# showing differential equation solution

• Feb 5th 2009, 01:49 PM
crafty
showing differential equation solution
for distinct roots (k1, k2) of the equation k^2 + 2Bk + w^2 show that x(t) = Ae^(k1t) + Be^(k2t) is a solution of the following differential equation: (d^2)x/dt^2 + 2B(dx/dt) + (w^2)x = 0
• Feb 5th 2009, 02:10 PM
Rincewind
All you need to do is take the general solution and sub it into the d.e. EG

$x = Ae^{k_1t} + Be^{k_2t}$

$\frac{dx}{dt} = Ak_1e^{k_1t} + Bk_2e^{k_2t}$

$\frac{d^2x}{dt^2} = Ak_1^2e^{k_1t} + Bk_2^2e^{k_2t}$

Substituting into the d.e.

$LHS = Ak_1^2e^{k_1t} + Bk_2^2e^{k_2t} + 2B(Ak_1e^{k_1t} + Bk_2e^{k_2t}) + w^2(Ae^{k_1t} + Be^{k_2t})$

Now collecting like terms

$LHS = Ae^{k_1t}\left(k_1^2 + 2Bk_1 + w^2\right) + Be^{k_2t}\left(k_2^2 +2Bk_2 + w^2\right)$

Now since $k_1$ and $k_2$ are roots of $(k^2 + 2Bk + w^2)$ the terms on the LHS are both zero therefore the d.e. is satisfied.
• Feb 5th 2009, 02:31 PM
crafty
thanks for that i was going about it in such a difficult way, this was much simpler than i had anticipated...
• Feb 5th 2009, 02:35 PM
crafty
theres another part to this question which i still can't seem to get use the initial conditions x(0) = 1 and x'(0) = 0 to determine the values of A and B, describe how the solution behaves as t -> infinity
• Feb 5th 2009, 04:17 PM
Rincewind
Quote:

Originally Posted by crafty
theres another part to this question which i still can't seem to get use the initial conditions x(0) = 1 and x'(0) = 0 to determine the values of A and B, describe how the solution behaves as t -> infinity

What you need to do here is substitute in the boundary conditions to determine A and B so...

$\begin{array}{rcl}
Ae^{k_10}+Be^{k_20} &=& 1\\
A + B &=& 1
\end{array}$

and

$\begin{array}{rcl}
Ak_1e^{k_10}+Bk_2e^{k_20} &=& 0\\
Ak_1 + Bk_2 &=& 0
\end{array}$

You can solve these simultaneous as 2 equations with 2 unknowns (A and B)

This problem is a bit tricky since throughout the B in the original equation is different from the B in the general solution.

Regarding the general behavior it depends on the sign of $k_1$ and $k_2$. If we assume the B in the original equation is real and positive and the w is real and less than B then the roots are

$k = -B\pm\sqrt{B^2 - w^2}$

which are both definitely negative and in that case the exponential terms go to zero as t goes to infinity.

Hope this helps.
• Feb 5th 2009, 06:17 PM
crafty
and how do i solve these simultaneous equations
• Feb 5th 2009, 07:03 PM
Rincewind
$\begin{array}{rclr}
A + B &=& 1 &(1) \\
k_1A + k_2B &=& 0 &(2)
\end{array}$

Step 1: $k_1 \times (1)$

$\begin{array}{rclr}
k_1A + k_1B &=& k_1 &(3)
\end{array}$

Step 2: $(3) - (2)$

$\begin{array}{rclr}
(k_1-k_2)B &=& k_1 &(4)
\end{array}$

Step 3: $(4) \div (k_1-k_2)$

$B = \dfrac{k_1}{k_1-k_2}$

Step 4: Substitute $B = \frac{k_1}{k_1-k_2}$ into $(2)$

$k_1A + \frac{k_1k_2}{k1-k_2} = 0$

$A = \frac{k_2}{k_2 - k_1}$
• Feb 5th 2009, 07:25 PM
crafty
by the way for that part about infinity i forgot to mention that W>B and that means the roots are complex, so what happens then??? sorry about forgetting to mention this earlier
• Feb 5th 2009, 10:20 PM
Rincewind
Quote:

Originally Posted by crafty
by the way for that part about infinity i forgot to mention that W>B and that means the roots are complex, so what happens then??? sorry about forgetting to mention this earlier

OK so if W > B and B > 0 then you have the roots equal to

$k_1 = -B + iC$ and $k_2 = -B - iC$

where $C = \sqrt{W^2 - B^2}$

$x = \frac{k_1 e^{k_2t}-k_2e^{k_1t}}{k_1 - k_2}$
$x = \frac{(-B+iC) e^{(-B-iC)t}-(-B-iC) e^{(-B+iC)t}}{2iC}$
$x = e^{-Bt}\left[\frac{1}{2}\left(e^{iCt}+e^{iCt}\right)+\frac{B}{2 iC}\left(e^{iCt}-e^{iCt}\right)\right]$
$x = e^{-Bt}\left[\cos(Ct)+\frac{B}{C}\sin(Ct)\right].$
So when t goes to infinity the exponential term sends the function to zero. The other terms just cause the function to wiggle around but the overall trend is $x \rightarrow 0$.