If y = f(x) is the solution of the differential equaiton `e^(3y) ((dy)/(dx) - 1) = e^(2x)` and y(0) = 0 then `y(x) = log (Ae^(3x) - Be^(2x))^((1)/(3))` where the value of (A + B) is:

Options

A. 1
B. `-1`
C. 7
D. `-7`

Answer

C

Explanation

`(dy)/(dx) = e^(2x-3y) + 1`
Put `2x - 3y = t`
Differentiating both sides wr.t. x
` 2-3 (dy)/(dx) = (dt)/(dx)`
`(dy)/(dx) = (1)/(3)((dt)/(dx) -2) rArr -(1)/(3) (dt)/(dx) + (2)/(3) + e^(t+1)`
`(dt)/(dx) - 2 = - 3e^(t) - 3 , " " (dt)/(dx) = - 3e^(t) -1`
`int (dt)/(-3e^(t) -1) = int dx " " rArr - int (e^(-t)dt)/(3+e^(-t)) = int 1dx`
`rArr log (3+e^(-t)) = x + C, " " log (3+e^(3y-2x)) = x+ c`
Given at x = 0 , y = 0 , log 4= c
`log (3 + e^(3y - 2x)) - log 4 = x , " " (3+e^(3y - 2x))/(4) = e^(x)`
` 3e^(2x) + e^(3y) = 4e^(3x) rArr y = lo g(4e^(3x) - 3e^(2x))^(1//3) rArr A + B = 7`