Particular Solutions If the differential equation is actually modeling something (like the cost of milk as a function of time) it is likely that you will know a specific value (like the fact that milk was $3.29/gallon on January 1, 2010.) This is what is called an initial condition. So if we are trying to solve y 4y = 0 and we know that y = 3 if x = 0 then we can use the general solution y = Ae 4x together with the initial condition y(0) = 3 to find a single solution (i.e. we can find out what A is). y = Ae 4x and y = 3 at x = 0 3 = Ae 4 0 3 = A y = 3e 4x
More separation of variables: In general you might be trying to solve an equation that looks like dy dx = f(x)g(y). For example y = xy, where f(x) = x, and g(y) = y. So we get all the x s on one side and all the y s on the other side and integrate: 1 g(y) dy = f(x)dx. Of course this relies on our ability to do these integrals, and in our example we can and we have 1 y dy = xdx ln y = x2 2 + C y = /2+C ex2 = Ae x2 /2
Interpreting word problems: Examples of differential equations come from all over the place, from economics, biology, chemistry, business, etc. You must learn to interpret certain words in problems. Most commonly if you see the phrase A is proportional to B all that means is A = kb for some number k. Also, all differential equations will involve rates of change because that is exactly what the derivative is.
Investments The rate of growth of an investment is proportional to the amount in the investment at any time t. That is, da dt = ka. What is A? In this problem A represents the amount (in dollars say) of the investment and A = A(t) i.e. A is a function of time, A may increase or decrease with time (depending on the investment and in this case the rate of growth k.) What is the general solution? 1 A da = kdt ln A = kt + c A(t) = Ce kt.
Investments cont. We have A(t) = Ce kt represents the rate of growth of an investment. If we know that an initial investment of $1000 grew to 3320.12 in 10 years, what is the particular solution? Now we have to find both C and k, what do we know? Initially the investment was worth $1000 so A = 1000 at t = 0 or A(0) = 1000. This tells me C : A(0) = Ce k 0 = Ce 0 = C C = 1000. Now we know that at t = 10 we have A = 3320.12, this will tell us what k is. Before we try to figure out what k is, can you tell me if k should be positive or negative?
Investments cont. We have A(t) = 1000e kt and we know A(10) = 3320.12, we are expecting k > 0, so let s figure out what it is. 3320.12 = 1000e k 10 3320.12 1000 = e 10k ln 3320.12 1000 = 10k k = ln 3320.12 1000 10 = 0.12 So A(t) = 1000e 0.12t
Newton s Law of Cooling Newton s law of cooling states that the rate of change in the temperature T of an object is proportional to the difference between the temperature of the object (T ) and the surrounding temperature T 0. This can be expressed by the differential equation: dt dt = k(t T 0). Note that here T 0 is the ambient, or surrounding, temperature. k is the rate of cooling and t is time. Since the object is going from higher temperature to cooler temperature the rate of change will be negative, so k < 0.
Newton s Law of Cooling cont. A room is kept at a constant temperature of 68 F. A pie is taken out of a 350 oven and placed on the counter. If the pie has reduced in temperature to 150 in 45 minutes, when will the pie reach 80? First we must find the general solution: 1 (T T 0 ) dt = kdt. ln(t T 0 ) = kt + c T T 0 = e kt+c = Ae kt T (t) = T 0 + Ae kt
Newton s Law of Cooling cont. We have the general solution T (t) = T 0 + Ae kt We will measure time in hours, though we could use minutes if we wanted. The ambient temperature here is 68. We know T (0) = 350 and T (.75) = 150 this will help us find k : T (0) = 68 + Ae k 0 = 68 + A = 350 A = 350 68 = 282. So now we have T (t) = 68 + 282e kt, and we know T (.75) = 150 so 150 = 68 + 282e.75k 150 68 = e.75k ln 82 282 282 =.75k k = ln 82 282.75 = 1.65 or k = 1.65
Relation to slope and graphs: We can also solve a problem like this: 1 At each point (x, y) on the graph, the slope is 2x/y. 2 The graph passes through the point (1, 1). The equation to solve is dy dx = 2x/y, and the additional information we know is that y(1) = 1, in other words when x = 1, then y = 1.
Relation to slope and graphs cont.: dy dx = 2x/y, First separate variables, then integrate: ydy = 2xdx. y2 2 = x2 + C Writing it a little more simply we have y 2 2x 2 = C, note that we don t have to write 2C we can just use C again for a different constant. Now plug in the known values (x, y) = (1, 1) to find the exact value for C : 1 2 2(1) 2 = 1 = C y 2 2x 2 + 1 = 0 is the curve we are trying to find.