\( \newcommand{\sech}{\mathrm{sech} } \) \( \newcommand{\csch}{\mathrm{csch} } \) \( \newcommand{\arcsinh}{\mathrm{arcsinh} } \) \( \newcommand{\arccosh}{\mathrm{arccosh} } \)
17calculus 17calculus
First Order Second Order Laplace Transforms Additional Topics Applications, Practice
Separation of Variables
Linear
Integrating Factors (Linear)
Substitution
Exact Equations
Integrating Factors (Exact)
Linear
Constant Coefficients
Substitution
Reduction of Order
Undetermined Coefficients
Variation of Parameters
Polynomial Coefficients
Cauchy-Euler Equations
Chebyshev Equations
Laplace Transforms
Unit Step Function
Unit Impulse Function
Square Wave
Shifting Theorems
Solve Initial Value Problems
Classify Differential Equations
Fourier Series
Slope Fields
Wronskian
Existence and Uniqueness
Boundary Value Problems
Euler's Method
Inhomogeneous ODE's
Resonance
Partial Differential Equations
Linear Systems
Exponential Growth/Decay
Population Dynamics
Projectile Motion
Chemical Concentration
Fluids (Mixing)
Practice Problems
Practice Exam List
Exam A1
Exam A3
Exam B2

You CAN Ace Differential Equations

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Separation Of Variables

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.

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.

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. Let's do an example to demonstrate the separation of variables technique.

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.

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

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? [3min54secs]

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.

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

first-order, linear differential equations →

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Practice Problems

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

Level A - Basic

Practice A01

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

solution

Practice A02

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

solution

Practice A03

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

solution

Practice A04

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

solution

Practice A05

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

solution

Practice A06

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

solution

Practice A07

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

solution

Practice A08

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

solution

Practice A09

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

solution

Practice A10

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

solution

Practice A11

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

solution

Practice A12

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

solution

Practice A13

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

solution

Practice A14

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

solution

Practice A15

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

solution

Practice A16

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

solution

Practice A17

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

solution

Practice A18

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

solution

Practice A19

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

answer

solution


Level B - Intermediate

Practice B01

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

answer

solution

Practice B02

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

solution

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