# Thread: 2nd order of differential equations

1. laplace transform can be converted using this formula right?

$\displaystyle {e}^{at}\leftrightarrow \frac{1}{(s+a)}$
that is where I got $\displaystyle {e}^{-10s}= \frac{1}{s+10}$

2. Originally Posted by jenjen
laplace transform can be converted using this formula right?
$\displaystyle {e}^{at}\leftrightarrow \frac{1}{(s+a)}$

that is where I got $\displaystyle {e}^{-10s}= \frac{1}{s+10}$
You mean,
$\displaystyle \mathcal{L}\{e^{at}\}=\frac{1}{s-a}$
You mistake is,
$\displaystyle \frac{1}{s+a}$

3. yeah that is correct. THANK YOU. so then how do i suppose to approach this problem altogether? do i have to use partial expansion?

4. Originally Posted by ThePerfectHacker
You mean,
$\displaystyle \mathcal{L}\{e^{at}\}=\frac{1}{s-a}$
You mistake is,
$\displaystyle \frac{1}{s+a}$
Except we are in the $\displaystyle s$ domain, we want $\displaystyle \mathcal{L}^{-1} e^{-10s}$.

RonL

5. so if that is true, then do we use partial fraction expansion to solve the problem? I am still confused about why we don't use partial fractions. because since
$\displaystyle {e}^{at}\leftrightarrow \frac{1}{(s-a)}$

$\displaystyle {e}^{-10s}= \frac{1}{s+10}$

$\displaystyle \left( \frac {1}{s+10} \right) \left( \frac {2}{s^2+4} \right)$

$\displaystyle \frac {2}{(s+10)(s^2+4)}$

$\displaystyle \frac {A}{s+10}+ \frac {B}{s^2+4}$

from here, I am stuck. Is this the right method?

6. Originally Posted by jenjen
so if that is true, then do we use partial fraction expansion to solve the problem? I am still confused about why we don't use partial fractions. because since
$\displaystyle {e}^{at}\leftrightarrow \frac{1}{(s-a)}$

$\displaystyle {e}^{-10s}= \frac{1}{s+10}$

$\displaystyle \left( \frac {1}{s+10} \right) \left( \frac {2}{s^2+4} \right)$

$\displaystyle \frac {2}{(s+10)(s^2+4)}$

$\displaystyle \frac {A}{s+10}+ \frac {B}{s^2+4}$

from here, I am stuck. Is this the right method?
If you want to decompse
$\displaystyle \frac{2}{(s+10)(s^2+4)}$
Then you need to write it as, (because of the quadradic )
$\displaystyle \frac{A}{s+10}+\frac{Bs+C}{s^2+4}$

7. can you help me solve that?? I tried many times but I can't seem to get it to work.

I also have another problem I couln't do.

problem: the absent-minded Dr. R claims that $\displaystyle {y_1}=cosh(t), {y_2}=sinh(t)$ form a linearly independent set of solutions for the differential equation $\displaystyle {y"-y=0}$. Is he right? If so: how can we conclude that we can solve the IVP, $\displaystyle {y"-y=0}, {y(5)}={7}, {y'(5)}={10}$ with a function of the form $\displaystyle {y}={c_1y_1+c_2y_2}$ where $\displaystyle {c_1}$ and $\displaystyle {c_2}$ are constants.

8. Originally Posted by jenjen
can you help me solve that?? I tried many times but I can't seem to get it to work.

I also have another problem I couln't do.

problem: the absent-minded Dr. R claims that $\displaystyle {y_1}=cosh(t), {y_2}=sinh(t)$ form a linearly independent set of solutions for the differential equation $\displaystyle {y"-y=0}$. Is he right? If so: how can we conclude that we can solve the IVP, $\displaystyle {y"-y=0}, {y(5)}={7}, {y'(5)}={10}$ with a function of the form $\displaystyle {y}={c_1y_1+c_2y_2}$ where $\displaystyle {c_1}$ and $\displaystyle {c_2}$ are constants.
Remember that,
$\displaystyle \cosh x=\frac{e^x+e^{-x}}{2}$
$\displaystyle \sinh x=\frac{e^x-e^{-x}}{2}$
They only differ by signs joining exponents. We need to show that $\displaystyle y''-y=0$ for $\displaystyle \frac{e^x\pm e^{-x}}{2}$
Thus,
$\displaystyle y=\frac{e^x\pm e^{-x}}{2}$
$\displaystyle y'=\frac{e^x\mp e^{-x}}{2}$
$\displaystyle y''=\frac{e^x\pm e^{-x}}{2}$
In other words $\displaystyle y''=y$ thus, $\displaystyle y''-y=0$
Notice that $\displaystyle \cosh x\not =\sinh x$ for all $\displaystyle x$ for some $\displaystyle k$. Thus, they are linearly independent.

Thus, the general solution to this differencial equation is,
$\displaystyle y=C_1\cosh x+C_2\sinh x$

Please next time start a new thread, if it is off topic.

9. how come the y(5)=7 and y'(5)=10 isn't included in the answer?

10. Originally Posted by jenjen
so if that is true, then do we use partial fraction expansion to solve the problem? I am still confused about why we don't use partial fractions. because since
$\displaystyle {e}^{at}\leftrightarrow \frac{1}{(s-a)}$
This post is an orphan, because you have not quoted anything form the
previous post we are guessing what your question is at this point.

RonL

11. Originally Posted by ThePerfectHacker
You mean,
$\displaystyle \mathcal{L}\{e^{at}\}=\frac{1}{s-a}$
You mistake is,
$\displaystyle \frac{1}{s+a}$
No his mistake is that he wants:

$\displaystyle \mathcal{L}^{-1} e^{as}$

which I have already dealt with in an earlier post in this thread.

RonL

12. well, thank you so much Captainblack for all the help. They were very helpful. I am going to school to take my final now.

Jen

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