Separable Differential Equations
Consider the differential equation $$${y}'={f{{\left({t},{y}\right)}}}$$$, or $$$\frac{{{d}{y}}}{{{d}{t}}}={f{{\left({t},{y}\right)}}}$$$.
If the function $$${f{{\left({t},{y}\right)}}}$$$ can be written as the product of the function $$${g{{\left({t}\right)}}}$$$ (function that depends only on $$${t}$$$) and the function $$${u}{\left({y}\right)}$$$ (function that depends only on $$${y}$$$), such a differential equation is called separable.
Let's see how it is solved.
$$$\frac{{{d}{y}}}{{{d}{t}}}={g{{\left({t}\right)}}}{u}{\left({y}\right)}$$$, or $$$\frac{{{d}{y}}}{{{u}{\left({y}\right)}}}={g{{\left({t}\right)}}}{d}{t}$$$.
Integrating both sides yields $$$\int\frac{{{d}{y}}}{{{u}{\left({y}\right)}}}=\int{g{{\left({t}\right)}}}{d}{t}+{C}$$$, where $$${C}$$$ is an arbitrary constant.
Example 1. Solve $$${\left({t}+{1}\right)}{d}{t}-\frac{{3}}{{{y}}^{{4}}}{d}{y}={0}$$$, $$${y}{\left({1}\right)}={3}$$$.
This equation is separable and can be rewrritten as $$$\frac{{3}}{{{y}}^{{4}}}{d}{y}={\left({t}+{1}\right)}{d}{t}$$$.
Integrating both sides yields $$$\int\frac{{3}}{{{y}}^{{4}}}{d}{y}=\int{\left({t}+{1}\right)}{d}{t}$$$, or $$$-\frac{{1}}{{{{y}}^{{3}}}}=\frac{{{{t}}^{{2}}}}{{2}}+{t}+{C}$$$, where $$${C}$$$ is an arbitrary constant.
Rewriting it a bit, we obtain that $$${y}=-\frac{{1}}{{\sqrt[{{3}}]{{\frac{{1}}{{2}}{{t}}^{{2}}+{t}+{C}}}}}$$$. This is the general solution. To find the particular solution, plug in the initial values and find the constant $$${C}$$$.
$$${y}{\left({1}\right)}={3}=-\frac{{1}}{{\sqrt[{{3}}]{{\frac{{1}}{{2}}{{1}}^{{2}}+{1}+{C}}}}}\to\frac{{3}}{{2}}+{C}=-\frac{{1}}{{27}}$$$. So, $$${C}=-\frac{{83}}{{54}}$$$. Thus, the particular solution is $$${y}=-\frac{{1}}{{\sqrt[{{3}}]{{\frac{{1}}{{2}}{{t}}^{{2}}+{t}-\frac{{83}}{{54}}}}}}$$$.
Now, let's take a look at another example.
Example 2. Solve $$${y}'={{y}}^{{2}}{{t}}^{{2}}$$$.
This equation is separable. We can rewrite it as $$$\frac{{{d}{y}}}{{{d}{t}}}={{y}}^{{2}}{{t}}^{{2}}$$$, or $$$\frac{{{d}{y}}}{{{y}}^{{2}}}={{t}}^{{2}}{d}{t}$$$.
Integrating both sides gives $$$\int\frac{{{d}{y}}}{{{y}}^{{2}}}=\int{{t}}^{{2}}{d}{t}$$$, or $$$-\frac{{1}}{{y}}=\frac{{1}}{{3}}{{t}}^{{3}}+{C}$$$, where $$${C}$$$ is an arbitrary constant.
Thus, the general solution is $$${y}=-\frac{{1}}{{\frac{{1}}{{3}}{{t}}^{{3}}+{C}}}$$$.
There are also 2 particular cases: when $$${g{{\left({t}\right)}}}={1}$$$ (the differential equation doesn't contain $$${t}$$$) or $$${f{{\left({y}\right)}}}={1}$$$ (the differential equation doesn't contain $$${y}$$$).
Let's do some more practice.
Example 3. Solve $$${y}'={{e}}^{{-{y}}}$$$.
We can rewrite it as $$$\frac{{{d}{y}}}{{{d}{t}}}={{e}}^{{-{y}}}$$$, or $$${{e}}^{{y}}{d}{y}={d}{t}$$$.
Integrating both sides gives $$${{e}}^{{y}}={t}+{C}$$$, where $$${C}$$$ is an arbitrary constant.
So, the general solution is $$${y}={\ln{{\left({t}+{C}\right)}}}$$$.
And one more final example.
Example 4. Solve $$${y}'={\sin{{\left({t}\right)}}}$$$, $$${y}{\left({0}\right)}={5}$$$.
We can rewrite it as $$$\frac{{{d}{y}}}{{{d}{t}}}={\sin{{\left({t}\right)}}}$$$, or $$${d}{y}={\sin{{\left({t}\right)}}}{d}{t}$$$.
Integrating both sides gives $$${y}=-{\cos{{\left({t}\right)}}}+{C}$$$, where $$${C}$$$ is an arbitrary constant.
To find the particular solution, use the initial condition $$${y}{\left({0}\right)}={5}$$$:
$$${y}{\left({0}\right)}={5}=-{\cos{{\left({0}\right)}}}+{C}\to{C}={6}$$$
So, the particular solution is $$${y}=-{\cos{{\left({t}\right)}}}+{6}$$$.