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If y = f(x) is the solution of the diffe...

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:

A

1

B

`-7`

C

7

D

`-7`

Text Solution

Verified by Experts

The correct Answer is:
C
`(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) implies - (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 implies - int (e^(-t) dt)/(3 + e^(-t)) - int 1dx`
`implies 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) = 4 e^(3x) implies y = log (4e^(3x) - 3e^(2x))^(1//3) implies A + B = 7`
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