# Initial conditions

• Jun 9th 2010, 04:39 PM
Hibachi
Initial conditions
Find the particular solution for the differential equation $\displaystyle \dfrac{d^2x}{dt^2}-\dfrac{dx}{dt}-20x = 0$ that satisfies the initial conditions $\displaystyle x = 0$, $\displaystyle \dfrac{dx}{dt} = 2$ when $\displaystyle t = 0$.

Substituting $\displaystyle x = Ae^{n_{1}x}+Be^{n_{2}x}$, I got $\displaystyle n = 5$ and $\displaystyle n = -4$, so the general solution is $\displaystyle x = Ae^{5t}+Be^{-4t}.$

What I don't know is how to make use of the initial conditions to find $\displaystyle A$ and $\displaystyle B$.
• Jun 9th 2010, 04:51 PM
slider142
Find (dx/dt)(t): $\displaystyle \frac{dx}{dt}(t) = 5Ae^{5t} - 4Be^{-4t}$
Now you can get two equations in two unknowns using the two initial conditions: (dx/dt)(0) = 2 and x(0) = 0.
• Jun 9th 2010, 05:11 PM
Hibachi
Quote:

Originally Posted by slider142
Find (dx/dt)(t): $\displaystyle \frac{dx}{dt}(t) = 5Ae^{5t} - 4Be^{-4t}$
Now you can get two equations in two unknowns using the two initial conditions: (dx/dt)(0) = 2 and x(0) = 0.

$\displaystyle 5A-4B = 2$, $\displaystyle 5A = 2+4B$, $\displaystyle A = \dfrac{2+4B}{5}.$

$\displaystyle A = -B$

$\displaystyle \dfrac{2+4B}{5} = -B$

$\displaystyle 2+4B = -5B$

$\displaystyle B = -\dfrac{2}{9}$

$\displaystyle \therefore \;\;\; A = \dfrac{9}{2}$ and $\displaystyle B = -\dfrac{9}{2}.$

Therefore the particular equation is $\displaystyle x = -\dfrac{2}{9}e^{5t}+\dfrac{2}{9}e^{-4t}.$