INSTRUCTOR: The mass of 2 kilograms is attached to a spring with constant 32 newtons per meter and comes to rest in the equilibrium position. Beginning at time t equals 0, an external force equal to f of t equals 68 times e to the negative 2t end exponent, times cosine open parentheses 4t close parentheses, is applied to the system. Find the equation of motion if there is no damping. What is the transient solution? What is the steady state solution?

Let's first write the equation we are working to solve based on a mass of 2 kilograms, a spring constant of 32, and the external force we're given, and no damping. That's also something. So that tells us our equation is 2 times the second derivative of x plus 32 times x equals 68 e to the negative 2t end exponent, times cosine open parentheses 4t close parentheses.

Considering the complementary equation and finding the characteristic equation for that, we see we're going to have 2 lambda squared plus 32 equals 0, which means that lambda is equal to plus or minus 4i. So this tells us our general solution is of the form c sub 1 times cosine open parentheses 4t close parentheses, plus c sub 2 times sine, open parentheses 4t close parentheses.

Now also because of the external force that we see applied, that gives us an estimate, a guess as to the form of our particular solution. So we're going to call this x sub p of t is equal to A e to the negative 2t end exponent, cosine open parentheses 4t close parentheses, plus B times e to the negative 2t end exponent, times sine open parentheses 4t close parentheses. Now what we need to do-- now I'm going to do this very much line by line. We need to find the second derivative of this so that we can evaluate it into the original equation, we have.

So the first derivative, x prime, so x sub p prime of t will be equal to negative 2A e to the negative 2t end exponent cosine, open parentheses 4t close parentheses, minus e to the negative 2t end exponent, sine open parentheses 4t close parentheses minus 2B times e to the negative 2 end exponent, times sine open parentheses 4t close parentheses plus 4B times e to the negative 2t end exponent, times cosine open parentheses 4t close parentheses. Because of the product rule, the number of terms should double. Every term should produce 2.

The second derivative is going to be quite long. And I'm not going to combine terms until the end because I make less mistakes that way. So x sub p, and the second derivative of t is equal to 4A e to the negative 2t end exponent times cosine, open parentheses 4t close parentheses, plus 8A times e to the negative 2t end exponent times sine, open parentheses 4t close parentheses, plus 8A e to the negative 2t end exponent sine, open parentheses 4t close parentheses, minus 16A e to the negative 2t end exponent cosine, open parentheses 4t close parentheses, plus 4B times e to the negative 2t end exponent times sine, open parentheses 4t close parentheses, minus 8B times e to the negative 2t end exponent times cosine, open parentheses 4t close parentheses, minus 8 B e to the negative 2t end exponent times cosine, open parentheses 4t close parentheses, minus 16B times e to the negative 2t end exponent, multiplied by sine, open parentheses 4t close parentheses.

8 terms-- we had 4, and we started with 2. So that works. Now we can combine terms in each of these. So I'm going to go ahead and rewrite the second derivative of x sub p as open parentheses negative 12A minus 16B close parentheses, e to the negative 2t end exponent, cosine, open parentheses 4t close parentheses, plus open parentheses 16A minus 12B closed parentheses, times e to the negative 2t end exponent, sine, open parentheses 4t close parentheses.

I may also combine terms in just the function itself as well. Well actually, no. I don't need to do that. So now that we have these things, we'll substitute these into the equation we had-- 2 times the second derivative of x plus 32 times x equals 68 times e to the negative 2t end exponent times cosine, open parentheses 4t close parentheses.

And that produces this equation. It's going to be a little bit long. It's going to be open parentheses negative 24A minus 32B close parentheses times e to the negative 2t end exponent, times cosine, open parentheses 4t close parentheses, plus open parentheses 32A minus 24B B close parentheses, times e to the negative 2t end exponent, times sine, open parentheses 4t close parentheses, plus 32A times e to the negative 2t end exponent, cosine, open parentheses 4t close parentheses plus 32B times e to the negative 2t end exponent, sine open parentheses 4t close parentheses equals 68 e to the negative 2t end exponent, cosine, open parentheses 4t close parentheses.

From this, we can equate coefficients to see that, if we go and combine some of these, we see that 32A plus 8B must be equal to 0. And also 8A minus 32B must be equal to 68. And we can solve this by substitution or elimination or something like that to see that A will be equal to 1/2 B will be equal to negative 2.

So this gives us an equation of this-- x of t equals c sub 1 times cosine, open parentheses 4t close parentheses, plus c sub 2 times sine, open parentheses 4t close parentheses, plus 1/2 times e to the negative 2t end exponent, cosine, open parentheses 4t minus 2 times e to the negative 2t end exponent, sine, open parentheses 4t. Now, what we need to do is take our two known values. That is that x of 0 equals 0, going back to this, it comes to rest in equilibrium position. So at time 0, the position is at 0 in relation to equilibrium.

And also it comes to rest there. So that also means it's x prime of 0 equals 0. So let's go about finding what c1 and c2 are. So x of 0 equals 0 means that c1 plus 1/2 equals 0, which of course means that c1, c sub 1 equals negative 1/2. And that's actually all we can determine from x of 0 equals 0.

However, x prime of 0 equals 0 gives us a much more complicated equation. We have to find the derivative. But once we find that derivative, our equation is 4 times c2 minus 1 minus 8 equals 0, which tells us that c sub 2 is equal to 9/4. So our final particular equation of motion is x of t equals negative 1/2 times cosine, parentheses 4t close parentheses, plus 9/4 times sine, open parentheses 4t close parentheses, plus 1/2 times e to the negative 2t end opponent, times cosine, open parentheses 4t close parentheses, minus 2 times e to the negative 2t end opponent, times sine, open parentheses 4t close parentheses.

That's our equation of motion. Now, two things we have to answer still. And that is what is the transient solution? What's the steady state solution?

The steady state solution is the portion of this equation that had a c1 and c2 from the general solution. So this is a steady state. So that's the first two terms of our final answer here. And the transient solution is the last two terms of that equation that we stated just a second ago, the second to last two terms of x of t.