\( \newcommand{\abs}[1]{\left| \, {#1} \, \right| } \) \( \newcommand{\cm}{\mathrm{cm} } \) \( \newcommand{\sec}{ \, \mathrm{sec} \, } \) \( \newcommand{\units}[1]{\,\text{#1}} \) \( \newcommand{\vhat}[1]{\,\hat{#1}} \) \( \newcommand{\vhati}{\,\hat{i}} \) \( \newcommand{\vhatj}{\,\hat{j}} \) \( \newcommand{\vhatk}{\,\hat{k}} \) \( \newcommand{\vect}[1]{\boldsymbol{\vec{#1}}} \) \( \newcommand{\norm}[1]{\|{#1}\|} \) \( \newcommand{\arccot}{ \, \mathrm{arccot} \, } \) \( \newcommand{\arcsec}{ \, \mathrm{arcsec} \, } \) \( \newcommand{\arccsc}{ \, \mathrm{arccsc} \, } \) \( \newcommand{\sech}{ \, \mathrm{sech} \, } \) \( \newcommand{\csch}{ \, \mathrm{csch} \, } \) \( \newcommand{\arcsinh}{ \, \mathrm{arcsinh} \, } \) \( \newcommand{\arccosh}{ \, \mathrm{arccosh} \, } \) \( \newcommand{\arctanh}{ \, \mathrm{arctanh} \, } \) \( \newcommand{\arccoth}{ \, \mathrm{arccoth} \, } \) \( \newcommand{\arcsech}{ \, \mathrm{arcsech} \, } \) \( \newcommand{\arccsch}{ \, \mathrm{arccsch} \, } \)

17Calculus Differential Equations - Separation Of Variables

17Calculus
Single Variable Calculus
Derivatives
Integrals
Multi-Variable Calculus
Precalculus
Functions

classification: this technique applies to first-order, separable differential equations

An equation that allows us to separate the variables is called a separable differential equation.

Resources

This is one of the simplest techniques when solving differential equations and sometimes it is introduced in first year calculus while studying integration. This technique is used when it is possible to isolate the variables, using algebra and differentials, on separate sides of the equal sign. This allows us to integrate both sides of the equation individually with only one variable on each side.

The idea with this technique is that the differential equation is in a form where we can isolate the two variables to each side of the equal sign. Then we can integrate each side separately. It's really that easy. The trick is to use algebra to get the equation into the right form.

Separable Equation Form

Some textbooks write separable equations as \(\displaystyle{ N(y)\frac{dy}{dx}=M(x) }\). This notation tells you that \(N(y)\) is some function that contains only \(y\)'s and maybe some constants. Similarly, \(M(x)\) is some function that contains only \(x\)'s and maybe some constants.

A better way to write a separable equation is an equation of the form \(\displaystyle{ \frac{dy}{dx}=g(x)h(y) }\). In this equation, it is easy to see what needs to done. We need to have two functions multipled together, with one having only \(x\)'s and the other with only \(y\)'s.

Before we go on, here is a short video clip explaining this and showing some examples of separable equations and other equations that are NOT separable.

Michael Penn - Differential Equations | Separation of Variables

video by Michael Penn

General Technique To Solve Separable Differential Equations

Solving these types of differential equations are quite easy. You may even have seen them in calculus. Here is a quick video clip on how to solve them in general.

Michael Penn - Differential Equations | Separation of Variables

video by Michael Penn

Okay, let's do an example to demonstrate the separation of variables technique. As we work this example, we will give you a heads up on something. Some instructors (most, from what I've seen), do a trick with constants that confuse a lot of people. We show you where it comes from.

Example

Find \(y\) for \( y' = x \).

For this separable equation, using the form \(\displaystyle{ \frac{dy}{dx}=g(x)h(y) }\), we have \(g(x)=x\) and \(h(y)=1\)

\(\displaystyle{ y' = x ~~~ \to ~~~ \frac{dy}{dx} = x }\)

First, we rewrite the derivative so that it is more obvious what we need to do.

\(\displaystyle{ \frac{dy}{dx} = x ~~~ \to ~~~ dy = x~dx}\)

Now we write the differential equation by 'moving' the \(dx\) to the other side. It looks like we are multiplying \(dx\) on both sides but that's not what is really happening. However, it helps me remember what to do by thinking of it this way. What we are doing is writing the equation in differential form.

\(\begin{array}{rcl} \int{dy} & = & \int{x~dx} \\ y + c_1 & = & \frac{x^2}{2} + c_2 \end{array}\)

Next, we integrate both sides. On the left we integrate with respect to \(y\), on the right with respect to \(x\).

\(\displaystyle{ y = \frac{x^2}{2} + c_2-c_1 }\)

Okay, so now we have two constants, one from each integration (don't forget these constants; they are extremely important when solving differential equations). We are going to move \(c_1\) to the other side and combine it with \(c_2\).

\(\displaystyle{ y = \frac{x^2}{2} + C }\)

Let \( C = c_2 - c_1 \) to get our final answer.

Notes
1. This is a general solution. To get the particular solution, we need initial conditions to determine the value of \(C\).
2. Many instructors skip the steps showing the two constants and combining them, i.e. \( C = c_2 - c_1 \) . They just jump to one constant. This can be confusing at first, if you don't know what is happening. Here is a video explaining this in more detail.

Krista King Math - How can the constant C absorb all that other stuff? [3mins-54secs]

video by Krista King Math

And, that's it! That's all there is to it. Just separate the variables and integrate. There is nothing fancy going on and no tricks. But this is a basic technique that you need to understand how to do and practice thoroughly since, later on in your study of differential equations, you will learn other techniques to convert more complicated equations into this form.

Here is an in-depth video discussing first-order linear equations, separation of variables and steady-state and transient solutions. If you are first starting to learn differential equations, this may be a bit above you but you can still get a lot out of it and being exposed to a little more advanced techniques will help you learn them later.

MIT OCW - Solving First-order Linear ODE's; Steady-state and Transient Solutions.

video by MIT OCW

After you work through the practice problems we recommend learning about first-order, linear differential equations next.

How to Read and Do Proofs: An Introduction to Mathematical Thought Processes

Practice

Unless otherwise instructed, solve these separable differential equations and find the particular solution, if possible. Give your answers in exact terms.

\(\displaystyle{ \frac{dy}{dx} = xy }\)

Problem Statement

\(\displaystyle{ \frac{dy}{dx} = xy }\)

Solution

Dr Chris Tisdell - 559 video solution

video by Dr Chris Tisdell

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\(\displaystyle{ \frac{dy}{dx} = y^2 (1+x^2) }\); \( y(0) = 1 \)

Problem Statement

\(\displaystyle{ \frac{dy}{dx} = y^2 (1+x^2) }\); \( y(0) = 1 \)

Solution

Dr Chris Tisdell - 560 video solution

video by Dr Chris Tisdell

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\(\displaystyle{ \frac{dy}{dx} = \cos(x) }\); \( y(0) = -1 \)

Problem Statement

\(\displaystyle{ \frac{dy}{dx} = \cos(x) }\); \( y(0) = -1 \)

Solution

PatrickJMT - 562 video solution

video by PatrickJMT

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\(\displaystyle{ \frac{dy}{dx} = x/y^2 }\)

Problem Statement

\(\displaystyle{ \frac{dy}{dx} = x/y^2 }\)

Solution

PatrickJMT - 563 video solution

video by PatrickJMT

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\(\displaystyle{ \frac{dy}{dx} \cdot \frac{y^3+y}{x^2+3x} = 1 }\)

Problem Statement

\(\displaystyle{ \frac{dy}{dx} \cdot \frac{y^3+y}{x^2+3x} = 1 }\)

Solution

PatrickJMT - 564 video solution

video by PatrickJMT

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\(\displaystyle{ \frac{dy}{dx} = \frac{\cos(x)}{y-1} }\)

Problem Statement

\(\displaystyle{ \frac{dy}{dx} = \frac{\cos(x)}{y-1} }\)

Solution

PatrickJMT - 565 video solution

video by PatrickJMT

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\(\displaystyle{ \frac{dy}{dx} = e^{4x-y} }\); \( y(0) = 5 \)

Problem Statement

\(\displaystyle{ \frac{dy}{dx} = e^{4x-y} }\); \( y(0) = 5 \)

Solution

PatrickJMT - 566 video solution

video by PatrickJMT

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\(\displaystyle{ \frac{dy}{dx} = 2x \sqrt{y-1} }\)

Problem Statement

\(\displaystyle{ \frac{dy}{dx} = 2x \sqrt{y-1} }\)

Solution

Krista King Math - 568 video solution

video by Krista King Math

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\(\displaystyle{ \frac{y+2}{x^2-x+2} \frac{dy}{dx} = \frac{x}{y} }\); \( y(1) = 2 \)

Problem Statement

\(\displaystyle{ \frac{y+2}{x^2-x+2} \frac{dy}{dx} = \frac{x}{y} }\); \( y(1) = 2 \)

Solution

PatrickJMT - 567 video solution

video by PatrickJMT

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\(\displaystyle{ y \frac{dy}{dx} = x^2 + \sech^2(x) }\); \( y(0) = 2 \)

Problem Statement

\(\displaystyle{ y \frac{dy}{dx} = x^2 + \sech^2(x) }\); \( y(0) = 2 \)

Solution

Dr Chris Tisdell - 561 video solution

video by Dr Chris Tisdell

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\( 2x + 3 + (2y-2)y' = 0 \)

Problem Statement

\( 2x + 3 + (2y-2)y' = 0 \)

Final Answer

\( x^2 + 3x + y^2 - 2y = C \)

Problem Statement

\( 2x + 3 + (2y-2)y' = 0 \)

Solution

\( 2x + 3 + (2y-2)y' = 0 \)

\( (2y-2)y' = -2x-3 \)

\( (2y-2)~dy = (-2x-3)~dx \)

\( \int{ 2y-2 ~dy } = \int{ -2x-3 ~dx } \)

\( y^2 - 2y = -x^2-3x + C \)

\( x^2+3x+y^2-2y = C \)

Final Answer

\( x^2 + 3x + y^2 - 2y = C \)

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\(\displaystyle{ \frac{dy}{dx} = \frac{x^2+1}{x^2(3y^2+1)} }\)

Problem Statement

\(\displaystyle{ \frac{dy}{dx} = \frac{x^2+1}{x^2(3y^2+1)} }\)

Solution

Krista King Math - 569 video solution

video by Krista King Math

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\(\displaystyle{ \frac{dy}{dx} = \frac{4-2x}{3y^2-5} }\); \( y(1) = 3 \)

Problem Statement

\(\displaystyle{ \frac{dy}{dx} = \frac{4-2x}{3y^2-5} }\); \( y(1) = 3 \)

Solution

Krista King Math - 570 video solution

video by Krista King Math

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\(\displaystyle{ \frac{dy}{dx} = 2 \sqrt{y} }\); \( y(0) = 9 \)

Problem Statement

\(\displaystyle{ \frac{dy}{dx} = 2 \sqrt{y} }\); \( y(0) = 9 \)

Solution

Krista King Math - 571 video solution

video by Krista King Math

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\(\displaystyle{ \frac{dy}{dx} = \frac{x^2}{1-y^2} }\)

Problem Statement

\(\displaystyle{ \frac{dy}{dx} = \frac{x^2}{1-y^2} }\)

Final Answer

\( x^3 + y^3 - 3y = C \)

Problem Statement

\(\displaystyle{ \frac{dy}{dx} = \frac{x^2}{1-y^2} }\)

Solution

\(\displaystyle{ \frac{dy}{dx} = \frac{x^2}{1-y^2} }\)

\( (1-y^2)~dy = x^2~dx \)

\( \int{ 1-y^2 ~dy } = \int{ x^2 ~dx } \)

\(\displaystyle{ y - \frac{y^3}{3} + c_1 = \frac{x^3}{3} }\)

\( 3y - y^3 + 3c_1 = x^3 \)

\( 3c_1 = x^3 + y^3 - 3y \)

Let \( C = 3c_1 \)

\( C = x^3 + y^3 - 3y \)

Since we were not given an initial condition, we cannot get a value for \(C\). So this is the general solution.

Final Answer

\( x^3 + y^3 - 3y = C \)

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\(\displaystyle{ \frac{dy}{dx} = \frac{y \cos(x)}{1+2y^2} }\); \( y(0) = 1 \)

Problem Statement

\(\displaystyle{ \frac{dy}{dx} = \frac{y \cos(x)}{1+2y^2} }\); \( y(0) = 1 \)

Final Answer

\( \ln |y| + y^2 = \sin x + 1 \)

Problem Statement

\(\displaystyle{ \frac{dy}{dx} = \frac{y \cos(x)}{1+2y^2} }\); \( y(0) = 1 \)

Solution

\(\displaystyle{ \frac{dy}{dx} = \frac{y \cos(x)}{1+2y^2} }\)

\(\displaystyle{ \frac{1+2y^2}{y} ~dy = \cos x ~ dx }\)

\(\displaystyle{ \left[ \frac{1}{y} + 2y \right] ~dy = \cos x ~ dx }\)

\(\displaystyle{ \int{ \left[ \frac{1}{y} + 2y \right] ~dy } = \int{ \cos x ~ dx } }\)

\( \ln |y| + y^2 = \sin x + C \)

Use the initial condition \( y(0) = 1 \) to find the constant \(C\)

\( \ln 1 + 1 = \sin 0 + C \to C = 1 \)

\( \ln |y| + y^2 = \sin x + 1 \)

Final Answer

\( \ln |y| + y^2 = \sin x + 1 \)

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\(\displaystyle{ \frac{du}{dt} = \frac{2t+\sec^2(t)}{2u} }\); \( u(0) = -5 \)

Problem Statement

\(\displaystyle{ \frac{du}{dt} = \frac{2t+\sec^2(t)}{2u} }\); \( u(0) = -5 \)

Solution

Krista King Math - 609 video solution

video by Krista King Math

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\(\displaystyle{ \frac{dy}{dx} = 2x \sqrt{y} }\); \( y(0)=1 \)

Problem Statement

\(\displaystyle{ \frac{dy}{dx} = 2x \sqrt{y} }\); \( y(0)=1 \)

Solution

Dr Chris Tisdell - 1842 video solution

video by Dr Chris Tisdell

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\(\displaystyle{ \frac{dy}{dx}=e^{x-y} }\); \( y(0)=\ln(2) \)

Problem Statement

\(\displaystyle{ \frac{dy}{dx}=e^{x-y} }\); \( y(0)=\ln(2) \)

Solution

Dr Chris Tisdell - 1843 video solution

video by Dr Chris Tisdell

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\(\displaystyle{ \frac{dy}{dt} = \frac{t}{t^2y+y} }\), \( y(0) = 3 \), \( y > 0 \)

Problem Statement

\(\displaystyle{ \frac{dy}{dt} = \frac{t}{t^2y+y} }\), \( y(0) = 3 \), \( y > 0 \)

Final Answer

\( \sqrt{ \ln(t^2+1) + 9 } \)

Problem Statement

\(\displaystyle{ \frac{dy}{dt} = \frac{t}{t^2y+y} }\), \( y(0) = 3 \), \( y > 0 \)

Solution

MIP4U - 2082 video solution

video by MIP4U

Final Answer

\( \sqrt{ \ln(t^2+1) + 9 } \)

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\( dv/ds = (s+1)/(sv+s) \)

Problem Statement

\( dv/ds = (s+1)/(sv+s) \)

Final Answer

\( v = -1 \pm \sqrt{ 2s + 2\ln(s) + C } \)

Problem Statement

\( dv/ds = (s+1)/(sv+s) \)

Solution

From the \(dv/ds\) term, we can tell that s is the independent variable and v is the dependent variable, so that the solution will be \(v(s)\). Let's try separation of variables.

\(\displaystyle{ dv/ds = \frac{(s+1)}{(sv+s)} }\)

\(\displaystyle{ dv/ds = \frac{(s+1)}{s(v+1)} }\)

\(\displaystyle{ (v+1)\frac{dv}{ds} = \frac{(s+1)}{s} }\)

\(\displaystyle{ (v+1) dv = \frac{(s+1)}{s} ds }\)

\(\displaystyle{ \int{v+1 ~dv} = \int{ \frac{s+1}{s} ds } }\)

\(\displaystyle{ \int{v+1 ~dv} = \int{ 1 + \frac{1}{s} ds } }\)

\(\displaystyle{ \frac{v^2}{2} + v = s + \ln(s) + c_1 }\)

\( v^2 + 2v = 2s + 2\ln(s) + 2c_1 \)

Adding 1 to both sides after integrating allows us to complete the square.

\( v^2 + 2v + 1 = 2s + 2\ln(s) + 1 + 2c_1 \)

\( (v+1)^2 = 2s + 2\ln(s) + 1 + 2c_1 \)

\( v = \pm \sqrt{ 2s + 2\ln(s) + 1 + 2c_1 } - 1 \)

We can combine all constants under the square root into one by letting \(C=1+2c_1\). This is not required but cleans up the equation.

\( v = \pm \sqrt{ 2s + 2\ln(s) + C } - 1 \)

Most of this should be easy to follow. There just a few clarifications in order.
1. When taking the square root, don't assume your answer will be positive. Use \( \pm \) unless you can show that the result will always be positive (or negative).
2. You may not need to solve for \(v\). Check to see what your instructor requires.

Final Answer

\( v = -1 \pm \sqrt{ 2s + 2\ln(s) + C } \)

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\(\displaystyle{ \frac{dy}{dx} = \frac{3x^2+4x+2}{2(y-1)} }\); \( y(0) = -1 \)

Problem Statement

\(\displaystyle{ \frac{dy}{dx} = \frac{3x^2+4x+2}{2(y-1)} }\); \( y(0) = -1 \)

Final Answer

\( y^2 - 2y = x^3 + 2x^2 + 2x + 3 \)

Problem Statement

\(\displaystyle{ \frac{dy}{dx} = \frac{3x^2+4x+2}{2(y-1)} }\); \( y(0) = -1 \)

Solution

\(\displaystyle{ \frac{dy}{dx} = \frac{3x^2+4x+2}{2(y-1)} }\)

\( 2(y-1) ~ dy = (3x^2+4x+2)dx \)

\( \int{ 2(y-1) ~ dy } = \int{ 3x^2+4x+2 ~ dx } \)

\( y^2 - 2y = x^3 + 2x^2 + 2x + C \)

Use the initial condition \( y(0) = -1 \) to find the constant \(C\)

\( 1 + 2 = 0 + 0 + 0 + C \to C = 3 \)

\( y^2 - 2y = x^3 + 2x^2 + 2x + 3 \)

Although technically, we could solve for \(y\) in the last equation, the form would be more complicated. However, check with your instructor to see what they expect.

Final Answer

\( y^2 - 2y = x^3 + 2x^2 + 2x + 3 \)

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Practice Search
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Practice Instructions

Unless otherwise instructed, solve these separable differential equations and find the particular solution, if possible. Give your answers in exact terms.

Do NOT follow this link or you will be banned from the site!

When using the material on this site, check with your instructor to see what they require. Their requirements come first, so make sure your notation and work follow their specifications.

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