# Diff EQ, did i mess up on a natural log rule?wee!

1. Jan 18, 2006

### mr_coffee

Hello everyone the problem im messing up is:
. Solve the following initial value problem:

When pluggin in answers and using my calculator to evaluate, its saying NON-REAL numbers, i'm messing up hardcore somewhere but not sure where anyone know what rules i'm breaking, i'm sure its a few!
The answer I submitted was the following:

Which was wrong! wee!

2. Jan 18, 2006

### Benny

I haven't checked your partial fraction decomposition but your working (apart from the exponentiation of the sum of logs) looks right.

It would be much easier to just write A = exp(c). That way you can avoid taking logs and the expressions will be neater. It also allows you to 'fudge' an answer. Also, when you look logs of both sides to solve for c, you can't just insert a modulus sign just because it's convenient.

If you write the DE in the form $\frac{{dy}}{{dx}} = f\left( {x,y} \right)$ you will find $f\left( {x,y} \right) = \frac{y}{{t^2 - 14t + 13}}$ where $\frac{{\partial f}}{{\partial y}} = \frac{1}{{t^2 - 14t + 13}}$ is not continous at (x,y) = (1,7). So you are not guaranteed to obtain a unique solution to the IVP. This question looks a little strange. Hopefully someone with more knowledge in this area can help you out.

Edit: $$\exp \left( {\frac{1}{{12}}\log \left| {t - 13} \right| - \frac{1}{2}\log \left| {t - 1} \right| + c} \right)$$

$$= \exp \left( {\log \left( {\frac{{\left| {t - 13} \right|^{\frac{1}{{12}}} }}{{\sqrt {\left| {t - 1} \right|} }}} \right) + c} \right)$$

$$= \exp \left( c \right)\exp \left( {\log \left( {\frac{{\left| {t - 13} \right|^{\frac{1}{{12}}} }}{{\sqrt {\left| {t - 1} \right|} }}} \right)} \right)$$

$$= \exp \left( c \right)\left( {\frac{{\left( {t - 13} \right)^{\frac{1}{{12}}} }}{{\sqrt {t - 1} }}} \right)$$

Last edited: Jan 18, 2006
3. Jan 18, 2006

### HallsofIvy

Staff Emeritus
Did you notice you had ln|y|= ... and on the next line y= ...

What happened to the absolute value? Here's a suggestion: instead of eC, just write C' (actually, I would just replace eC with C reminding myself that the two "C"s represent different values).
Since C is an unknown constant, eC is also an unknown constant. Technically, eC must be positive. But allowing C' to take any value allows for the absolute value.

Also you do have an error in your partial fractions decomposition: if
$$\frac{1}{t^2- 14t+ 13}= \frac{A}{t-13}+ \frac{B}{t-1}$$
then A= $\frac{1}{12}$ and B= $-\frac{1}{12}$.

That means that your general solution should be
$$ln|y|= \frac{1}{12}ln(t- 13)- \frac{1/12}ln(t-1)+ C= ln\left(\frac{t-13}{t-1}\right)^{\frac{1}{12}}+C$$
so that
$$y= C'\left(\frac{t-13}{t-1}\right)^\frac{1}{12}$$
Now put x= 7 and y= 1.

4. Jan 21, 2006

### mr_coffee

hey thanks guys, sorry it took so long for me to reply. I tried your method Ivey but i got:

I solved for C', and ended up getting C' = 1. But I had to fudge some answers, becuase i ended up with:
1 = C'(-1)^(1/12) which isn't allowed.
so i assumed
1 = C'|-1|^(1/12), in the end i screwed up because it says its inncorrect. This is what I did:
y = C'((t-13)/(t-1))^(1/12);
1 = C' ((7-13)/(7-1))^(1/12);
1 = C'(-1)^(1/12);

I plugged in Y = 1, and t = 7, any ideas? Thanks!

5. Jan 21, 2006

### HallsofIvy

Staff Emeritus
Looks like you have done the partial fractions correctly.

But don't forget that
$$\int \frac{dx}{x}= ln\left|x\right|+ C$$
NOT ln(x)!!

The integral, then is
$$\int \frac{dy}{y}= \frac{1}{12}\left(\int\frac{dt}{t-13}-\frac{dt}{t-1}\right)$$
so
$$ln|y|= \frac{1}{12}\left(ln|t-13|- ln|t-1|\right)+ C= \frac{1}{12}ln\left|\frac{t-13}{t-1}\right|+ C$$
(don't forget the absolute value signs)
Taking exp of both sides:
$$y= C'\left|\frac{t-13}{t-1}\right|^{\frac{1}{12}}$$.

When x= 7, y= 1 so
$$1= C'\left|\frac{-6}{6}\right|^{\frac{1}{12}}= C'$$

Last edited: Jan 21, 2006
6. Jan 21, 2006

### mr_coffee

thanks for the explanation Ivey, so the answer would be, y = |(t-13)/(t-1)|^(1/12) right? i submitted that and it was wrong or am i misunderstanding somthing? I do that alot hah.

7. Jan 22, 2006

### HallsofIvy

Staff Emeritus
You are submitting that to an auto-site? I hate those things- the don't distinguish alternate ways of writing the same answer!

The initial value is at x= 7 which is between 1 and 13. For all x between 7 and 13, $\frac{t-13}{t-1}$ is negative so you might try $y= -\frac{t-13}{t-1}$ or $\frac{13-t}{t-1}$ or $\frac{t-13}{1-t}$.