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}
$
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?
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,Originally Posted by jenjen
$\displaystyle \cosh x=\frac{e^x+e^{-x}}{2}$
$\displaystyle \sinh x=\frac{e^x-e^{-x}}{2}$
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.