homework2bb

Homework 2b:

Read the following and turn in on paper answers to the 3 problems below that are identified as homework problems (some problems have multiple parts):

1.2 = 2.2: Separation of variables.

A very useful method to solve mathematics problems is to convert a problem into one you already know how to solve.
To solve the differential equations in section 1.2 and 2.2, you can covert these problems into single variable calculus integration problems.

Example: $~~x \frac{dy}{dx} = y$

To make this look more like an integration problem from calculus, we will multiply both sides by $dx$ since integration problems do not have $dx$ (or $dy$) in the denominator:

$$x dy = ydx$$

But $xdy$ (and $ydx$) are not single variable calculus problems since they both involve 2 variables. Thus we need to separate the variables: $$\frac{dy }{y}= \frac{dx}{x}$$

Note to separate the variables, I moved the $x$ to the side with the $dx$ by dividing both sides by $x$ and the $y$ to the side with the $dy$ by dividing both sides by $y$. I did not want to move $dx$ or $dy$ since they are already on opposite sides, and I don't want to divide by either of them since we don't want $dx$ or $dy$ in the denominator.

To get rid of the derivative, we can integrate both sides:

$$\int \frac{dy }{y}= \int \frac{dx}{x}$$

Both the LHS and RHS are now single variable calculus 1 problem. Thus we have changed our original differential equation into a problem we know how to do. After integrating:

$$ln|y|= ln|x| + C$$

We now have an algebra problem where we can solve for $y$:

$$e^{ln|y|}= e^{ln|x| + C} = e^{ln|x|}e^C = |x|e^C = xC$$

Thus $y = Cx$ since $e^C$ is a constant.

FYI: In the above I lost solution the solution $y = 0$ when I divided by $y$, but I regained it when I replaced $e^C$ with $C$. In the real world you should be careful about losing solutions and replacing a positive number like $e^C$ with an arbitrary constant $C$. We won't be that careful in this class.

HW problem 1: Use separation of variables to solve $y' = x^2y + y$

Hint 1: To separate the variables, you need to see all the variables, so rewrite $y'$ as $\frac{dy}{dx}$

Hint 2: To separate the variables, $x$ from $y$, sometimes you will need to factor. For example, in the above on the RHS, you can factor out the $y$ before moving it to the other side (using the appropriate algebraic operation).

2.1: Solving linear differential equations using an integrating factor.

A first order differential equation is linear if it can be written in the form: $a(t)y' + b(t)y = g(t)$.

It is called "linear" since the LHS of the equation is a linear combination of $y$ and $y'$ (where the coefficients are functions of $t$) and there are no $y$'s (including $y'$'s) on the RHS.

Simple motivating example: $~~e^{3t}y' + 3e^{3t}y = 1$

Note the LHS comes from the product rule: $~~e^{3t}y' + 3e^{3t}y = [e^{3t}y]'$

Thus we have: $~~[e^{3t}y]' = 1$

We can now integrate both sides using the Fundamental Theorem of Calculus: $\int F'(t)dt = F(t) +C$. In other words the integral of the derivative of a function returns the function, offset by a constant $C$.

Integrating both sides: $~~\int[e^{3t}y]'dt = \int 1dt$

Hence $~~e^{3t}y = t + C$ and thus $y = te^{-3t} + Ce^{-3t}$

Please avoid common errors: Remember the $+C$ and if you divide one side by $e^{3t}$, you must divide the entire other side (both the $t$ and the $C$ by $e^{3t}$).

Real example: $~~ty' + 2y = \frac{e^t}{t}$

We want to create the product rule: $u(t)y' + u'(t)y = [u(t) y]'$ since $[u(t) y]'$ is easy to integrate using the FTC. To do so, we will use an integrating factor.

For the differential equation $y' + p(t)y = g(t)$, the integrating factor is $u(t) = e^{\int p(t)dt}$

Optional thought question: After reading the below, do you know why we chose $u(t) = e^{\int p(t)dt}~$?

Step 0: Divide both sides by $t$ since to use the above formula for the integrating factor, the coefficient in front of $y'$ must be a one.

$$1y' + \frac{2}{t}y = \frac{e^t}{t^2}$$

Step 1: Find the integrating factor: The coefficient of $y$ is $p(t) = \frac{2}{t}$ and thus the integrating factor is $u(t) = e^{\int p(t)dt}$ is

$$u(t) = e^{\int p(t)dt}= e^{\int\frac{2}{t}dt} = e^{ 2 ln(t)} = e^{ln(t^2)} = t^2$$

Note in the above, I was sloppy: I did not use absolute value signs or the +C since I just need one integrating factor that works.

We also used the log rule: $~~~ln(a^r) = r ln(a)$.
We had to use this log rule to move the 2 as we needed the exponential function applied directly to the logarithmic function for these functions to cancel each other out.

Step 2: Multiply both sides of the differential equation by the integrating factor $u(t) = t^2$.

$$y' + \frac{2}{t}y = \frac{e^t}{t^2}$$$$t^2y' + {2}{t}y = e^t$$

You should now see the product rule on the LHS:

$$[u(t)y]' = [t^2y]' = t^2y' + {2}{t}y$$

Thus $~~~~~~~~t^2y' + {2}{t}y = e^t$

simplifies to $~~~[t^2y]' = e^t$

We can now integrate both sides, using the fundamental theorem of calculus to easily integrate the RHS. In this case I have fixed the problem to make it easy to also integrate the LHS, but often integrating the LHS is the most time-consuming part of the problem -- make sure you know how to do integration by substitution, by parts, and by partial fractions.

$$\int [t^2y]' = \int e^t$$$$t^2y = e^t + C$$

We can now easily solve for $y$ by dividing both sides by the integrating factor $u(t) = t^2$.

$$y = \frac{e^t}{t^2} + \frac{C}{t^2}$$

HW problem 2: Solve the following differential equation. Clearly indicate the step where you use the product rule

$$ty' + 3y = \frac{4}{t^3(t+2)}$$

Hint 1: Sometimes (but not always) you will need to use a log rule to simplify the integrating factor. For this problem you will need to use a log rule.

Hint 2: We will use partial fractions in chapter 6, so make sure you remember how to do integration by partial fractions. This problem involves integration by partial fractions.

Hint for one of the 2.1 WP homework problems: To find the limit, you do not need to integrate the RHS. Instead, you can use l'hopital's rule.

Hint for one of the 2.3 WP homework problems: To find the amplitude of $Acos(wt) + Bsin(wt)$, use that $Acos(wt) + Bsin(wt) = Rcos(wt - \delta)$ where $R = \sqrt{A^2 + B^2}$.

FYI: This formula comes from converting $(Acos(wt), Bsin(wt))$ into polar coordinates $(R, \delta)$ where $R$ is the radius of the circle centered at (0, 0) and containing the point $(Acos(wt), Bsin(wt))$.

We will use this trig formula in sections 3.7 and 3.8 when we discuss mechanical and electrical vibrations.

2.3: Modeling.

The simplest differential equation model is one where the rate of change is constant: $$\frac{dp}{dt} = k$$

For example, suppose you put \$10 in a piggy bank every month. Since you are not earning interest, the amount of money you have in this piggy bank changes by \\$10 every month. In other words, if $p(t)$ is the amount of money you have at time $t$, the rate of change (in \$/month) is $$\frac{dp}{dt} = 10$$

The next simplest differential equation is where the rate of change of $p$ is proportional to $p$ (where we will use $r$ as the proportionality constant): $$\frac{dp}{dt} = rp$$

Suppose you have \$500 in the bank earning 1\% annual interest. Assuming you make no additional payments, the amount you have increases by 1\% each year. Note the amount you earn is proportional to the amount you have. In other words if you have $p(t)$ dollars in the bank at time $t$, then the rate of change of $p$ is proportional to $p$ with a proportionality constant $r = 0.01$ since you are earning 0.01\% interest. Thus the rate of change (in \\$/year) is $$\frac{dp}{dt} = 0.01p$$

Note the above differential equation does not fully model this scenario as it does not include the information that we started out the \$500. Thus the full model should also include this initial value: $p(0) = 500$.

An initial value problem includes BOTH the differential equation and initial value(s) -- for a first order differential equation we only need one initial value.

Thus the initial value problem (IVP) modeling the above scenario is

$$\frac{dp}{dt} = 0.01p, ~~p(0) = 500$$

To determine the amount of money in the bank at any time $t$, where $t$ is measured in years, we need to know both the differential equation and the initial value.

In many cases, you have a rate of change that is a combination of the rate of change proportional to your variable and a constant rate of change:

$$\frac{dp}{dt} = rp + k$$

Suppose you initially deposit \$500 in the bank earning 1\% annual interest. Suppose you also make deposits of \\$120 each year into this account. Thus the rate of change (in \$/year) is $$\frac{dp}{dt} = 0.01p + 120, ~~p(0) = 500$$

WARNING: Check your units to make sure they are all consistent. For example,

Suppose you initially deposit \$500 in the bank earning 1\% **annual** interest. Suppose you also make deposits of \\$10 each month into this account. You need to choose whether to measure time in years or months. If you measure in years, you get the equation in the previous example. If you measure in months, the rate of change (in \$/month) $$\frac{dp}{dt} = \frac{0.01}{12}p + 10, ~~p(0) = 500$$ where we converted the 1\% annual interest to a $\frac{1}{12}$\% monthly interest.

Many of your 2.3 homework problems will involve a differential equation of the form $\frac{dp}{dt} = rp + k$ where $r, k$ are real numbers.

HW problem 3: State the initial value problem (IVP) that models the following scenarios. Note you only need to state the IVP, you do not need to solve it for this homework problem.

HW Problem 3a: The population of rabbits doubles every 20 days. If you start with 500 male rabbits and 500 female rabbits, state the IVP that models this scenario where the units of $\frac{dp}{dt} =$ change in rabbits/day.

Hint: The rate of increase is proportional to the number of rabbits. To find the proportionality constant, convert the doubling rate to a daily increase instead of the 20 day increase.

HW Problem 3b: A field contains 500 male rabbits and no female rabbits. Foxes eat 3 rabbits each day. State the IVP modeling this scenario. Hint: The rate of change is negative since the number of rabbits is decreasing. Also, does the rate of change depend on the population or is it constant?

HW Problem 3c: A field contains 500 male rabbits and 500 female rabbits. The population of rabbits doubles every 20 days. But the population decreases via predation as foxes eat 3 rabbits each day. State the IVP modeling this scenario.