If `y_(1)(x)` is a solution of the differential equation `(dy)/(dx)-f(x)y = 0`, then a solution of the differential equation `(dy)/(dx) + f(x) y = r(x)` is

Options

A. `y = (1)/(y_(1)(x))int r(x) y_(1)(x) dx +(c)/(y_(1)(x))`
B. `y = y_(1)(x) int (r(x))/(y_(1)(x))dx + c`
C. `y = int r(x) y_(1)(x) dx + c`
D. None of these

Answer

A

Explanation

`(dy)/(dx)-f(x)*y=0`
`therefore" "(dy)/(y) = f(x) dx`
`therefore" "In y = int f(x) dx`
`therefore" "y_(1)(x) = e^(int f(x) dx)`
Then for given equation, I.F `= e^(int f(x) dx)`
Hence, solution is `y*y_(1)(x) = int r(x)*y_(1)(x) dx + c`
`y = (1)/(y_(1)(x)) int r(x) * y_(1)(x)dx + (c)/(y_(1)(x))`