Suppose your friend is walking down University Avenue at a speed of $2$ blocks per minute. You want to know where she will be in five minutes so that you can surprise her. This situation can be modeled by the pure-time differential equation $\frac{d x}{d t} = 2$. The state variable $x$, is a function of time so is sometimes written as $x(t)$. The value $x(t)$ represents the number of blocks your friend is down University Avenue from Central Avenue , where $t$ represents time in minutes. Finding the solution to the differential equation $\frac{d x}{d t}=f(t)$ means explicitly finding a function $x(t)$ whose derivative is equal to $f(t)$.
How can we find $x$ as a function of $t$? There are several approaches to solving pure-time differential equations, and one is “guess and check.” Use your intuition to guess a solution to $\frac{d x}{d t} = 2$: $x{\left (t \right )}=$
Now verify that this is a solution by taking the derivative: $\frac{d x}{d t}=$
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Is the solution you found in part a the only solution to the differential equation? To help you find another function that has the same derivative, consider: what is the derivative of a constant? What happens to the derivative of a function if you add a constant to the function? It decreases. It stays the same. It increases. Find a different solution to the differential equation: $x{\left (t \right )} = $ .
If we want to represent all possible solutions to this differential equation, we have to add an arbitrary constant. Let's call this constant $C$. Then $x{\left (t \right )}=$ is a solution for any choice of $C$. This is called the general solution.
If we can add an arbitrary constant to our solution to get another solution, how many solutions are there to the differential equation $\frac{d x}{d t} = 2$? infinitely many two five zero Only one of these solutions will actually tell you where your friend is at any given time, though. How can you decide which one? Ask a friend. Pull one out of a hat. Use an initial condition.
We can't figure out where your friend will be in five minutes without knowing where she is at some point in time. Suppose that right now ($t=0$), she is already $4$ blocks from Central Ave; in other words, we suppose we have the initial condition $x(0)=4$. What is the value of $C$ such that $x{\left (0 \right )} = 4$? To determine $C$, first plug in $t=0$ into the general solution from part b to determine that $x(0)=$ . Therefore, to match this condition, we need to set $C=$.
The particular solution that describes your friend's location with the initial condition $x{\left (0 \right )} = 4$ is $x{\left (t \right )} = $ . How many blocks from Central Ave will your friend be in five minutes? $x(5)=$ .
What if the previous information about your friend was incorrect and, instead, the correct initial condition is that your friend is $7$ blocks from Central Ave at $t=0$? The new function describing her location is $x{\left (t \right )}=$ . In five minutes, she will be blocks from Central Ave.
Write a pure-time differential equation to model this scenario. Start by choosing a state variable to represent the volume of the cell: . (Online, the remaining equations will fill in with your chosen variable. On paper, you can write in the variable in each small blank ＿, below.)
With this state variable, the differential equation is $\frac{d ＿ }{dt} =$ . What is $t$ measured in? days minutes hours seconds
The right hand side of the differential equation is just a function of $t$ that gives the rate of increase in volume, $\frac{d ＿ }{dt}$, at time $t$.
Use your intuition to guess a solution to the differential equation you found in part a: $＿(t) =$
Now verify that this is a solution by taking the derivative: $\frac{d ＿ }{dt} =$
Again, this is not the only solution. What is the general solution to this differential equation? $＿(t) =$
Suppose we know that at $t=0$, the cell has volume $800$ $\mu {\rm m}^3$. What is the particular solution describing the volume of the cell? $＿(t) =$
What is the volume of the cell after $5$ seconds? (The second answer blank is for units.) After $10$ seconds? As $t$ gets large, does this solution make sense?
mu
μ
What if at $t=0$, the cell has volume $550$ $\mu {\rm m}^3$? What is the particular solution describing the volume of the cell? $＿(t) =$