# 2nd order differentiation query

• Dec 2nd 2010, 09:38 PM
bugatti79
2nd order differentiation query
Hi All,

Given $u (x) = Ae^x$ and $u"(x)=k^2 u (x)$

then u"(x) is calculated as $u"(x) = A^2 u (x)$

I calculate both u'(x) and u"(x) to be equal to Ae^x. I dont know how the A^2 appears in front of u(x)!!!.....

Thanks
• Dec 3rd 2010, 04:30 AM
HallsofIvy
Why do you say that $u''(x)$ "is calculated as $u''(x)= A^2u(x)$? If $u(x)= Ae^x$ then $u'= Ae^x$ and $u''= Ae^x$. That can be written as $u''= k^2u(x)$ if and only if $k^2= A$ so that $k= \sqrt{A}$.
• Dec 3rd 2010, 06:52 AM
bugatti79
$u"(x)=k^2(x)$

$u(x)=Ae^x$ therefore

$u"(x)=(u'(x))'=(Au(x))'=Au'(x)=A^2u(x)$ ????

Therefore $A^2u(x)=k^2u(x)$ implies

$A_1=k and A_2=-k$

I dont understand the third line!!...I dont think its the standard way of getting solution for a nd order equation?
Thanks
• Dec 3rd 2010, 07:05 AM
hmmmm
so you have $u'(x)=Au(x)$ and $u(x)=Ae^x$ ?? In the third line, that seems strange?
• Dec 3rd 2010, 09:16 AM
bugatti79
I dont know what his method is. I think I will just stick with the standard way of getting the solution ie,

$u"(x)=k^2u(x)$ is of the form

$u"(x)-n^2u(x)=0$ giving roots
$m^2-n^2=0$ therefore

$m=+ - n$

therefore solution is $u(x)=Ae^{nx} +Be^{-nx}$

This ok?
• Dec 4th 2010, 04:24 AM
HallsofIvy
$u"(x)=k^2(x)$

$u(x)=Ae^x$ therefore

$u"(x)=(u'(x))'=(Au(x))'$[/quote
This is wrong. If $u= Ae^x$ then $u'= A(e^x)$ not $Au$.

Quote:

$=Au'(x)=A^2u(x)$ ????

Therefore $A^2u(x)=k^2u(x)$ implies

$A_1=k and A_2=-k$

I dont understand the third line!!...I dont think its the standard way of getting solution for a nd order equation?
Thanks
Are you sure you don't have something like $e^{Ax}$ instead?
If $u(x)= e^{Ax}$, then $u'= A e^{Ax}= Au$ so $u''= Au'= A(Ae^{Ax})= A^2 e^{Ax}$.
• Dec 4th 2010, 12:13 PM
bugatti79
aha, thats what it was. It mustve been a typo on behalf or the lecturers :-)
But the standard derivation in post 5is correct? Thanks
• Dec 5th 2010, 04:29 AM
HallsofIvy
Quote:

Originally Posted by bugatti79
I dont know what his method is. I think I will just stick with the standard way of getting the solution ie,

$u"(x)=k^2u(x)$ is of the form

$u"(x)-n^2u(x)=0$ giving roots
$m^2-n^2=0$ therefore

$m=+ - n$

therefore solution is $u(x)=Ae^{nx} +Be^{-nx}$

This ok?

Why did you change from k to n? If the differential equation is $u''= k^2u$, then the characteristic equation is $m^2= k^2$ so that $m= \pm k$ and the general solution to the differential equation is $u(x)= Ae^{kx}+ Be^{-kx}$. Note that this could also be written as $u(x)= Ccosh(kx)+ Dsinh(kx)$.

By the way, using two consecutive single quotes in LaTex rather than a double quote will give $u''$ rather than $u"$.
• Dec 5th 2010, 10:46 AM
bugatti79
Quote:

Originally Posted by HallsofIvy
Why did you change from k to n? If the differential equation is $u''= k^2u$, then the characteristic equation is $m^2= k^2$ so that $m= \pm k$ and the general solution to the differential equation is $u(x)= Ae^{kx}+ Be^{-kx}$. Note that this could also be written as $u(x)= Ccosh(kx)+ Dsinh(kx)$.

By the way, using two consecutive single quotes in LaTex rather than a double quote will give $u''$ rather than $u"$.

The k to n, just a typo. Thanks.