# Thread: showing differential equation solution

1. ## 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

2. All you need to do is take the general solution and sub it into the d.e. EG

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

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

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

Substituting into the d.e.

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

Now collecting like terms

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

Now since $\displaystyle k_1$ and $\displaystyle k_2$ are roots of $\displaystyle (k^2 + 2Bk + w^2)$ the terms on the LHS are both zero therefore the d.e. is satisfied.

3. thanks for that i was going about it in such a difficult way, this was much simpler than i had anticipated...

4. 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

5. 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...

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

and

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

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 $\displaystyle k_1$ and $\displaystyle k_2$. 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

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

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

Hope this helps.

6. and how do i solve these simultaneous equations

7. $\displaystyle \begin{array}{rclr} A + B &=& 1 &(1) \\ k_1A + k_2B &=& 0 &(2) \end{array}$

Step 1: $\displaystyle k_1 \times (1)$

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

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

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

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

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

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

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

$\displaystyle A = \frac{k_2}{k_2 - k_1}$

8. 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

9. 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

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

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

$\displaystyle x = \frac{k_1 e^{k_2t}-k_2e^{k_1t}}{k_1 - k_2}$

$\displaystyle x = \frac{(-B+iC) e^{(-B-iC)t}-(-B-iC) e^{(-B+iC)t}}{2iC}$

$\displaystyle x = e^{-Bt}\left[\frac{1}{2}\left(e^{iCt}+e^{iCt}\right)+\frac{B}{2 iC}\left(e^{iCt}-e^{iCt}\right)\right]$

$\displaystyle x = e^{-Bt}\left[\cos(Ct)+\frac{B}{C}\sin(Ct)\right].$

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 $\displaystyle x \rightarrow 0$.

If B < 0 then the function x goes to infinity.