# Differential equation IVP

1. Sep 29, 2011

### fishingspree2

1. The problem statement, all variables and given/known data
Find a particular solution to this IVP:

dy/dx = 1 - 2y
y(0) = 5/2

2. The attempt at a solution
I find -0.5 *ln (1- 2y) = x + C

However, y = 5/2 gives ln(-4), which is a problem....Have I done something wrong? Any suggestions please? How to find a family of solutions that would be defined for the given point? Thank you.

2. Sep 29, 2011

### HallsofIvy

The integral of 1/x is ln|x|, not ln(x)!

Often the distinction is not important but this is one problem where it is crucial.

3. Sep 29, 2011

### dynamicsolo

If we look at the differential equation, y' = 1 - 2y , we see that y' = 0 for y = 1/2 . This is what is sometimes called a "stationary solution" (also referred to as the "trivial" solution). If the initial value for y (for any time choice) were 1/2 , the value of y would remain at 1/2 forever. For any initial value y(0) < 1/2 , the function for y will have y' > 0 , with y' becoming smaller as y increases toward 1/2 . By the same token, for any initial value y(0) > 1/2 , y' < 0 , meaning that y will decline toward 1/2 , and at ever slower rates as y approaches that value.

This results in two cases for the IVP then:

for y(0) < 1/2 , the integration gives -0.5 ln ( 1 - 2y ) = x + C , which leads to an acceptable value for C because ( 1 - 2y ) is always positive;

but for y(0) > 1/2 , the integration gives -0.5 ln ( 2y - 1 ) = x + C , which makes the argument of the logarithm positive and maintains the values y will pass through within the domain of the function (and so will give a sensible value for C).

This is what is contained in the standard result for this problem, y = -0.5 ln | 1 - 2y | = x + C : the presence of the absolute value signals that there are two cases to consider (sadly, a matter often inadequately covered in introductory DE courses).