# Initial Value Problem

• Aug 24th 2005, 05:03 AM
mactev
Initial Value Problem
abcdefg
• Aug 25th 2005, 02:01 AM
ticbol
Explicit form?

Is it like this?

dy/dx = 1 -4y^2
dy/(1 -4y^2) = dx
dy/[1^2 -(2y)^2] = dx
[(1/2)(2dy)]/[1^2 -(2y)^2] = dx --------(i)

Using the Table, where u^2 is less than a^2,
INT.[du / (a^2 -u^2)] = [1/(2a)]*ln{(a+u)/(a-u)} -------(ii)

The left side of (i) falls into (ii) because
>>>maximum (2y)^2 is (2 * 1/2)^2 or 1^2. But (y = -1/2) and (y = 1/2) are not part of the interval of y, so maximum (2y)^2 is a little less than 1^2.
>>>a^2 is 1^2, so u^2 is less than a^2.

Integrate both sides of (i),
(1/2)[1/(2*1)]*ln{(1 +2y)/(1 -2y)} +C = x
(1/4)ln{(1 +2y)/(1 -2y)} +C = x
Clear the fraction, multiply both sides by 4,
ln{(1 +2y)/(1 -2y)} +4C = 4x -------(iii)

We can find C, or 4C, when y = 1/4 and x=0.
ln{(1 +2(1/4))/(1 -2(1/4))} +4C = 4(0)
ln{(1.5)/(0.5)} +4C = 0
ln(3) +4C = 0
4C = -ln(3) ----***

Substitute that into (iii)
ln{(1 +2y)/(1 -2y)} -ln(3) = 4x
Combine the two ln's,
ln{(1 +2y)/(1 -2y) / 3} = 4x
ln{(1/3)(1 +2y)/(1 -2y)} = 4x
Antilogs,
(1/3)(1 +2y)/(1 -2y) = e^(4x)
Multiply both sides by 3(1 -2y),
1 +2y = 3(1 -2y)e^(4x)
1 +2y = 3e^(4x) -2y(3e^(4x))
1 +2y +2y(3e^(4x)) = 3e^(4x)
1 +2y(1 +3e^(4x)) = 3e^(4x)
2y(1 +3e^(4x)) = 3e^(4x) -1
y = (1/2)[(3e^(4x) -1) / (3e^(4x) +1)] --------answer.

Check, when y=1/4 and x=0,
1/4 =? (1/2)[(3e^0 -1)/(3e^0 +1)]
1/4 =? (1/2)[(3-1)/(3+1)]
1/4 =? (1/2)[2/4]
1/4 =? 1/4
Yes, so, OK.
• Aug 25th 2005, 05:56 AM
mactev
That makes sense. Explicit form just means y in terms of x (ie the conventional way) as oppose to implicit form.

However, if i took C to be on the left hand side of the differential equation, that is 1/4 ln (1+2y/1-2y)=x+C, would the value of the constant not then be ln(3) rather than negative ln(3)?

Or does it not matter which side the constant is added to since it is absorbed into the equation?
• Aug 25th 2005, 06:07 AM
ticbol
It doesn't matter.
As you know, constant is constant, whether it is positive, negative, C1, C2, C, 4C, C+24, 53-2C, etc...
In fact, there are supposed to be two constants after integrating both sides of (i).

Integrate both sides of (i),
(1/2)[1/(2*1)]*ln{(1 +2y)/(1 -2y)} +C1 = x +C2
But [C1 -C2] is still a constant. I wrote it as C only on the left side.
Then I solved for 4C, still a constant.
• Aug 25th 2005, 07:49 AM
mactev
I thought as much.

Chris
• Aug 26th 2005, 12:31 AM
ticbol
"However, if i took C to be on the left hand side (you mean righthand side) of the differential equation, that is 1/4 ln (1+2y/1-2y)=x+C, would the value of the constant not then be ln(3) rather than negative ln(3)?
Or does it not matter which side the constant is added to since it is absorbed into the equation?"

(1/4)ln{(1 +2y)/(1 -2y)} = x +C --------(iv)
When y=1/4 and x=0,
(1/4)ln(3) = 0 +C
C = (1/4)ln(3) ------***
Substitute that into (iv),
(1/4)ln{(1 +2y)/(1 -2y)} = x +(1/4)ln(3)
Multiply both sides by 4,
ln{(1 +2y)/(1 -2y)} = 4x +ln(3)
Collect the ln's,
ln{(1 +2y)/(1 -2y)} -ln(3) = 4x ----***

That is exactly the same as the one when C was put at the left side in the beginning.

Therefore,
>>>No, C is positive (1/4)ln(3). [In my answer, 4C = -ln(3). Not C. You said "constant", so it should be the 4C. Still, that constant is positive.]
>>>Yes, it doesn't matter where you put your C after integration. The positive constant ln(3) became negative when transposed to the lefthand side.
• Aug 26th 2005, 02:39 AM
mactev
So infact a constant would be added to the left hand side, b say, and one to the right, a say, thus giving an overall constant b-a=C say. The truism is that this constant C is absorbed back into the differential equation and thus does not matter which side it is placed.

Is that right?
• Aug 26th 2005, 02:59 AM
ticbol
Yes.

(I was not allowed to post the "Yes" only, so this.)