4.3 Separable equations (Page 2/8)

Calculus volume 2 Page 2 / 8

Use the method of separation of variables to find a general solution to the differential equation $y' = 2 x y + 3 y - 4 x - 6 .$

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

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Solving an initial-value problem

Using the method of separation of variables, solve the initial-value problem

y' = (2 x + 3) (y^{2} - 4), y (0) = −3 .

Follow the five-step method of separation of variables.

In this example, $f (x) = 2 x + 3$ and $g (y) = y^{2} - 4 .$ Setting $g (y) = 0$ gives $y = \pm 2$ as constant solutions.
Divide both sides of the equation by $y^{2} - 4$ and multiply by $d x .$ This gives the equation

$\frac{d y}{y^{2} - 4} = (2 x + 3) d x .$
Next integrate both sides:

$\int \frac{1}{y^{2} - 4} d y = \int (2 x + 3) d x .$

To evaluate the left-hand side, use the method of partial fraction decomposition. This leads to the identity

$\frac{1}{y^{2} - 4} = \frac{1}{4} (\frac{1}{y - 2} - \frac{1}{y + 2}) .$

Then [link] becomes

$\begin{matrix} \frac{1}{4} \int (\frac{1}{y - 2} - \frac{1}{y + 2}) d y & = & \int (2 x + 3) d x \\ \frac{1}{4} (ln | y - 2 | - ln | y + 2 |) & = & x^{2} + 3 x + C . \end{matrix}$

Multiplying both sides of this equation by $4$ and replacing $4 C$ with $C_{1}$ gives

$\begin{matrix} ln | y - 2 | - ln | y + 2 | & = & 4 x^{2} + 12 x + C_{1} \\ ln | \frac{y - 2}{y + 2} | & = & 4 x^{2} + 12 x + C_{1} . \end{matrix}$
It is possible to solve this equation for y . First exponentiate both sides of the equation and define $C_{2} = e^{C_{1}} :$

$| \frac{y - 2}{y + 2} | = C_{2} e^{4 x^{2} + 12 x} .$

Next we can remove the absolute value and let $C_{2}$ be either positive or negative. Then multiply both sides by $y + 2 .$

$\begin{matrix} y - 2 = C_{2} (y + 2) e^{4 x^{2} + 12 x} \\ y - 2 = C_{2} y e^{^{4 x^{2} + 12 x}} + 2 C_{2} e^{^{4 x^{2} + 12 x}} . \end{matrix}$

Now collect all terms involving y on one side of the equation, and solve for $y :$

$\begin{matrix} y - C_{2} y e^{4 x^{2} + 12 x} & = & 2 + 2 C_{2} e^{4 x^{2} + 12 x} \\ y (1 - C_{2} e^{4 x^{2} + 12 x}) & = & 2 + 2 C_{2} e^{4 x^{2} + 12 x} \\ y & = & \frac{2 + 2 C_{2} e^{4 x^{2} + 12 x}}{1 - C_{2} e^{4 x^{2} + 12 x}} . \end{matrix}$
To determine the value of $C_{2},$ substitute $x = 0$ and $y = −1$ into the general solution. Alternatively, we can put the same values into an earlier equation, namely the equation $\frac{y - 2}{y + 2} = C_{2} e^{4 x^{2} + 12} .$ This is much easier to solve for $C_{2} :$

$\begin{matrix} \frac{y - 2}{y + 2} & = & C_{2} e^{4 x^{2} + 12 x} \\ \frac{−1 - 2}{−1 + 2} & = & C_{2} e^{4 {(0)}^{2} + 12 (0)} \\ C_{2} & = & −3 . \end{matrix}$

Therefore the solution to the initial-value problem is

$y = \frac{2 - 6 e^{4 x^{2} + 12 x}}{1 + 3 e^{4 x^{2} + 12 x}} .$

A graph of this solution appears in [link] .

Graph of the solution to the initial-value problem $y' = (2 x + 3) (y^{2} - 4), y (0) = −3 .$

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Find the solution to the initial-value problem

6 y' = (2 x + 1) (y^{2} - 2 y - 8), y (0) = −3

using the method of separation of variables.

$y = \frac{4 + 14 e^{x^{2} + x}}{1 - 7 e^{x^{2} + x}}$

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Applications of separation of variables

Many interesting problems can be described by separable equations. We illustrate two types of problems: solution concentrations and Newton’s law of cooling.

Solution concentrations

Consider a tank being filled with a salt solution. We would like to determine the amount of salt present in the tank as a function of time. We can apply the process of separation of variables to solve this problem and similar problems involving solution concentrations .

Determining salt concentration over time

A tank containing $100 L$ of a brine solution initially has $4 kg$ of salt dissolved in the solution. At time $t = 0,$ another brine solution flows into the tank at a rate of $2 L/min .$ This brine solution contains a concentration of $0.5 kg/L$ of salt. At the same time, a stopcock is opened at the bottom of the tank, allowing the combined solution to flow out at a rate of $2 L/min,$ so that the level of liquid in the tank remains constant ( [link] ). Find the amount of salt in the tank as a function of time (measured in minutes), and find the limiting amount of salt in the tank, assuming that the solution in the tank is well mixed at all times.

A diagram of a cylinder filled with water with input and output. It is a 100 liter tank which initially contains 4 kg of salt. The input is 0.5 kg salt / liter and 2 liters / minute. The output is 2 liters / minute. — A brine tank with an initial amount of salt solution accepts an input flow and delivers an output flow. How does the amount of salt change with time?

First we define a function $u (t)$ that represents the amount of salt in kilograms in the tank as a function of time. Then $\frac{d u}{d t}$ represents the rate at which the amount of salt in the tank changes as a function of time. Also, $u (0)$ represents the amount of salt in the tank at time $t = 0,$ which is $4$ kilograms.

The general setup for the differential equation we will solve is of the form

\frac{d u}{d t} = INFLOW RATE - OUTFLOW RATE .

INFLOW RATE represents the rate at which salt enters the tank, and OUTFLOW RATE represents the rate at which salt leaves the tank. Because solution enters the tank at a rate of $2$ L/min, and each liter of solution contains $0.5$ kilogram of salt, every minute $2 (0.5) = 1 kilogram$ of salt enters the tank. Therefore INFLOW RATE = $1 .$

To calculate the rate at which salt leaves the tank, we need the concentration of salt in the tank at any point in time. Since the actual amount of salt varies over time, so does the concentration of salt. However, the volume of the solution remains fixed at 100 liters. The number of kilograms of salt in the tank at time $t$ is equal to $u (t) .$ Thus, the concentration of salt is $\frac{u (t)}{100}$ kg/L, and the solution leaves the tank at a rate of $2$ L/min. Therefore salt leaves the tank at a rate of $\frac{u (t)}{100} \cdot 2 = \frac{u (t)}{50}$ kg/min, and OUTFLOW RATE is equal to $\frac{u (t)}{50} .$ Therefore the differential equation becomes $\frac{d u}{d t} = 1 - \frac{u}{50},$ and the initial condition is $u (0) = 4 .$ The initial-value problem to be solved is

\frac{d u}{d t} = 1 - \frac{u}{50}, u (0) = 4 .

The differential equation is a separable equation, so we can apply the five-step strategy for solution.

Step 1. Setting $1 - \frac{u}{50} = 0$ gives $u = 50$ as a constant solution. Since the initial amount of salt in the tank is $4$ kilograms, this solution does not apply.

Step 2. Rewrite the equation as

\frac{d u}{d t} = \frac{50 - u}{50} .

Then multiply both sides by $d t$ and divide both sides by $50 - u :$

\frac{d u}{50 - u} = \frac{d t}{50} .

Step 3. Integrate both sides:

\begin{matrix} \int \frac{d u}{50 - u} & = & \int \frac{d t}{50} \\ - ln | 50 - u | & = & \frac{t}{50} + C . \end{matrix}

Step 4. Solve for $u (t) :$

\begin{matrix} ln | 50 - u | & = & - \frac{t}{50} - C \\ e^{ln | 50 - u |} & = & e^{− (t / 50) - C} \\ | 50 - u | & = & C_{1} e^{− t / 50} . \end{matrix}

Eliminate the absolute value by allowing the constant to be either positive or negative:

50 - u = C_{1} e^{− t / 50} .

Finally, solve for $u (t) :$

u (t) = 50 - C_{1} e^{− t / 50} .

Step 5. Solve for $C_{1} :$

\begin{matrix} u (0) & = & 50 - C_{1} e^{−0 / 50} \\ 4 & = & 50 - C_{1} \\ C_{1} & = & 46 . \end{matrix}

The solution to the initial value problem is $u (t) = 50 - 46 e^{− t / 50} .$ To find the limiting amount of salt in the tank, take the limit as $t$ approaches infinity:

\begin{matrix} lim_{t \to \infty} u (t) & = 50 - 46 e^{− t / 50} \\ = 50 - 46 (0) \\ = 50 . \end{matrix}

Note that this was the constant solution to the differential equation. If the initial amount of salt in the tank is $50$ kilograms, then it remains constant. If it starts at less than 50 kilograms, then it approaches 50 kilograms over time.

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Source: OpenStax, Calculus volume 2. OpenStax CNX. Feb 05, 2016 Download for free at http://cnx.org/content/col11965/1.2

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