classical methods

[logistic_guy]

Solve the following differential equation using classical methods and the given initial conditions.

$\displaystyle \frac{d^2 x}{dt^2} + 2\frac{dx}{dt} + 2x = \sin 2t$

$\displaystyle x(0) = 2; \ \ \ \frac{dx}{dt}(0) = -3$

 

[logistic_guy]

First we solve the homogeneous.

$\displaystyle \frac{d^2 x}{dt^2} + 2\frac{dx}{dt} + 2x = 0$

$\displaystyle r = \frac{-2\pm\sqrt{2^2 - 4(1)(2)}}{2(1)} = -1\pm i$

Then, the solution is:

$\displaystyle x(t) = c_1 e^{-t}\cos t + c_2 e^{-t}\sin t$

 

[logistic_guy]

Now we will look for a particular solution of the form $\displaystyle x_p(t) = A\cos 2t + B\sin 2t$

$\displaystyle \frac{dx_p}{dt} = -2A\sin 2t + 2B\cos 2t$

$\displaystyle \frac{d^2x_p}{dt^2} = -4A\cos 2t - 4B\sin 2t$

When we substitute this in the original differential equation, we get:

$\displaystyle -4A\cos 2t - 4B\sin 2t + 2(-2A\sin 2t + 2B\cos 2t) + 2(A\cos 2t + B\sin 2t) = \sin 2t$

$\displaystyle -4A\cos 2t - 4B\sin 2t - 4A\sin 2t + 4B\cos 2t + 2A\cos 2t + 2B\sin 2t = \sin 2t$

$\displaystyle (4B - 2A)\cos 2t + (-2B - 4A)\sin 2t = \sin 2t$

If we compare the left side with the right side, we see that:

$\displaystyle 4B - 2A = 0$
$\displaystyle -2B - 4A = 1$

This gives:

$\displaystyle A = 2B$
$\displaystyle -2B -4(2B) = 1$
$\displaystyle -2B - 8B = 1$
$\displaystyle -10B = 1$
$\displaystyle B = -\frac{1}{10}$

And

$\displaystyle A = 2\left(-\frac{1}{10}\right) = -\frac{1}{5}$

Then,

$\displaystyle x_p(t) = -\frac{1}{5}\cos 2t - \frac{1}{10}\sin 2t$

 

[logistic_guy]

The general solution is:

$\displaystyle x(t) = c_1e^{-t}\cos t + c_2e^{-t}\sin t - \frac{1}{5}\cos 2t - \frac{1}{10}\sin 2t$

All remained is just to apply the initial conditions. And the idea of those conditions is to find the two constants $\displaystyle c_1$ and $\displaystyle c_2$.

That's what we'll do in the next post.