Given equation is `(dy)/(dx) - y= e^(x)`
The equation is of the type
`(dy)/(dx) +Py= Q`
where `" "P = - 1 and Q = e^(x)`
`therefore" "I.F = e^(int - 1dx) dx = e^(-x) = (1)/(e^(x))`
Multiplying both side of the differential equation by I.F., we get
`(1)/(e^(x)).[(dy)/(dx) - y] = (1)/(e^(x)).e^(x)`
`(1)/(e^(x)) (dy)/(dx) - (y)/(e^(x)) = 1`
`(d)/(dx) ((y)/(e^(x))) = 1`
Integrating both sides w.r.t.x, we get
`(y)/(e^(x)) = int 1 dx + c`
`(y)/(e^(x)) = x + c`
`ye^(-x) = x + c" "........(1)`
Which is the general soultion of the given equation.
Putting x = 0 and y = 1 in equations (1) , we get
`1 = 0 + c`
` c = 1`
Putting c = 1 in equations (1) , we get
`ye^(-x) = x + 1`
which is the particular solution.