{"id":11951,"date":"2018-06-15t12:18:59","date_gmt":"2018-06-15t19:18:59","guid":{"rendered":"\/\/www.catharsisit.com\/hs\/?p=11951"},"modified":"2022-06-14t19:24:25","modified_gmt":"2022-06-15t02:24:25","slug":"ap-calculus-review-differential-equations","status":"publish","type":"post","link":"\/\/www.catharsisit.com\/hs\/ap\/ap-calculus-review-differential-equations\/","title":{"rendered":"ap calculus review: differential equations"},"content":{"rendered":"
a differential equation<\/strong> is a relation that involves an unknown function and its derivative. there are many kinds of differential equations and tons of specialized techniques we may use to solve them. fortunately, on the ap calculus exams you will only encounter a handful of the most basic kinds.<\/p>\n you have probably worked out hundreds of differential equations without even realizing it!<\/p>\n <\/p>\n it’s true! let me give you an example.<\/p>\n <\/p>\n what makes this a differential equation<\/em>?<\/p>\n well think about what the notation means. you know that integration is the opposite of differentiation. so what we’re looking for here is a function f<\/em>(x<\/em>) whose derivative<\/em> is equal to 2x<\/em>. in other words, we have to solve:<\/p>\n f<\/em>\u00a0‘(x<\/em>) = 2x<\/em><\/p>\n that’s a relation involving an unknown function f<\/em> and its derivative.<\/p>\n but this differential equation is trivial to solve! just use the power rule for integrals (or guess-and-check).<\/p>\n f<\/em>(x<\/em>) = x<\/em>2<\/sup> + c<\/em><\/p>\n (for a full review of integration, check out: ap calculus exam review: antidifferentiation<\/a> or ap calculus exam review: integrals<\/a>.)<\/p>\n now suppose you have more information about an unknown function, such as its value at a certain point. then you may be able to solve for the function explicitly, rather than getting stuck with an unknown constant of integration<\/em> at the end.<\/p>\n an initial value problem<\/strong> typically gives a derivative expression along with a function value. the goal is to produce the original function.<\/p>\n if f<\/em>\u00a0‘(x<\/em>) = 1\/x<\/em>2<\/sup> for x<\/em> > 0, and f<\/em>(1) = 5, find the expression for f<\/em>(x<\/em>) for x<\/em> > 0.<\/p>\n because the derivative expression is given, we integrate to find the original function. don’t forget about your constant of integration!<\/p>\n <\/p>\n next, plug in the known info and solve for the unknown constant of integration.<\/p>\n <\/p>\n this implies that the original function must be: \u00a0\u00a0 f<\/em>(x<\/em>) = -1\/x<\/em> + 6.<\/p>\n of course not every differential equation (or de<\/strong>) problem can be handled by simple integration. often more advanced techniques must be used, especially if both x<\/em> and y<\/em> show up in the equation.<\/p>\n for example, the following de cannot be cracked by integration (…not in its current form, that is).<\/p>\n <\/p>\n the key is to somehow break up the dy<\/em> and dx<\/em>.<\/p>\n the derivative notation dy<\/em>\/dx<\/em> looks like a fraction. however, it’s not really<\/em> a fraction, because the two quantities dy<\/em> and dx<\/em> (called differentials<\/strong>) are supposed to represent the idea of taking \u0394x<\/em> to zero in the limit. in other words, if you had to identify the numerical “value” of a differential, it would make sense to say dx<\/em> = dy<\/em> = 0. and we all know that 0\/0 is undefined.<\/p>\n but those differential gadgets are more subtle than that. they really don’t carry a numerical value. instead, they stand for a limiting process. and if we’re very careful, we can work with dx<\/em> and dy<\/em> individually.<\/p>\n the method of separation of variables<\/strong> works by manipulating dy<\/em>\/dx<\/em> as if it were a fraction, and then using integration to get rid of the differentials.<\/p>\n now let’s take another look at our example above.<\/p>\n if \u00a0\u00a0\u00a0, and y<\/em>(0<\/em>) = -3, find an equation for y<\/em> in terms of x<\/em>.<\/p>\n first multiply dx<\/em> to both sides. then multiply or divide as necessary so that only expressions of y<\/em> are on the left, and only those with x<\/em> are on the right.<\/p>\n <\/p>\n next, we use the magic of calculus! all you have to do is to apply the integral symbol to each side of the equation. now you have two separate integrals to work out.<\/p>\n <\/p>\n now even though both sides generated a constant of integration, those can be combined into a single constant on the right hand side. the next few steps are just for isolating y<\/em>.<\/p>\n <\/p>\n notice that the quantity ec<\/sup><\/em> is replaced by another constant k<\/em>. this is because c<\/em> is still arbitrary (and unknown). so just think of k<\/em> as a related, but still arbitrary constant.<\/p>\n finally we solve for k<\/em> using the known information, y<\/em>(0<\/em>) = -3.<\/p>\n (by the way, we will drop the “\u00b1” notation at this point because k<\/em> itself can take care of that choice.)<\/p>\n <\/p>\n therefore, replacing k<\/em> by its computed value -3, we obtain the final form of the function.<\/p>\n <\/p>\n fortunately you won’t encounter any de problems on the ap calculus exam that can’t be handled by either integration or separation of variables.<\/p>\n however, you may be asked to set up and\/or analyze a de (without solving it).<\/p>\n a certain bacteria culture has p<\/em> cells at time t<\/em> and grows in proportion to the square root of the amount present. set up a differential equation that models a<\/em>(t<\/em>).<\/p>\n because the culture grows at a certain rate with respect to time, we know we’ll be working with the derivative da<\/em>\/dt<\/em>.<\/p>\n let k<\/em> be the constant of proportionality. then translate the given information into mathematics:<\/p>\n da<\/em>\/dt<\/em> = k<\/em> \u00d7 \u221a(a<\/em>)<\/p>\n remember, we don’t have to solve for a<\/em>(t<\/em>); just set up<\/em> the de that could be used to solve for it. so we’re done at this step!<\/p>\n the number y<\/em> of people infected with the flu in a certain town at time t<\/em> can be modeled by the differential equation, dy<\/em>\/dt<\/em> = 0.00002y<\/em>(21500-y<\/em>), where y<\/em>(0) = 1. determine the limit of y<\/em> as t<\/em> \u2192 \u221e.<\/p>\n it would be difficult to solve this de for y<\/em> explicitly. instead, look closely at the equation. it has the form of a logistics equation<\/strong>.<\/p>\n <\/p>\n that means we can answer this question with no work at all! you just have to remember that logistics equations always have solutions that tend toward the carrying capacity<\/strong> m<\/em> as t<\/em> \u2192 \u221e. here, m<\/em> = 21500. that’s all you need to answer the question!<\/p>\n the limit will be 21500.<\/p>\nbasic differential equations: integration<\/h2>\n
initial value problems<\/h3>\n
example 1: initial value problem<\/h3>\n
solution<\/h4>\n
separation of variables<\/h2>\n
example 2: separation of variables<\/h3>\n
solution<\/h4>\n
setting up differential equations<\/h2>\n
example 3: setting up a de<\/h3>\n
solution<\/h4>\n
example 4: analyzing a de<\/h3>\n
solution<\/h4>\n