Anonymous

# 1/5 ln |y-2|- 1/5 ln |y+3| = x+c .... find y?

By using the method of separation of variables, find the general solution of the following initial value equations:

y'=y^2+y-6, y(5)=10

I reached this step :

1/5 ln |y-2|- 1/5 ln |y+3| = x+c

I still don't know how can I arrange the equation to make LHS of the equation equals to Y so that I'll be able to find the constant c, hence the initial value...

Anyone thinks he's good :-P

Thanks before hand :D

### 1 Answer

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- Anonymous8 years agoFavorite Answer
(1/5)*ln[abs(y - 2)] - (1/5)*ln[abs(y + 3)] = x + c

I couldn't get a solution closed form but I got this:

for Y=< -3 or Y => 2

Y = {2 + 3*e^[5*(x + c)]}/{1 - e^[5*(x + c)]}

for -3 =< Y =< 2

Y = {2 - 3*e^[5*(x + c)]}/{1 + e^[5*(x + c)]}

Maybe this can be useful. Good luck on finding 'c' !

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