Separable Differential Equations

Section 7.1 Separable Differential Equations

In this section, we start our study of differential equations. A differential equation is an equation that relates an unknown function and one or more of its derivatives.

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Remark 7.1.1.

You are encouraged to peruse the Wikipedia page on differential equations (in particular the introduction and the Applications section).

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Of interest is to determine the unknown function, also known as the solution of the differential equation. It is not always possible (or necessary) to determine the solution of the differential equation explicitly (in closed form, or in terms of elementary functions).

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In this section, we focus on separable differential equations, for which solutions can be determined relatively easily.

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Subsection 7.1.1 How to Read a Differential Equation

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Consider the differential equation

$\frac{dy}{dx} = -k y^2,$ where

$k$ is a constant. In this differential equation, the independent variable is

$x\text{,}$ and the dependent variable is

$y\text{.}$ We refer to the constant

$k$ as a parameter. That is, we are seeking the unknown function

$y(x)$ that satisfies the given differential equation.

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Remark 7.1.2.

A parameter is a constant that has a specific value in a specific setting, but may have a different value in a different setting. A well-known example of a parameter is the constant of acceleration due to gravity,

$g\text{.}$ Its value on Earth is

$g=9.81$ m/s

$^2\text{,}$ but the value is different on the moon or on Mars.

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Checkpoint 7.1.3.

Consider the differential equation $$\frac{dN}{dt} = rN \left( 1 - \frac{N}{K} \right).$$ Identify the independent and dependent variables and the parameters, and identify the function we are seeking.

Solution

The independent variable is $t\text{,}$ the dependent variable is $N\text{,}$ the parameters are $r$ and $K\text{,}$ and the unknown function we are seeking is $N(t)\text{.}$

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Checkpoint 7.1.4.

Consider the differential equation $$\frac{dc}{dt} = k.$$ Identify the independent and dependent variables and the parameters, and identify the function we are seeking.

Solution

The independent variable is $t\text{,}$ the dependent variable is $c\text{,}$ the parameter is $k\text{,}$ and the unknown function we are seeking is $c(t)\text{.}$

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Subsection 7.1.2 Two 'Simple' Differential Equations Compared

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Consider the two differential equations

$\frac{dy}{dt} = t$ and

$\frac{dy}{dt}=y.$ The essential difference is that the right-hand side of the first differential equation involves the independent variable

$t\text{,}$ while the right-hand side of the second differential equation involves the dependent variable

$x\text{.}$

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For the first differential equation, we seek the function

$y(t)$ that satisfies

$y'(t) = t\text{,}$ and we readily determine

$\begin{alignat*}{1} y(t) &= \int y'(t) \ dt \\ &= \int t \ dt\\ &= \frac{1}{2} t^2 + C, \end{alignat*}$

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that is, the solution of the first differential equation is readily obtained by integrating the right-hand side.

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The first differential equation is not so exciting. Even if we replace the right-hand side of the differential equation with a more complicated function of

$t\text{,}$ we can readily obtain the solution by simply integrating the right-hand side with respect to

$t\text{.}$

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The second differential equation is much more interesting, and foundational for much of the work coming up. For this differential equation, we seek the function

$y(t)$ that satisfies

$y'(t) = y\text{.}$ By inspection, we observe that

$y(t) = e^t$ is a solution.

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Checkpoint 7.1.5.

Confirm that $y(t) = e^t$ satisfies the differential equation $\displaystyle \frac{dy}{dt}=y\text{.}$

Solution

We differentiate $y(t) = e^t$ with respect to $t\text{,}$ and find

\begin{alignat*}{1} \frac{d}{dt} y(t) &= \frac{d}{dt} e^t\\ &= e^t\\ &= y(t), \end{alignat*}

that is, $\displaystyle \frac{d}{dt} y(t)=y(t)$ or, when writing $y(t)$ simply as $y\text{,}$ $\displaystyle \frac{dy}{dt}=y\text{,}$ as required.

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But observe that

$y(t) = 5e^t$ also is a solution, and

$y(t) = \pi e^t\text{,}$ and

$y(t) = \sqrt{2} e^t\text{,}$ etc. In general,

$y(t) = C e^t,$ where

$C$ is a real number, is a solution (we will see later that

$C$ stems from a constant of integration).

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Checkpoint 7.1.6.

Confirm that $y(t) = Ce^t$ satisfies the differential equation $\displaystyle \frac{dy}{dt}=y\text{.}$

Solution

We differentiate $y(t) = Ce^t$ with respect to $t\text{,}$ and find

\begin{alignat*}{1} \frac{d}{dt} y(t) &= \frac{d}{dt} C e^t\\ &= C e^t\\ &= y(t), \end{alignat*}

that is, $\displaystyle \frac{d}{dt} y(t)=y(t)$ or, when writing $y(t)$ simply as $y\text{,}$ $\displaystyle \frac{dy}{dt}=y\text{,}$ as required.

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Subsection 7.1.3 The World's Most Important Differential Equation

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The World's Most Important Differential Equation simply is the generalized version of the second differential equation from the previous section, namely

$\frac{dy}{dt} = \lambda y,$ where

$\lambda \ne 0$ is a constant.

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By inspection,

$y(t) = e^{\lambda t}$ is a solution, but so is

$y(t) = C e^{\lambda t}\text{.}$

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Checkpoint 7.1.7.

Confirm that $y(t) = Ce^{\lambda t}$ satisfies the differential equation $\displaystyle \frac{dy}{dt}=\lambda y\text{.}$

Solution

We differentiate $y(t) = Ce^{\lambda t}$ with respect to $t\text{,}$ keeping the Chain Rule in mind, and find

\begin{alignat*}{1} \frac{d}{dt} y(t) &= \frac{d}{dt} C e^{\lambda t}\\ &= \lambda C e^{\lambda t} \cdot \lambda \\ &= \lambda \left( C e^{\lambda t} \right)\\ &= \lambda y(t), \end{alignat*}

that is, $\displaystyle \frac{d}{dt} y(t) = \lambda y(t)$ or, when writing $y(t)$ simply as $y\text{,}$ $\displaystyle \frac{dy}{dt} = \lambda y\text{,}$ as required.

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We refer to the solution

$y(t) = Ce^{\lambda t}$ as the general solution. The constant

$C$ can take on any value, so we have a family of solutions. Members of the family include

$y(t) = e^{\lambda t}\text{,}$

$y(t) = 4 e^{\lambda t}\text{,}$

$y(t) = -\frac{1}{\pi} e^{\lambda t}\text{,}$ etc.

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If we know that the solution passes through the point

$(t_0, y_0)\text{,}$ generally representing an initial condition, then we can identify one member from this family. This one member is called the particular solution.

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Checkpoint 7.1.8. Determining the Particular Solution from a Given Initial Condition at 0.

Find the particular solution for the differential equation $\displaystyle \frac{dy}{dt}=- y$ subject to the initial condition $y(0) = 5\text{.}$

Solution

The general solution is $y(t) = C e^{-t}\text{.}$

Substituting $t=0$ gives $y(0) = C e^{0} = C\text{.}$ But we are given that $y(0) = 5\text{.}$ Therefore $C = 5\text{,}$ and the particular solution is $$y(t) = 5 e^{-t}.$$

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Checkpoint 7.1.9. Determining the Particular Solution to the World's Most Important Differential Equation.

Find the particular solution for the differential equation $\displaystyle \frac{dy}{dt} = \lambda y$ subject to the initial condition $y(t_0) = y_0\text{.}$

Solution

The general solution is $y(t) = C e^{\lambda t}\text{.}$

Substituting $t=t_0$ gives $y(t_0) = C e^{\lambda t_0}\text{.}$ But we are given that $y(t_0) = y_0\text{.}$ Therefore, $$ C e^{\lambda t_0}= y_0, $$ that is, $$ C = y_0 e^{-\lambda t_0}. $$

Substituting this result into the general solution gives

\begin{alignat*}{1} y(t) &= C e^{\lambda t}\\ &= \left( y_0 e^{-\lambda t_0} \right) e^{\lambda t} \\ &= y_0 e^{\lambda t} e^{-\lambda t_0}\\ &= y_0 e^{\lambda t - \lambda t_0}\\ &= y_0 e^{\lambda (t - t_0)}. \end{alignat*}

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Subsection 7.1.4 Separable Differential Equations

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A separable equation is a first-order differential equation in the form

$\frac{dy}{dt} = f(t) g(y)$ because we can separate the independent and dependent variables.

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Remark 7.1.10.

A first-order differential equation is a differential equation that involves only the first derivative of the unknown function

$y(t)\text{.}$

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In particular, we can rewrite a separable differential equation in the following differential form

$\frac{dy}{g(y)} = f(t) \ dt.$ Note that the dependent variable

$y$ is only on the left-hand side of the equation, while the independent variable

$t$ is only on the right-hand side of the equation.

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Integrating both sides of the equation gives

$\int \frac{dy}{g(y)} = \int f(t) \ dt.$

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At this point, we use the integration techniques from the first few weeks of this course to determine the antiderivatives of both sides. In general, the result is an equation that defines

$y$ implicitly as a function of

$t\text{.}$ In some cases, it is possible to solve for

$y$ explicitly in terms of

$t\text{.}$

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The World's Most Important Differential Equation is a separable differential equation. In the section above, we determined its solution by inspection. In the following example, we determine the solution by following the steps outlined for separable differential equations.

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Example 7.1.11. Solving the World's Most Important Differential Equation by Treating it as a Separable Differential Equation.

Determine the general solution for the World's Most Important Differential Equation $$\displaystyle \frac{dy}{dt} = \lambda y.$$

Solution

We separate the differential equation to give $$ \frac{dy}{y} = \lambda \ dt. $$ Next, we determine the antiderivatives on both sides to give $$ \ln|y| = \lambda t + C_1. $$ Note that one constant of integration suffices (here called $C_1\text{,}$ so as not to confuse it with the eventual $C$ in the solution).

Exponentiating both sides gives

\begin{alignat*}{1} |y| &= e^{ \lambda t + C_1}\\ &= e^{\lambda t} e^{C_1}. \end{alignat*}

Then $$ y = \pm e^{C_1} e^{\lambda t}. $$ Letting $C = \pm e^{C_1}\text{,}$ we obtain the familiar general solution $$ y(t) = C e^{\lambda t}. $$

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Checkpoint 7.1.12.

Solve the separable differential equation $\displaystyle \frac{dy}{dx} = y^2 e^{-x}$ subject to the initial condition $\displaystyle y(0) = \frac{1}{2}\text{.}$

Answer

The particular solution is $\displaystyle y(x) = \frac{1}{e^{-x}+1}\text{.}$

Solution

We separate the variables and integrate to give

\begin{alignat*}{1} \int \frac{dy}{y^2} &= \int e^{-x} \ dx \\ -\frac{1}{y} &= -e^{-x} + C\\ \frac{1}{y} &= e^{-x} - C \\ y &= \frac{1}{e^{-x} - C}. \end{alignat*}

Note that we are able to solve explicitly for $y\text{.}$ Keep in mind that this is not always possible. But here, we can conclude that the general solution is $$ y(x) = \frac{1}{e^{-x} - C}. $$ Now we apply the initial condition $\displaystyle y(0) = \frac{1}{2}$ to determine $C\text{.}$ Setting $x=0$ in the general solution gives $$y(0) = \frac{1}{e^{0} - C} = \frac{1}{1-C} = \frac{1}{2}.$$ Solving for $C$ (exercise) gives $$C=-1.$$ Substituting $C=-1$ into the general solution gives

\begin{alignat*}{1} y(x) &= \frac{1}{e^{-x} - C} \\ &= \frac{1}{e^{-x} - (-1)} \\ &= \frac{1}{e^{-x} + 1}. \end{alignat*}

Therefore the particular solution is $\displaystyle y(x) = \frac{1}{e^{-x}+1}\text{.}$

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Subsection 7.1.5 Summary

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Differential Equations.

A differential equation is an equation that contains an unknown function and one or more of its derivatives.
The solution to a differential equation is a function.
The form of the derivative(s) in a differential equation gives information about the independent and dependent variables. For example:
- If the equation involves $dy/dx\text{,}$ then we seek a function $y(x)$ that solves the differential equation.
- If the equation involves $dv/dt\text{,}$ then we seek a function $v(t)$ that solves the differential equation.
The solution to a differential equation always involves a constant of integration.
- If no further information is provided that allows us to determine the value of the constant of integration, then we have the general solution to a differential equation, which represents a family of solutions.
- If we have information that does allow us to determine the value of the constant of integration, then we have the particular solution, which represents one member of the family of solutions from above.

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The World's Most Important Differential Equation.

The general solution to the World's Most Important Differential Equation $\frac{dy}{dt} = \lambda y$ is $y(t) = C e^{\lambda t},$ where $C$ is an arbitrary constant.
The particular solution to the World's Most Important Differential Equation $\frac{dy}{dt} = \lambda y$ subject to initial condition $y(t_0) = y_0$ is $y(t) = y_0 e^{\lambda (t-t_0)}.$

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Separable Differential Equations.

A separable equation is a first-order differential equation in which the expression for $dy/dt$ can be factored as a function of $t$ times (or divided by) a function of $y\text{.}$ In other words, it can be written in the form $\frac{dy}{dt} = f(t) g(y).$
To solve a separable differential equation:
- Write the differential equation in differential form: $\frac{dy}{g(y)} = f(t) dt.$
- Integrate both sides of the equation in differential form: $\int \frac{dy}{g(y)} = \int f(t) dt.$ In general, the result is an equation that defines $y$ implicitly as a function of $t\text{.}$
- If possible, solve for $y$ explicitly in terms of $t\text{.}$

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Subsection 7.1.6 Don't Forget

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Don't forget to return to eClass to complete the pre-class quiz.

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Subsection 7.1.7 Further Study

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Remember that the notes presented above only serve as an introduction to the topic. Further study of the topic will be required. This includes working through the pre-class quizzes, reviewing the lecture notes, and diligently working through the homework problems.

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As you study, you should reflect on the following learning outcomes, and critically assess where you are on the path to achieving these learning outcomes:

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Learning Outcomes

Recognize the World's Most Important Differential Equation, recall its solution, and interpret the solution in terms of exponential growth or decay.
Recognize a separable differential equation and determine its solution.

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The following references provide a good start for review and further study:

Learning Outcome	Video	Textbook Section
1	N/A	7.4
2	N/A	7.4