<p>At any time t>=0 the velocity of a particle traveling along the x-axis is given by the differential equation dx/dt - 10x = 60e^(4t)</p>
<p>Find the general solution x(t) for the position of the particle</p>
<p>My answer is x(t)=15e^(4t) +10x is that correct?</p>
<p>That's a strange problem. I got y(t) = -10e^(4t) + Ce^(10t).</p>
<p>I might have made a mistake doing that, but are you sure you entered the problem in correctly? This isn't something you'd generally study until an elementary differential equations course, unless I'm seriously forgetting my old AP Calc material. If that -10x is actually a -10t, things suddenly make a lot more sense. Regardless though, I can tell you that your solution is definitely not correct.</p>
<p>i entered the equation correctly it is a 10x NOT 10t, i suspected that too</p>
<p>Hmm... Well, that's strange, because this is really a little beyond the scope of BC Calc, but I guess it's an opportunity to look really smart. Here's how you would go about solving a problem like this:</p>
<p>First, multiple the entire equation by an integrating factor mu(t). That's mu the Greek letter, by the way, not m*u or anything like that.</p>
<p>Then we have mu(t) x(t) + -10mu(t) x(t) = mu(t) *60e^(4t)
Focusing on the left side for now, we have mu(t) x(t) + -10mu(t) x(t).
If we can find some way to rewrite it in the form d/dt[f(t)], the differential equation is easy to solve, since only a simple integration is necessary to find f.</p>
<p>Recall that, by the product rule, d/dt[mu(t) x(t)] = mu(t) y'(t) +mu'(t) x(t). Comparing this to the left hand side of our old equation, we want mu(t) x'(t) +mu'(t) x(t) = mu(t) x(t) + -10mu(t) x(t). </p>
<p>The only way for this to happen is if mu'(t) = -10mu(t). If this is the case, it isn't difficult to observe that mu(t) e^(-10t).</p>
<p>Thus, our old equation now becomes
e^(-10t) x'(t) - 10 e^(-10t) x = 60e^(-6t)
or
d/dt[e^(-10t) y(t)] = 60e^(-6t)</p>
<p>Integrate to get
e^(-10t) y(t) = -10e^(-6t) +C</p>
<p>Divide by e^(-10t) to get
y(t) = -10e^(4t) + Ce^(10t)</p>
<p>I hope that helps. It's a little messy to look at, but it's not really all that complicated. This is what we spent two lectures covering in my college course, so this is a somewhat condensed version. There's really just one major trick you have to be able to use to solve the problem. Then everything becomes pretty much regular integral calculus.</p>