Please help me regarding on this maths?

Question

Given that y is always positive and y(0) = 2. By direct integration method, find the particular solution of the differential equation.

dy/dx = y(1-y)/y+1

many thanks!

Answer 1

Yes, jgmb, technically the author used an ambiguous expression, but I think we know what was meant. To the author: I'm ready for bed, but I'll give this a shot, and sorry if I foul it up. Please look over the steps and make sure that I didn't.

dy/dx = y (1-y) / (y+1)

Otherwise, the y in the numerator cancels out the y in the denominator. OK, this becomes

( (y+1) / (y(1-y))) dy / dx = 1

and we're going to integrate both sides with respect to dx. Looks like a mess on the left side, but we can clean it up a little with partial fractions. Looking at the quadratic in the denominator:

1 / (y(1-y)) = a/y + b/(1-y)

1 = a(1-y) + by
1 = a + (b-a)y

1+0y = a + (b-a)y

so: a=1, b-a=0, so b=a=1

and 1/(y(1-y)) = 1/y + 1/(y-1)

Our factor to the left of dy/dx now becomes

(y+1)/y + (y+1)/(y-1)

= (y+1)/y + ((y-1)+2)/(y-1)

= 1 + 1/y + 1 + 2/(y-1)

= 2 + 1/y + 2/(y-1)

y being considered a function of x, what we are left with is the integral with respect to x of

(2 + 1/y + 2/(y-1)) dy / dx

Courtesy of the chain rule and the fundamental theorem of calculus, this should give us

2y + ln (y) + 2 ln (y-1) + C1

where C1 is some indeterminate constant, the right side integrating out to x plus another indeterminate constant C2, yielding the general solution:

2y+ ln (y) + 2 ln (y-1) = x + C

where C = C2 - C1 (difference between two indeterminate constants, giving us a third indeterminate constant).

I'm fairly sure that this is as good as you can do, in deriving the general solution. This looks like a transcendental equation, meaning that the only solution that you're likely to see for it given a particular value of x is a numerical analytical (approximate) one.

Calculate the specific value of the indeterminate constant that came out of our integration to obtain the specific solution corresponding to our intial condition: y(0)=2, so

2y(0) + ln y(0) + 2 ln (y(0) -1) = 0 + C
2*2 + ln 2 + 2 ln (2 -1 ) = C
4 + ln 2 + 2 ln 1 = C
4 + ln 2 = C since ln 1 = 0

Thus, we get

2y+ ln (y) + 2 ln (y-1) = x + 4 + ln 2

Answer 2

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Answer 3

y(0) = 2
y(t) > 0 for t>0

y' = y(1-y)/y + 1= (1-y) + 1 = 2 - y

arghhhh why dont you put ( ) if the order of operations is unclear, i dont expect that this is what you want but i dont know either what you want ...