# ode's and constants

• Feb 6th 2007, 04:27 PM
dopi
ode's and constants
i want to find the values of the real constants a,b,c for which x(t) = ae^(bt) +ct is a solution to the ODE d^2/dt^2 = 16e^(4t) / xThis is what i have done so far

i differentiated x to get abe^(bt) + c
then i differentialiated that again to get (ab^2)*e^(bt)

then i substituted x and x" into the ODE given to get

(ab^2)*e^(bt) = 16e^(4t) / ae^(bt) +ct

this is where i got stuck..am i on the right path..from this part i need to find a b and c..any help thankz
• Feb 6th 2007, 04:34 PM
dopi
i just worked out that a=1 b=4 and c=0
is that right ..did anyone else get that?
• Feb 6th 2007, 05:02 PM
dopi
Quote:

Originally Posted by dopi
i just worked out that a=1 b=4 and c=0
is that right ..did anyone else get that?

acutally..i just went through it again..and it didnt work out...i tried new values for a,b,c

a=2 b=2 and c=0...........did anyone else get that???
thankz
• Feb 6th 2007, 06:55 PM
ThePerfectHacker
Quote:

Originally Posted by dopi
i want to find the values of the real constants a,b,c for which x(t) = ae^(bt) +ct is a solution to the ODE d^2/dt^2 = 16e^(4t) / xThis is what i have done so far

The differencial equation is,

$x''=\frac{16 e^{-4t}}{x}$
$x\cdot x''=16e^{-4t}$.
Now we are told that,
$x=ae^{bt}+ct$
Is some solution.
Then by substituting it into the differencial equation we have equality.
First we find the first derivative,
$x'=abe^{bt}+c$
$x''=ab^2e^{bt}$
Thus,
$(ae^{bt}+ct)(ab^2e^{bt})=16e^{-4t}$
Multiply through,
$a^2b^2e^{2bt}+ab^2cte^{bt}=16e^{-4t}$
Note the left hand side has some multiple of $tb^{bt}$ while the right does not. Thus to match we need it to be zero. Thus, one of the constants (at least) is zero, that is $a \mbox{ or } b \mbox{ or } c=0$.
Thus it dies,
$a^2b^2e^{2bt}=16e^{-4t}$.
We see that it cannot be $a,b$ because they must be non-zero in this equation. Thus, $c=0$.
The exponents need to match, for equality, that is,
$2b=-4$
$b=-2$.
Then by substituting for this value we have,
$a^2 (-2)^2 e^{-4t}=16e^{-4t}$
We immediately see that $a=\pm 2$.

Thus, there are two possibilities,
$(a,b,c)=(-2,2,0)$
$(a,b,c)=(2,2,0)$
• Feb 7th 2007, 03:15 AM
chogo
sorry can u explain again why one of the constants has to be zero? i dont quite understand that

thanks
• Feb 7th 2007, 05:39 AM
dopi
[QUOTE=ThePerfectHacker;38412]The differencial equation is,

$x''=\frac{16 e^{-4t}}{x}$

you made a mistake on e^(-4t), its actually a postive 4, but thats alright, still came out with the same answer, thanks
• Feb 7th 2007, 07:00 AM
ThePerfectHacker
Quote:

Originally Posted by chogo
sorry can u explain again why one of the constants has to be zero? i dont quite understand that

thanks

It is basicaly the same thing you do when you do the Method of Undetermined coefficients, I assume you heard of it. If note you assume a solution has a certain form like here and substitute it and solve for the constants.
For example,
$2\sin t +t\cos t=A\sin t +2Bt\cos t +Ce^t$
For the right and left hand sides to be equal you make the coefficients equal to each other.

1)There is a $2\sin t$ on left side and there is $A\sin t$ on right side. To be equal the coefficient must match, that is, $A=2$.

2)There is $t\cos t$ on left side and there is $2B\cos t$ on right side. To be equal the coefficient must match, that is, $2B=1$. Thus, $B=1/2$.

3)And there is not $e^t$ on left hand side. Thus, $C=0$
• Feb 7th 2007, 07:34 AM
chogo
yes thanks alot never heard of Method of Undetermined coefficients. But read up on it, thanks alot it was very interesting.