Question

Find the solution of `"dy"/"dx"` = 2^y–x.

Answer 1

The given differential equation is

`"dy"/"dx"` = 2^y–x 

⇒ `"dy"/"dx" = 2^y/2^x`

Separating the variables, we get

`"dy"/2^y = "dx"/2^x`

⇒ `2^-y "d"y = 2^-x "d"x`

Integrating both sides, we get

`int 2^-y "d"y = int 2^-x "d"x`

`(-2^-y)/log2 = (-2^-x)/log2 + "c"`

⇒ `-2^-y = -2^-x + "c" log 2`

⇒ `-2^-y + 2^-x = "c" log 2`

⇒ `2^-x - 2^-y` = k  .....[Where c log 2 = k]