A 1st order differential equation is separable if it can be written as \(h(y)y'=g(x)\).

From this form, integrate both sides with respect to the independent variable (and use \(dy=y\,'\,dx\)) to solve for the general solution:

\(\displaystyle h(y)y'=g(x)\rightarrow \)

\(\displaystyle \int h(y)\,y\,' \,dx=\int g(x)\,dx\rightarrow \)

\(\displaystyle \int h(y)\,dy=\int g(x)\,dx\rightarrow \)

\(H(y)=G(x)+C\) where \(H(y)\) is an antiderivative of \(h(y)\) and \(G(x)\) is an antiderivative of \(g(x)\).

\(H(y)=G(x)+C\) is called an \(``\)implicit" general solution; if you can solve for \(y\), the result is called an \(``\)explicit" general solution.

Example: Let’s solve the initial value problem (IVP) \(\displaystyle y\,'=\frac {2\cos {(2t)}}{y^2}\), \(y(0)=1\).

First we \(``\)separate" the variables: \(y^2 \,y\,'=2\cos {(2t)}\).

Then we integrate both sides with respect to \(t\): \(\displaystyle \frac {1}{3}y^3=\sin {(2t)}+C\).

Then we solve for \(y\): \(y=\sqrt [3]{3\sin {(2t)}+C}\) (Note: \(``3C\)" can be relabeled as \(``C\)")

This is an explicit general solution. To find the solution that matches the IVP we use the initial condition \(y(0)=1\): \(1=\sqrt [3]{0+C}\), so \(C=1\).

The IVP solution is \(y=\sqrt [3]{3\sin {(2t)}+1}\).

Another example: Let’s solve the equation \(\displaystyle y\,'=\frac {y+y^3}{x}\).

First observe that \(y=0\) is a solution. Let’s look for non-trivial solutions.

Separate the variables: \(\displaystyle \frac {1}{y+y^3}y\,'=\frac {1}{x}\).

Integrate: \(\displaystyle \int \frac {1}{y(1+y^2)}\,y\,' dx=\int \frac {1}{x}\,dx\).

Using partial fractions, \(\displaystyle \int \frac {1}{y}-\frac {y}{1+y^2} \, dy=\ln {|x|}\).

So \(\displaystyle \ln {|y|}-\frac {1}{2}\ln {(1+y^2)}=\ln {|x|}+C\) (OR \(y=0\)).

That’s an implicit solution. By log rules, we can rewrite it as \(\displaystyle \ln {\left (\frac {|y|}{\sqrt {1+y^2}}\right )}=\ln {|x|}+C\).

Exponentiating both sides, \(\displaystyle \frac {|y|}{\sqrt {1+y^2}}= e^{\ln {|x|}+C}=e^C |x|\) which can be expressed as

\(\displaystyle \frac {y}{\sqrt {1+y^2}}=Cx\) (Note: \(``\pm e^C\)" can be relabeled as \(``C\)" and we even allow for \(C=0\)).

Finally, \(\displaystyle y^2 = Cx^2(1+y^2)\rightarrow \) \(y^2 - Cx^2 y^2=Cx^2\rightarrow \) \(y^2(1-Cx^2)=Cx^2\rightarrow \) \(y=\pm \sqrt {\frac {Cx^2}{1-Cx^2}}\) where \(C\) is a non-negative constant.

We can even express this as \(y=0\) OR \(\displaystyle y=\pm \sqrt {\frac {x^2}{C-x^2}}=\pm \frac {x}{\sqrt {C-x^2}}\) where \(C\) is a positive constant.

Note: Sometimes we cannot get to an explicit solution, and we just leave it implicit.

One can show that the IVP \(\displaystyle y\,'=\frac {y+y^3}{x}\), \(y(3)=-0.75\) has the solution \(\displaystyle y=-\frac {x}{\sqrt {25-x^2}}\).

But on what intervals would this be a valid solution?

First of all, by looking at the de, \(x\neq 0\), so \(x>0\) to include \(x=3\). Next by looking at the domain of this given solution, we see that \(x<5\).

Putting this all together, this IVP solution is valid on any open interval contained in \((0,5)\), including \((0,5)\).