If you want to describe the world around you, be it the forces acting on a body, the growth of a virus or the temperature of the coffee in your cup, you will be dealing with differential equations. On this page we will look at the simplest type: differential equations in which we can separate the variables. To be successful with the topic on this page, you will need to have strong foundations in the following areas: integration techniques, manipulation of logarithms and exponents and partial fractions. It is recommended that you refresh your techniques in those areas if you are not feeling 100% confident.
Key Concepts
On this page, you should learn to
recognise differential equations with separable variables - in the form \(\frac{dy}{dx}=f(x)g(y)\)
solve these differential equations using integration
Essentials
The following videos will help you understand all the concepts from this page
Example 1 - Finding a General Solution
In the following video we look at how we can solve separable differential equations
In the following video, we will see an example of finding a particular solution to a differential equation. The variables in this differential equation are separable
To solve differential equations, we are often required to manipulate exponents and logarithms. Here is a quiz that will give you some practice in what you need
Which of the following is equivalent to \(e^{x+y}\)?
From the first law of indices \(a^{x+y}=a^x \cdot a^y\)
Which if the following is equivalent to \(e^{x-y}\)
From the first law of indices \(a^{x-y}=a^{x+(-y)}=a^x \cdot a^{-y}\)
From the second law of indices \(a^{x-y}=a^x \div a^y\)
Select the correct answers to the following
01x2xx²
\(\large e^x \cdot e^{-x}=\)
\(\large e^{lnx}=\)
\(\large ln(e^x)=\)
Select the correct answers to the following
01x2xx²
\(\large e^{lnx²}=\)
\(\large e^{2lnx}=\)
\(\large ln(e^{x²})=\)
Select the correct answers to the following
cosxcos²x2cosxln(cosx)
\(\large e^{ln(cosx)}=\)
\(\large e^{2ln(cosx)}=\)
\(\large ln(e^{2cosx})=\)
When giving the general solution to differential equation, we have to manipulate arbitrary constants. Try this quiz to ensure that you know how to do this
Dealing with Arbitrary Constants of Integration 1/1
Following an integration, the result is simplified. Decide whether the second statement is a correct simplification of the first one, given that \(\large c\) and \(\large c_1\) are arbitrary constants
\(\large y=e^{x+c}\)
\(\large y=c_1e^{x}\)
We can write \(\large y=e^{x+c}\) as \(\large y=e^{x}\cdot e^{c}\)
\(\large e^c\) is an arbitrary constant
Following an integration, the result is simplified. Decide whether the second statement is a correct simplification of the first one, given that \(\large c\) and \(\large c_1\) are arbitrary constants
\(\large y=lnx+c\)
\(\large y=ln(c_1x)\)
We can write \(\large y=lnx+c\) as \(\large y=lnx+lnc_1\), since \(lnc_1\) is an arbitrary constant
This can be simplified to \(\large y=ln(c_1x)\)
Following an integration, the result is simplified. Decide whether the second statement is a correct simplification of the first one, given that \(\large c\) and \(\large c_1\) are arbitrary constants
\(\large y^2=x+c\)
\(\large y=\pm \sqrt{x}+c_1\)
The correct simplification is \(\large y=\pm \sqrt{x+c_1}\)
Following an integration, the result is simplified. Decide whether the second statement is a correct simplification of the first one, given that \(\large c\) and \(\large c_1\) are arbitrary constants
\(\large \sqrt {y}=sinx +c\)
\(\large y=sin^2x+c_1\)
The correct simplification is \(\large y=(sinx+c_1)^2\)
Following an integration, the result is simplified. Decide whether the second statement is a correct simplification of the first one, given that \(\large c\) and \(\large c_1\) are arbitrary constants
\(\large lny=lnx+c\)
\(\large y=c_1x\)
We can write \(\large lny=lnx+c\) as \(\large lny=lnx+lnc_1\), since \(lnc_1\) is an arbitrary constant
This can be simplified to \(\large lny=ln(c_1x)\)
which can be simplified to \(\large y=c_1x\)
Following an integration, the result is simplified. Decide whether the second statement is a correct simplification of the first one, given that \(\large c\) and \(\large c_1\) are arbitrary constants
\(\large lny=x^2+c\)
\(\large y=c_1e^{x^2}\)
We can write \(\large lny=x^2+c\) as \(\large y=e^{x^2+c}\)
This can be write as \(\large y=e^{x^2}\cdot e^{c}\)
\(\large e^c\) is an arbitrary constant
Therefore, \(\large y=c_1e^{x^2}\)
Following an integration, the result is simplified. Decide whether the second statement is a correct simplification of the first one, given that \(\large c\) and \(\large c_1\) are arbitrary constants
\(\large y^3=xlnx+c\)
\(\large y=\sqrt[3]{xlnx+c_1}\)
This is a correct simplification
Following an integration, the result is simplified. Decide whether the second statement is a correct simplification of the first one, given that \(\large c\) and \(\large c_1\) are arbitrary constants
\(\large siny=e^{-x}+c\)
\(\large y=c_1arcsin(e^{-x})\)
We can simplify to \(\large y=arcsin(e^{-x}+c)\)
Following an integration, the result is simplified. Decide whether the second statement is a correct simplification of the first one, given that \(\large c\) and \(\large c_1\) are arbitrary constants
\(\large \frac{1}{x}y=lnx+c\)
\(\large y=xlnx+c_1\)
\(\large \frac{1}{x}y=lnx+c\) can be simplified to \(\large \frac{1}{x}y=lnx+lnc_1\)
which can be written as \(\large \frac{1}{x}y=ln(c_1x)\)
And the general solution is \(\large y=xln(c_1x)\)
Following an integration, the result is simplified. Decide whether the second statement is a correct simplification of the first one, given that \(\large c\) and \(\large c_1\) are arbitrary constants
\(\large y=ln(x-1)-ln(x+1)+c\)
\(\large y=ln\frac{c_1(x-1)}{x+1}\)
We can write \(\large y=ln(x-1)-ln(x+1)+c\) as \(\large y=ln(x-1)-ln(x+1)+lnc_1\)
And use log laws to simplify this to \(\large y=ln\frac{c_1(x-1)}{x+1}\)
Here is a quiz that gets you to practise solving variables separable differential equations