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Solution of Bernoulliās Equations Academic Resource Center
Contents
ā¢ First order ordinary differential equation
ā¢ Linearity of Differential Equations
ā¢ Typical Form of Bernoulliās Equation
ā¢ Examples of Bernoulliās Equations
ā¢ Method of Solution
ā¢ Bernoulli Substitution
ā¢ Example Problem
ā¢ Practice Problems
First Order differential equations
ā¢ A differential equation having a first derivative as the highest derivative is a first order differential equation.
ā¢ If the derivative is a simple derivative, as opposed to a partial derivative, then the equation is referred to as ordinary.
Linearity of Differential Equations
ā¢ The terminology ālinearā derives from the description of a line.
ā¢ A line, in its most general form, is written as š“š„ + šµš¦ = š¶
ā¢ Similarly, if a differential equation is written as š š„šš¦
šš„+
š š„ š¦ = ā(š„)
ā¢ Then this equation is termed linear, as the highest power of šš¦
šš„
and y is 1. They are analogous to x and y in the equation of a line, hence the term linear.
Typical form of Bernoulliās equation ā¢ The Bernoulli equation is a Non-Linear differential equation of
the form šš¦
šš„+ š š„ š¦ = š(š„)š¦š
ā¢ Here, we can see that since y is raised to some power n where nā 1.
ā¢ This equation cannot be solved by any other method like homogeneity, separation of variables or linearity.
Examples
ā¢ Here are a few examples of Bernoulliās Equation
š„šš¦
šš„+ š¦ = š„2š¦2
š¦2š„šš¦
šš„= š„š¦3 + 1
š”2šš¦
šš”+ š¦2 = š”š¦
Method of Solution
ā¢ The first step to solving the given DE is to reduce it to the standard form of the Bernoulliās DE. So, divide out the whole expression to get the coefficient of the derivative to be 1.
ā¢ Then we make a substitution š¢ = š¦1āš
ā¢ This substitution is central to this method as it reduces a non-linear equation to a linear equation.
Bernoulli Substitution
ā¢ So if we have š¢ = š¦1āš,
then šš¢
šš„= (1 ā š)š¦āš
šš¦
šš„
ā¢ From this, replace all the yās in the equation in terms of u and
replace šš¦
šš„ in terms of
šš¢
šš„ and u.
ā¢ This will reduce the whole equation to a linear differential equation.
Example Problem
Let us try to solve the given differential equation
š„2šš¦
šš„ā š¦2 = 2š„š¦
First of all, we rearrange it so that the derivative and first power of y are on one side.
š„2šš¦
šš„ā 2š„š¦ = š¦2
Then we divide out by š„2 to make the co-efficient of the derivative equal to 1.
Contd.
Then, that leaves us with šš¦
šš„ā2
š„š¦ =
1
š„2š¦2
Here, we identify the power on the variable y to be 2. Therefore, we make the substitution
š¢ = š¦1ā2 = š¦ā1
Thus, šš¢
šš„= āš¦ā2
šš¦
šš„
Contd.
Then, knowing that: šš¢
šš„= āš¦ā2
šš¦
šš„ and š¢ = š¦ā1
We have š¦ =1
š¢ and ā
1
š¢2šš¢
šš„=šš¦
šš„
We substitute these expressions into our Bernoulliās equation and that gives us
ā1
š¢2šš¢
šš„ā2
š„
1
š¢=1
š„21
š¢2
Contd.
Once we have reached this stage, we can multiply out by āš¢2 and then the DE simplifies to
šš¢
šš„+2
š„š¢ = ā
1
š„2
This is now a linear differential equation in u and can be solved by the use of the integrating factor. Thatās the rational behind the use of the specialized substitution.
Contd.
Proceeding, we identify š š„ =2
š„, we have the integrating
factor
š¼ š„ = š 2š„šš„
š¼ š„ = š2ln (š„) = š„2
Then multiplying through by š¼ š„ , we get
š„2šš¢
šš„+ 2š„š¢ = ā1
Which simplifies to š
šš„š„2š¢ = ā1
Contd.
We now integrate both sides,
š
šš„š„2š¢ = ā1šš„
This gives š„2š¢ = āš„ + š¶
š¢ =ā1
š„+š¶
š„2
We have now obtained the solution to the simplified DE. All we
have to do now is to replace š¢ =1
š¦
Contd.
Our final answer will therefore look like
š¦ =ā1
š„+š¶
š„2
ā1
Practice Problems
Try practicing these problems to get the hang of the method. You can also try solving the DEs given in the examples section.
1. š„šš¦
šš„ā 1 + š„ š¦ = š„š¦2
2.šš¦
šš„= š¦(š„š¦3 ā 1)
3. 3 1 + š”2šš¦
šš„= 2š”š¦(š¦3 ā 1)
References
ā¢ A first Course in Differential Equations 9th Ed., Dennis Zill.
ā¢ Elementary Differential Equations, Martin & Reissner
ā¢ Differential and Integral Calculus Vol 2, N. Piskunov
ā¢ Workshop developed by Abhiroop Chattopadhyay