py = Q`, where `P = (-1)/( x ) and Q = ((x + 1 ) e ^(-x))/( x )` Thus, the given differential equation is linear. `IF = e ^(int Pdx) = e ^(-int (1)/(x) dx)= e ^(log (1//x)) = (1)/(x)` So, its solution is given by `y xx IF = int (Q xx IF ) dx + C`, i.e., `y xx (1)/(x) = int ((x + 1 ) ^(-x))/( x ^(2)) dx + C `. `rArr y xx (1)/(x) = int underset ("I")( (1)/(x)) underset("II") e ^(-x) dx = int (1)/(x ^(2)) e ^(-x) dx + C ` `rArr y xx (1)/(x) = (1)/(x) * (- e ^(-x)) + int (1)/(x ^(2) (-e^(-x)) dx) + int (1)/(x ^(2)) e ^(-x) dx + C` `rArr (y)/(x)= (-e^(-x))/( x )+C` `rArr y = - e ^(-x) +Cx" " `...(iii) It is being given that when ` x = 1, `then `y =0 `. Putting `x = 1 and y = 0` in (ii) , we get `C = e ^(-1)`. `therefore y = - e ^(-x) + xe^(-1) ` is the required solution.

Find the particular solution of the differential equation ` x(dy)/(dx) - y = (x + 1 ) e^(-x) ` given that `y = 0 ` when ` x = 1 `.