# Solving an IVP using a direction field I think

1. Feb 19, 2014

### castrodisastro

1. The problem statement, all variables and given/known data
Instructions: You will need to use software for these problems. The goal is to find a number x0 such that the solution to the initial value problem x'=ƒ(t,x) x(0)=x0 passes through the given "target". Attach figures that you used to answer the problems.

1. Consider the differential equation x'=x-t
(a) Find x0 such that the solution of the initial value problem x'=x-t, x(0)-x0 will pass through (4,0).

(b) Find x0 such that the solution of the initial value problem x'=x-t, x(0)-x0 will pass through (6,0).

2. Relevant equations

3. The attempt at a solution

So there are many places here where I "freeze up" before I really attempt to solve this problem. One is because I don't even know how to use the software they want us to use. I think it's something similar to matlab but this is only to visualize direction fields. The link we were given was

http://math.rice.edu/~dfield/dfpp.html

I went there but I wasn't sure how to get it to look like anything other than the default. I wrote the equation x'=x-t into the field for differential equation and hit graph but nothing happened.

I just want to double check, I am not supposed to start solving it by hand right? I tried to...

x'=x-t

subtract x to the other side since it is a linear equation, no higher powers involved.

x'-x=-t

Using the integrating factor e∫-1dt=e-t

now if i multiply everything by e-t

e-tx'-e-tx=-te-t

but the second term, the one containing the x to the power of one, does not contain any t term that I can use with x and x' so that I can use the product rule in reverse. This leads me to think it is not a linear equation? which then leads me back to my first question...am I supposed to not do anything by hand?

When I pull up the "DFIELD" applet there is a field for me to put my differential equation in but it is set to take in an x' by default so should I assume I should just put it into the field?

I am still not comfortable enough to look at these problems and visualize what it asks. I don't feel like I understand what goes on in every case. It's like I am writing an essay, and I am choosing to include long and sophisticated words that while I may know their definition if singled out of context, I have no idea when and where to use it.

Any help is appreciated.

2. Feb 20, 2014

### HallsofIvy

Staff Emeritus
When t= 4 and x= 0, x'= x- t= 0- 4= -4. That is the initial value of x' you need.

Similarly, for the second problem, with t= 6 and x= 0, x'= 0- 6= -6.