We solve separable differential equations and initial value problems.

### 1 Differential Equations

A differential equation is an equation that involves one or more derivatives of an unknown function. Solving a differential equation entails determining the unknown function.

We seek such that . By inspection is a solution of this equation. However, is also a solution. In fact, there are infinitely many solutions: , where is any constant (including 0). This family of solutions to the differential equation is called the

**general solution**.

An initial value problem is a differential equation along with other information about the solution, usually the value of the function at a point. The purpose of the initial value is to determine one specific solution of the differential equation, in the event that there was more than one solution.

We saw in the last example that the differential equation has infinitely many solutions, , where is any constant. The initial value can be used to determine the value of so that we will have a unique solution. Since , we can see that . Thus the initial value problem has the unique solution .

The general solution of the differential equation is

The solution to the initial value problem is

### 2 Separable Differential Equations

A **separable differential equation** is a differential equation that can be put in the
form . To solve such an equation, we separate the variables by moving the ’s to one
side and the ’s to the other, then integrate both sides with respect to and solve for .
In general, the process goes as follows: Let for convenience and we have

Integrating both sides with respect to using a the substitution , we have and so Now we solve for algebraically to get the final answer. To simplify the computations, we will use instead of and then make a slight abuse of notation to get to the same end result. From the beginning: again using for convenience. Now we write this last equation in differential form and put an integral sign on both sides(!): yielding as before.

Rewrite the equation as Move the to the left and the to the right to get Integrate both sides: This gives Solve for : Note that has been replaced by , as this makes no difference. Now take reciprocals: Note that was replaced by as this makes no difference in the general solution. We use the initial condition to find the particular solution. Substitute and into the general solution to get: which simplifies to and we conclude by inspection that . The particular solution is thus Note that we can check the general solution to the differential equation by calculating and observing that . From the quotient rule, we have

Separating the variables gives

Integrating gives

The general solution is

Separating the variables gives

Integrating gives

The general solution is

The solution to the initial value problem is

Rewrite the equation as Move the to the left and the to the right to get Integrate both sides: This gives Solve for : Note that is a positive constant. Remove the absolute values: where is any non-zero constant. Looking back at the original differential equation, we see that is a solution, so we can say that is actually any constant. We can check the differential equation by calculating and observing that . From the chain rule, we have

Separating the variables gives

Integrating gives

The general solution is

Separating the variables gives

Integrating gives

The general solution is

The solution to the initial value problem is

Let denote the population of the bacteria (in millions) after hours. Since the rate of growth is proportional to the number present, we have for some constant . Moving to the left and to the right and integrating gives Anti-differentiating gives: Solving for : Removing the absolute value bars, we have where is a nonzero constant. To determine the values of and we will use the given information about the size of the population at time and . The population was initially 2 million, so and in one hour it doubled, so . the first of these gives, Using the second condition gives: Thus is the formula for the population at time . Finally, after 90 minutes (or 1.5 hours), the population is million bacteria.

Let denote the population (in billions) years after the year 2000.

The differential equation which models human population is

for some constant .

The general solution to this differential equation is

The value of is .

The value of to 3 decimal places is .

The population in the year 2050 is projected to be

Let denote the amount of money in the account at time in years.

The differential equation which models the amount in the account is

The general solution to this differential equation is

To determine the doubling time, we should let

The doubling time is approximately

Let represent the amount of salt in the tank at time . Then represents the rate of change in the amount of salt at time . Salt is entering the tank at a rate of 0.6 lb/min and salt is leaving the tank at a rate of 0.03A lbs/min. This leads to the differential equation To solve this differential equation, we first rewrite the derivative on the left side: Separating variables gives Integrate both sides: Computing the anti-derivatives, we have Solving for , we obtain Removing the absolute value bars, we have where K is a nonzero constant. subtracting and dividing by we have We can find the value of since we know that at time there was no salt in the tank, so that : Since we can write the amount of salt in the tank as Note that as , pounds of salt. This makes sense since we are adding salt at a concentration of 0.2 lbs/gal to a 100 gallon tank and 0.2 lbs/gal 100 gal = 20 lbs.

Let represent the amount of salt in the tank at time minutes.

Salt is entering the tank at a rate of lb/min

Salt is leaving the tank at a rate of lbs/min

This leads to the differential equation Separating variables gives

Integrating gives

Solving for , we have

The value of is

The time it takes for the tank to contain 2 pounds of salt is approximately

The differential equation that models Newton’s Law of Cooling is

Separating variables gives

Integrating gives

Solving for , we have

The value of is .

The value of the constant of proportionality, is

The additional time until the bread is at is