# Wronskian problem

1. Feb 28, 2005

### Gale

$$w[f,g](t)= t^2\exp{t}\\f(t)=t$$

Thats what i get, the problem is to find g(t)

So, i start; f'(t)=1

$$w[f,g](t)= t^2\exp{t}=f(t)g'(t)-f'(t)g(t)\\t^2\exp{t}=tg'(t)-g(t)$$

divide by t,

$$t\exp{t}=g'(t)-\frac{g(t)}{t}$$

its a 1st order linear eq. I solve for the integrating factor and get t. i multiply through and reduce

$$(tg(t))'=t^2\exp{t}$$

then i integrate with the product rule and get

$$tg(t)=t^2\exp{t}+2t\exp{t}+C$$

divide by t and get

$$g(t)=t\exp{t}+2\exp{t}+\frac{C}{t}$$

which is wrong. The answer in the book is

[tex}t\exp{t}+Ct[/tex]

not sure where i went wrong, i know its probably something dumb, but its late, so i need help.

~gale~

2. Feb 28, 2005

### Hurkyl

Staff Emeritus
I don't like your integrating factor. When you expand out (t g(t))', you don't get the right thing.

3. Feb 28, 2005

### Gale

mk, well, the int factor is

$$e^{\int{1/t}dt}{$$

right? so the exponential and log cancel and t is all that's left... how's that wrong?

( i can't get the latex right on that, hope you get what i mean)
(also, i accidently posted twice, you can delete the other one... i dunno how)

Last edited: Feb 28, 2005
4. Feb 28, 2005

### Gale

oh i just realized, its supposed to be

$$e^{-\int{\frac{1}{t}dt}$$

i forgot about that negative sign.
which makes my int factor 1/t which changes absolutely everything and makes the problem right.... AUGH, DAMNED NEGATIVES.... grr. thanks...

5. Mar 1, 2005

### HallsofIvy

I really wish people would not post the same thing twice! I just posted a reply to the OTHER "Wronskian problem" before I saw that it had already been settled here.

6. Mar 1, 2005

### Gale

i said sorry... i didn't mean to, and i couldn't delete it....

7. Mar 1, 2005

### dextercioby

I hope you've got the point.No double posting...You should have PM-ed a (preferably online) mentor/admin.He would have deleted the thread.

Daniel.