9.1: Differential Equations

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Throughout this module, if something does not exist, write DNE in the answer box.

Recap Video

Take a look at the following video which recaps the ideas from the section.

Example Video

Below is a video showing a worked example.

Problems

Consider the differential equation $\displaystyle \frac {dy}{dx} = -3y$ . Which of the following is a solution for $y$ ? Select all that apply.

$y = 2e^{-3x}$ $y = 2e^{-3x}+1$ $y = 4e^{-3x}$ $y = x^2$

Which of the following gives a general solution to the differential equation?

$y = e^{-3x}+C$ $y = Ce^{-3x}$ $y = x^2+C$

Which of the following gives the solution to the initial-value problem $\displaystyle \frac {dy}{dx} = -3y$ , $y(0) = 2$ ?

$y = e^{-3x}+1$ $y = 2e^{-3x}$ $y = x^2+2$

Consider the differential equation $y'' = -4y$ .

What is the order of this differential equation? $\answer {2}$ .
For what positive value of $k$ is $y = \cos (kx)$ a solution to the differential equation?
We can plug in $y = \cos (kx)$ into the equation and try and solve for $k$ . Note that if $y = \cos (kx)$ , then $y'' = \answer {-k^2\cos (kx)}$ . Therefore, the equation reads $-k^2\cos (kx) = -4\cos (kx).$ For this to be true, we need $k^2 = \answer {4}$ , which means the positive value of $k$ which works is $k = \answer {2}$ .
For what positive value of $k$ is $y = \sin (kx)$ a solution? $k = \answer {2}$ .
Check that $y = A\cos (2x)+B\sin (2x)$ is a solution to the differential equation, where $A$ and $B$ are any constants.
We can check something is a solution by plugging it into the differential equation. Notice that if $y = A\cos (2x)+B\sin (2x)$ , then $y' = -2A\sin (2x)+2B\cos (2x),$ and $y'' =\answer {-4A\cos (2x)-4B\sin (2x)}.$ Also note that $-4y = \answer {-4A\cos (2x)-4B\sin (2x)}.$ Since $y''=-4y$ , this $y$ is a solution to the differential equation.
Suppose $y(0) = 1$ and $y'(0) = 2$ . Find the value of $A$ and $B$ such that $y = A\cos (2x)+B\sin (2x)$ solves the initial-value problem.
Notice that if $y = A\cos (2x)+B\sin (2x)$ , then plugging in $x=0$ gives $y(0) = \answer {A}$ . Since we want $y(0) = 1$ , this tells us $A = \answer {1}$ . Therefore, the solution looks like $y = \cos (2x)+B\sin (2x)$ . To get $B$ , we need to use $y'(0)=2$ . Taking a derivative gives $y' = -2\sin (2x)+2B\cos (2x).$ Plugging in $x=0$ into this expression gives $y'(0) = \answer {2B}$ . Since we want $y'(0) = 2$ , this tells us $B = \answer {1}$ .

Consider the differential equation $(x^2-1)y'+xy = 0.$

Which of the following gives the general solution to the differential equation?
$y = \frac {1}{2}\ln (x^2-1)+C$ $y = C(x^2-1)$ $y = C(x^2-1)^{-1/2}$
Using the above, find the solution to the initial-value problem $(x^2-1)y'+xy=0,\quad y(\sqrt {2}) = 2.$ The solution is $y = \answer {2(x^2-1)^{-1/2}}$ .
In the previous part, you got the general solution of $y = C(x^2-1)^{-1/2}$ . Now find $C$ so that $y(\sqrt {2}) = 2$ .

Suppose the number of bacteria at time $t$ is given by $N(t)$ . Initially, there are $10$ bacteria, and the number of bacteria grows at a rate proportional to the number of bacteria, with proportionality constant $k$ .

Write down an initial-value problem to model this growth.
The initial number of bacteria tells us $N(0) = \answer {10}$ . To say the rate of growth is proportional to the number of bacteria means that $dN/dt$ , which is the rate of change, is a constant times the number of bacteria. The constant is called the proportionality constant. Therefore, the equation reads $\frac {dN}{dt} = kN,\quad N(0) = 10.$
If the constant is $k = 2$ , which of the following gives the solution to the initial-value problem above:
$N = 10e^{2t}$ $y = 10t$ $y = 10/t$
As $t \to \infty$ , we see that the population:
Grows indefinitely Grows to a population of $1000$ , then stops growing Eventually dies out

A fish population (with the number given by $F(t)$ ) grows at rate proportional to the product of the number of fish and $1000$ minus the number of fish.

A differential equation to model this is $\frac {dF}{dt} = kF(\answer {1000-F}).$
Suppose $k>0$ . If the initial fish population is $F(0) = 2000$ , what will happen to the fish population?
The population begins to decrease. The population begins to increase. The population stays the same.
Suppose $k>0$ . If the initial fish population is $F(0) = 1000$ , what will happen to the fish population?
The population begins to decrease. The population begins to increase. The population stays the same.
Suppose $k>0$ . If the initial fish population is $F(0) = 500$ , what will happen to the fish population?
The population begins to decrease. The population begins to increase. The population stays the same.