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

The second kind we will look at are called \(``\)nonlinear homogeneous equations" and they can be transformed into separable equations. Note: Don’t confuse these equations with \(``\)linear homogeneous" equations.

A first order de is called nonlinear homogeneous if it is nonlinear and can be written in the form \(y'=q(y/x)\) where \(q(u)\) is some function (of a single variable).

To transform \(y'=q(y/x)\) into a separable equation, use the substitution \(u=\frac {y}{x}\), which is equivalent to \(y=ux\). Then:

\(y'=u'\,x+u\rightarrow \) \(u'\,x+u=q(u)\rightarrow \) \(x\,u'=q(u)-u\) which is separable!

Note: I’d encourage not memorizing this, but rather going through the steps.

Example: Let’s solve \(\displaystyle y'=\frac {y-x\sec {\left (\frac {y}{x}\right )}}{x}\).

It’s nonlinear homogeneous as it can be rewritten as \(\displaystyle y'=\frac {y}{x}-\sec {\left (\frac {y}{x}\right )}\).

Applying the substitution \(u=\frac {y}{x}\), or equivalently \(y=ux\), yields \(u'\,x+u=u-\sec {(u)}\).

We get \(u'\,x=-\sec {(u)}\) after subtracting \(u\) from both sides.

Separating variables gives \(-\cos {(u)}u'=\frac {1}{x}\). Integrating both sides gives \(-\sin {(u)}=\ln {|x|}+C\).

Hence \(u=\sin ^{-1}{\left (C-\ln {|x|}\right )}\) (with \(``-C\)" relabeled as \(C\)).

Now we use that \(u=\frac {y}{x}\) to get \(y=x\sin ^{-1}{\left (C-\ln {|x|}\right )}\).

Now consider the IVP \(\displaystyle y'=\frac {y-x\sec {\left (\frac {y}{x}\right )}}{x}\), \(y(e)=0\). By imposing the initial condition, we get \(C=1\), so \(y=x\sin ^{-1}{\left (1-\ln {(x)}\right )}\). Note: This works on any open interval contained in \((0,e^2)\).

Sometimes a little manipulation is needed.

Consider the de \((x+y)y\,'=2x\). It is nonlinear and nonseparable. It is not Bernoulli.

But it is nonlinear homogeneous: \(\displaystyle y\,'=\frac {2x}{x+y}=\frac {2}{1+\frac {y}{x}}\).

So with \(u=\frac {y}{x}\), equivalently \(y=ux\), we get \(\displaystyle u'\,x +u=\frac {2}{1+u}\).

Then \(\displaystyle xu'=\frac {2}{1+u}-u=\frac {2-u(1+u)}{1+u}=\frac {2-u-u^2}{1+u}=-\frac {u^2+u-2}{1+u}=-\frac {(u-1)(u+2)}{u+1}\).

Separating variables, \(\displaystyle \frac {u+1}{(u-1)(u+2)}u'=-\frac {1}{x}\). Integrating both sides, using partial fractions,

\(\displaystyle \int \frac {u+1}{(u-1)(u+2)}\;du=-\ln {|x|}+C\rightarrow \)    \(\displaystyle \int \frac {2/3}{u-1}+\frac {1/3}{u+2}\;du=-\ln {|x|}+C\rightarrow \)

\(\displaystyle \int \frac {2}{u-1}+\frac {1}{u+2}\;du=-3\ln {|x|}+C\rightarrow \)    \(\displaystyle 2\ln {|u-1|}+\ln {|u+2|}=-3\ln {|x|}+C\rightarrow \)

\(\displaystyle \ln {\left [(u-1)^2|u+2|\right ]}=\ln {|x^{-3}|}+C\rightarrow \)   \((u-1)^2(u+2)=Cx^{-3}\).

Since \(u=\frac {y}{x}\) we get \(\displaystyle \left (\frac {y}{x}-1\right )^2\left (\frac {y}{x}+2\right )=Cx^{-3}\), or equivalently \(\displaystyle \left (y-x\right )^2\left (y+2x\right )=C\), which we leave as an implicit general solution (as it would be hard to solve for \(y\) here).

Sometimes we create a substitution to suit. For example, let’s attempt to solve \(y'=\frac {1}{y-x}\).

Why not try \(u=y-x\)? If we do we get that \(u'=y'-1\) or equivalently \(y'=u'+1\).

Then \(\displaystyle u'+1=\frac {1}{u}\). So \(\displaystyle u'=\frac {1-u}{u}\), which is a separable! By separating variables,

\(\displaystyle \frac {u}{1-u}u'=1\). Then integrating both sides, using that \(\frac {u}{1-u}=-1+\frac {1}{1-u}\), gives

\(-u-\ln {|1-u|}=x+C\). So \(\displaystyle -y+x-\ln {\left |1-y+x\right |}=x+C\), or equivalently

\(y+\ln {|1-y+x|}=C\), which we leave implicit (as it’s impossible to solve for \(y\) here).