Solution of Bernoulli's Equations, Schemes and Mind Maps of Differential Equations

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Solution of

Bernoulli’s

Equations

Academic Resource Center

Contents

• First order ordinary differential equation
• Linearity of Differential Equations
• Typical Form of Bernoulli’s Equation
• Examples of Bernoulli’s Equations
• Method of Solution
• Bernoulli Substitution
• Example Problem
• Practice Problems

Linearity of Differential Equations

• The terminology ‘linear’ derives from the description of a line.
• A line, in its most general form, is written as 𝐴𝑥 + 𝐵𝑦 = 𝐶
• Similarly, if a differential equation is written as 𝑓 𝑥 𝑑𝑦𝑑𝑥 + 𝑔 𝑥 𝑦 = ℎ(𝑥)
• Then this equation is termed linear, as the highest power of 𝑑𝑦𝑑𝑥 and y is 1. They are analogous to x and y in the equation of a line, hence the term linear.

Typical form of Bernoulli’s

equation

• The Bernoulli equation is a Non-Linear differential equation of the form 𝑑𝑦𝑑𝑥 + 𝑃 𝑥 𝑦 = 𝑓(𝑥)𝑦𝑛
• Here, we can see that since y is raised to some power n where n≠1.
• This equation cannot be solved by any other method like homogeneity, separation of variables or linearity.

Method of Solution

• The first step to solving the given DE is to reduce it to the standard form of the Bernoulli’s DE. So, divide out the whole expression to get the coefficient of the derivative to be 1.
• Then we make a substitution 𝑢 = 𝑦1−𝑛
• This substitution is central to this method as it reduces a non- linear equation to a linear equation.

Bernoulli Substitution

• So if we have 𝑢 = 𝑦1−𝑛, then 𝑑𝑢𝑑𝑥 = (1 − 𝑛)𝑦−𝑛 𝑑𝑦𝑑𝑥
• From this, replace all the y’s in the equation in terms of u and replace 𝑑𝑦𝑑𝑥 in terms of 𝑑𝑢𝑑𝑥 and u.
• This will reduce the whole equation to a linear differential equation.

Contd.

Then, that leaves us with 𝑑𝑦 𝑑𝑥 −

𝑥^2 𝑦

2

Here, we identify the power on the variable y to be 2. Therefore, we make the substitution 𝑢 = 𝑦1−2^ = 𝑦− Thus, 𝑑𝑢 𝑑𝑥 = −𝑦

Contd.

Then, knowing that: 𝑑𝑢 𝑑𝑥 = −𝑦

−2 𝑑𝑦 𝑑𝑥 and^ 𝑢 = 𝑦

−

We have 𝑦 = (^1) 𝑢 and − (^) 𝑢^12 𝑑𝑢𝑑𝑥 = 𝑑𝑦𝑑𝑥

We substitute these expressions into our Bernoulli’s equation and that gives us

−

𝑢^2

𝑥^2

𝑢^2

Contd.

Proceeding, we identify 𝑃 𝑥 = (^2) 𝑥, we have the integrating factor 𝐼 𝑥 = 𝑒

2 𝑥𝑑𝑥 𝐼 𝑥 = 𝑒2ln(𝑥)^ = 𝑥^2 Then multiplying through by 𝐼 𝑥 , we get 𝑥^2

Which simplifies to 𝑑 𝑑𝑥 𝑥

Contd.

We now integrate both sides, 𝑑 𝑑𝑥 𝑥

This gives 𝑥^2 𝑢 = −𝑥 + 𝐶 𝑢 =

𝑥^2

We have now obtained the solution to the simplified DE. All we have to do now is to replace 𝑢 = (^) 𝑦^1

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. 𝑥 𝑑𝑦𝑑𝑥 − 1 + 𝑥 𝑦 = 𝑥𝑦^2
2. 𝑑𝑦𝑑𝑥 = 𝑦(𝑥𝑦^3 − 1)
3. 3 1 + 𝑡^2 𝑑𝑦𝑑𝑥 = 2𝑡𝑦(𝑦^3 − 1)

References

• A first Course in Differential Equations 9th^ Ed., Dennis Zill.
• Elementary Differential Equations, Martin & Reissner
• Differential and Integral Calculus Vol 2, N. Piskunov
• Workshop developed by Abhiroop Chattopadhyay