The family of curves passing through `(0,0)` and satisfying the differential equation `y_2/y_1=1` `("where", y_n=(d^ny)/dx^n)` is (A) `y=k` (B) `y=kx` (C) `y= k(e^x +1)` (C) `y= k(e^x -1)`

Options

A. y = k
B. y = kx
C. `y = k(e^(x) + 1)`
D. `y = k(e^(x)-1)`

Answer

D

Explanation

`(y_(2))/(y_(1))=1`
`therefore" "(dp)/(dx)=p("where p"=(dy)/(dx))`
`therefore" ""In p" = x + c`
`therefore" "p = e^(x+c)`
`therefore" "(dy)/(dx)=ke^(x)`
`therefore" "y = ke^(x) + lambda`
Satisfying (0, 0), So `lambda = -k`
`therefore" "y = k(e^(x) - 1)`