Sometimes a substitution is useful to transform a first order non-linear, non-separable differential equation into one we know how to solve.

The first kind we will look at are called \(``\)Bernoulli equations" and they can be transformed into either a separable de (Method A) or a linear de (Method B).

A non-linear first order de is called a Bernoulli equation if it can be written in the form \(y'+p(x)y=f(x)y^{r}\) where \(r\) is any real number other than \(0\) or \(1\) (those values of the \(r\) make the de linear).

Method A: Find a non-trivial solution \(y_1\) to the \(``\)complementary equation" \(y'+p(x)y=0\). Since you only need ONE solution, pick \(y_1=e^{-P(x)}\) where \(P(x)\) is ONE antiderivative of \(p(x)\). Then use the substitution \(\displaystyle u=\frac {y}{y_1}\) which is equivalent to \(y=uy_1\).

\(y=uy_1\rightarrow \) \(y\,'=u\,y_1\,'+u\,'\,y_1 \rightarrow \)    \((u\,y_1\,'+u\,'\,y_1)+p(x)uy_1=f(x)(uy_1)^r \rightarrow \)

\(u(y_1\,' +p(x)y_1)+u\,'\,y_1=f(x) u^r y_1^r\rightarrow \)    \(u\,'\,y_1=f(x) u^r y_1^r\), which is separable!

Solve for the general solution for \(u\), and then \(y=uy_1\).


Method B: One can use the substitution \(v=y^{1-r}\) to transform a Bernoulli equation into a linear equation:

\(v'=(1-r) y^{-r} y'\rightarrow \) \(y'=\frac {1}{1-r} y^{r} v'\rightarrow \) \(\frac {1}{1-r} y^{r} v' +p(x)y=f(x)y^{r}\rightarrow \) (dividing both sides by \(y^r\))

\(\frac {1}{1-r} v' +p(x)y^{1-r}=f(x)\rightarrow \) \(\frac {1}{1-r} v'+p(x)v=f(x)\), which is linear!

Solve for the general solution for \(v\), and then \(\displaystyle y=v^{\frac {1}{1-r}}\).

Example: Let’s solve \(y'-2xy=(xy)^{3}\), which is a Bernoulli de. Note: \(y=0\) is a solution to this equation. Let’s look for other solutions.

I prefer method A: First find a non-trivial solution \(y_1\) to the complementary de \(y'-2xy=0\), which you (the viewer) should be able to do!

One such solution is \(y_1=e^{x^2}\).

So the substitution \(u=\frac {y}{y_1}\), which is equivalent to \(y=uy_1\), in this example is \(y=ue^{x^2}\).

\(y'=u'e^{x^2}+u\left (e^{x^2}\right )'=u' e^{x^2}+u(2xe^{x^2})\rightarrow \)

\([u' e^{x^2}+u(2xe^{x^2})]-2xue^{x^2}=(xue^{x^2})^3\) by substituting \(y\) and \(y'\) into the de.

So \(u' e^{x^2}=x^3u^3e^{3x^2}\), which is separable! Of course, we have to solve the separable equation.

Separate the variables: \(u^{-3} u' =x^3e^{2x^2}\).

Integrate both sides: \(\displaystyle \int u^{-3}\;du=\int x^3e^{2x^2}\;dx\). (Integrate by parts on the right side.)

So \(\displaystyle -\frac {1}{2u^{2}}=\frac {x^{2}}{4}e^{2x^{2}}-\frac {1}{8}e^{2x^{2}}+C\rightarrow \)   \(\displaystyle \frac {1}{u^{2}}=-\frac {x^{2}}{2}e^{2x^{2}}+\frac {1}{4}e^{2x^{2}}+C\) (\(``-2C\)" relabeled as \(C\).)

By algebra, \(\displaystyle \frac {1}{u^{2}}=\frac {(1-2x^2)e^{2x^2}+C}{4}\) (\(``4C\) relabeled as \(C\)).

Then we get \(\displaystyle u^2=\frac {4}{(1-2x^2)e^{2x^2}+C}\rightarrow \)   \(\displaystyle u=\pm \frac {2}{\sqrt {(1-2x^2)e^{2x^2}+C}}\).

This (Or \(u=0\)) is the general solution to the transformed equation. Finally, using \(y=ue^{x^2}\) yields:

\(\displaystyle y=\pm \frac {2e^{x^2}}{\sqrt {(1-2x^2)e^{2x^2}+C}}\), OR \(y=0\), which is the general solution to our Bernoulli equation!