Converting To Standard Form
Since the Bernoulli and standard forms of DE are nearly identical, our goal is to make it such. We follow three steps to accomplish this:
- Remove \(y^n\)
- Introduce \(m\)
- Substitute \(y^m\)
Remove \(y^n\)
Recall that the standard form is \(y^\prime + a(x)y = b(x)\). Bernoulli's \(y^n\) at the right side is the only thing that makes the equation non-linear. So, the first step is to eliminate this. To do so, all we have to do is to divide the whole equation by \(y^n\)
- \(y’ + a(x)y = b(x)y^n\)
- \(\frac{y^{\prime}+a(x) y}{y^n}=\frac{b(x) y^n}{y^n}\)
- \(y^{-n} y^{\prime}+a(x) y^{1-n}=b(x)\)
Introduce m
After removing \(y^n\), our equation then becomes: \(y^{-n} y^{\prime}+a(x) y^{1-n}=b(x)\). If we look closely, this form still does not conform to the standard format because of the first term \(y^{-n}y^{\prime}\) in the left side of the equation.
For this reason, we will multiply the equation by a factor \(m\).
- \(y^{-n} y^{\prime}+a(x) y^{1-n}=b(x)\)
- \(m y^{-n} y^{\prime}+m a(x) y^{1-n}=m b(x)\)
Let’s investigate the first term \(m y^{-n} y^{\prime}\) together with the power rule:
\(\frac{d}{d x}\left(u^m\right)=m u^{m-1} \frac{d u}{d x}\)
There is a parallelism between these expressions:
- \(y^{-n}=u^{m-1}\)
- \(y^{\prime} = \frac{du}{dx}\)
With this, we can rewrite the first term as \(\frac{d}{d x}\left(y^m\right)\). Consequently, Bernoulli's equation becomes:
- \(\frac{d}{d x}\left(y^m\right)+m a(x) y^{1-n}=m b(x)\)
Let's go back and investigate the factor \(m\). What is it specifically?
Going back to the first parallelism: \(y^{-n}=u^{m-1}\), we can solve for \(m\):
- \(y^{-n}=u^{m-1}\)
- \(-n=m-1\)
- \(m=1-n\)
Notice that the \(y\) in the second term in the expression \(\frac{d}{d x}\left(y^m\right)+m a(x) y^{1-n}=m b(x)\) equals \(1-n\), we can further simplify it with \(m\); hence, the expression becomes:
- \(\frac{d}{d x}\left(y^m\right)+m a(x) y^{m}=m b(x)\)
Substitute \(y^m\)
After arriving at \(\frac{d}{d x}\left(y^m\right)+m a(x) y^{m}=m b(x)\), we substitute \(y^m\) with any random variable to arrive at a first-order linear differential equation.
The following steps are similar to finding the general solution of a first-order linear DE problem. After solving the DE, we substitute \(y^m\) back to it.
Example
Consider the differential equation: \(y^\prime + xy = xy^2\). Let's find a general solution for it.
Investigate DE
Before we proceed, let's investigate the DE if it obeys the form \(y^\prime + a(x)y = b(x)y^n\). In our example:
- \(a(x) = x\)
- \(b(x) = x\).
In addition, the form follows Bernoulli's format; hence, we need to change the equation into a first-order linear DE and then find a solution.
Apply Method
The following shows the whole process for solving the general solution of a Bernoulli Equation. There are two parts: (1) the conversion to a standard DE and (2) the solution through first-order linear DE.
To convert to its standard form, we remove first \(y^2\) on the right side:
- \(y^\prime + xy = xy^2\)
- \(\frac{y^{\prime}+x y}{y^2}=\frac{x y^2}{y^2}\)
- \(y^{\prime} y^{-2}+x y^{-1}=x\)
After, we multiply it with \(m\):
- \(m=1-n=1-(2)=-1\)
- \(y^{\prime} y^{-2}+x y^{-1}=x\)
- \((-1) y^{\prime} y^{-2}+(-1) x y^{-1}=(-1) x\)
- \(\frac{d}{d x}\left(y^{-1}\right)-x y^{-1}=-x\)
Finally, we substitute \(y^{-1}\) with a variable - say \(z\):
- \(\frac{d}{d x}(z)-x z=-x\)
We have converted Bernoulli's equation to a first-order differential equation. All that remains is to solve this:
- \(a(x)=-x\)
- \(b(x)=-x\)
- \(\mu=e^{\int a(x) d x}\)
- \(\mu=e^{\int-x d x}=e^{\frac{-x^2}{2}}\)
- \(y=\frac{1}{\mu} \int \mu b(x) d x\)
- \(z=1+C e^{\frac{x^2}{2}}\)
Remember though to return \(y^-1=z\) after solving it:
- \(y^{-1}=1+C e^{\frac{x^2}{2}}\)
- \(y=\left(1+C e^{\frac{x^2}{2}}\right)^{-1}\)
Summary
A Bernoulli Equation is a DE of the form \(y^\prime + a(x)y = b(x)y^n\). It's the non-linear format of the first-order linear differential equation. The expressions \(\frac{dy}{dx}\), \(y^\prime\) is the first derivative, \(a(x)\) and \(b(x)\) are functions, and \(y\) is the dependent variable
The value \(n\) is the degree of the \(y\) variable. It must NOT be equal to either zero or one.
To find the general solution to this DE, we must change this equation to its standard form and then solve it using the integrating factor and the general solution equation.
There are three general steps to accomplish this: (1) remove \(y^n\), (2) introduce \(m\), and (3) substitute \(y^m\).
To remove \(y^n\), divide the whole equation by \(y^n\)
Multiply the whole expression with \(m=1-n\) afterwards and convert the form into: \(\frac{d}{d x}\left(y^m\right)+m a(x) y^{m}=m b(x)\)
Substitute \(y^m\) with a random variable to get the standard form
Solve for the standard form using the integrating factor and the general solution equation.
Return \(y^m\) back after getting the general solution.
